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[Music] are you ready for the definite integral

so by now you’ve worked so hard to get to this point

with limits derivatives applications curve sketching and even anti-derivatives

in this video i first review the basic properties of summation

and area then we explore the definition of the definite integral

and then several of its properties so why is the definite integral

so important hi everyone welcome back i’m dave this is the calculus one explore

discovered learn series and in this video i’m going to talk

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about first summation and then review its basic properties and talk about area

and then we’ll talk about the riemann sums and how to use riemann sums to build

definite integrals and then stick around to the end where we’ll talk about some

exercises so let’s get started okay so up first is summation we’re

going to talk about sigma notation first now this is something that you probably

have seen in calculus one but i’m going to assume that maybe it’s

been a while since you’ve seen it or you just need a good review but we

definitely need to have a good uh hold or a good understanding of

sigma notation in the following video so let’s get started so

i’m going to talk about this sigma notation here and we have this sum

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um from i equals m all the way to n and we have a sub i so the way we do that

is we substitute in for the i and we start with the lower

m the lower index um so we have the m and then n plus one n plus two n plus

three and then the second to last term will be n minus one

and then the last one will be n so here’s an example of one

and let’s say i have i equals 1 to 5 and my ai here are given say by an

i squared and so what the sigma notation says

is that first you know here we first we plugged in the m

here the m is one so first i’m going to plug in the one

so i’m going to get one squared and then two squared

so we keep increasing by one so m n plus one

and plus two right so i’m using one and then one plus one

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and then one and then and then we’ll have a three squared

and then a four squared and where do we stop that’s the second to last one

that’s five minus one the last one will be five squared

so this right here is a handy way of writing all these all these

terms out using sigma notation now you don’t have to start at one and

you don’t have to go to five obviously i can

start here at four and just go to seven and now let’s say i’m doing um i minus

one so what is the sum right here well this says start at four plus

now use a five now use a six and then last i use a seven

and so we can go add up all those numbers if we want

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but the point is is that this is sigma notation and this is called

expanded form so you need we need to we’re going to

practice on some examples right now of writing out the sigma notation into

expanded form and then working backwards if someone gives you the expanded form

can you write it in sigma notation so we’re going to practice both of those

right now here we go so first up let’s write this in expanded form

so we’re starting at zero so here we go k equals zero to four

and we’re looking at two k minus one 2 k plus 1 so in this example

um we’re not using an i here we’re using a k

and so k is the variable so we’re going to go to 0 first so 2 0 minus 1 over

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two times zero plus one so i substitute in the zero everywhere

and now sigma of course means add now we’re going to substitute in one

everywhere add now we’re going to substitute in 2 everywhere

and now at and then so we did 0 1 2 the next one is 3. we’re going to stop

at 4 4 is the upper so the next one is three

and then the final one is to substitute the four everywhere

so this is the expanded form here now we

could go crunch all those numbers up and the number that you’ll get at the end

will be four four nine over three fifteen so

we can write this number out here and either

expanded form right here all of this or in sigma notation

so these are equivalent to each other it’s just that this one is

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called expanded form and this one is using sigma notation

okay so that’s don’t really need to do this here’s the expanded form right here

i just want to show you that it’s just equal to a number in other words all of

this right here is a number and you can find that number by writing out an

expanded form and then cranking out those numbers and finding out what it is

is what is one way to do it all right let’s look at the next example

uh write out the sum in expanded form so this time we have parentheses in here

and so we’re going to do here sum from 1 to 6 of

i over i plus 1 and then minus 40. so now our index here

is i that’s our variable here we’re going to start at 1 and we’re

going to go to 6. now the whole thing is in parenthesis so

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don’t think of this as the 40 is going with this here

so the summation is separate from the 40. that’s what the parentheses means all

right so when i is one we’re gonna get one over one plus one and then a two

two over two plus one and then a three and then a four and then a five

and then finally a six and so there is the um oh and then the minus 40.

all right so we written the sum out here

in expanded form we expanded the sum out

and then we just left the minus 40 alone

now it doesn’t really actually go to say what find that actual number if you

wanted to find it you could crank on all these numbers over here

use a calculator use you know scratch paper whatever

you get this number right here so this this whole thing right here on

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the left side is equal to this number so is all this right here

this is the expanded form and this is sigma notation here

okay next example so now we’re going to reverse the question

so now we’re given the expanded form and we’re going to write out here the

sigma notation so to do that need to recognize a pattern

so um when i’m looking at adding up all these things

i notice that the numerator is always a one

so i’m going to write my summation here and i’m going to say

the top is the numerator is one and that what’s happening here

it’s going one two three and all the way to something squared here and so

what you know can we think of seven two two five

in terms of a squared so it ends in 2 5 so it’s going to be something 5

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squared right and then i think it’s going to be

85 squared right remember how to square 85

you just add 1 to the 8 and say 9 times 8 so 72 25 right okay so

this looks like it’s going to be i squared and then i is going to go

from 1 to get the 1 squared which is the first one

and then it looks like it’s going to go to 85.

so this is how i would write out this and spin expanded form here and we can

check that one over one squared and then use a two so one over two squared

and then use a three one over three squared

and then the last one would be one over 85 squared

which is the seven two two five so yeah this is

this is the sigma notation that we’re looking for right here

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that is our sigma notation okay so let’s try another one here

so let’s look for a pattern and what are we gonna what are we gonna

see for our pattern here so i’m trying to write it in sigma notation so sigma

i’m gonna use an i for my variable why not use the k

well okay sure we can use the k [Music] i j k are common choices here

um so but whatever you choose here you have to use here obviously

that’s your variable so if i use a k then i have to have some k’s over here

all right so it looks like we’re starting at three and we’re just adding

one at the top and we’re going to 23. so i’m going to say here

k and i’m going to say k starts at three and then what’s the bottom the bottom

always looks like four more than the top doesn’t it

seven is four more than three eight is four more than

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four nine is four more than five so i’m gonna say k plus four

and we’re going to go to 23 it looks like

so the top starts at three and then goes to four and five and six

and seven and the last one is 23. so if we were to expand this out we would get

what we want we would get the 3 over 7 and then we would get the 4 over 8 until

we get to the last one which would be 23 over 23 plus 4 which is 27

so yeah we get the same sum that’s up here

now keep in mind though there’s there’s lots of ways to do it

we here’s one way here’s the sigma notation right there

what about if i choose say an i and i insist upon starting with a one

so if i have a one here how can i get to my three

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well i can add three to it so i can say three and then i can do here um a six

so actually that’ll be a two to get to a three i’ll just

you know start with one and i get a three and this will be four more than

the top so i’ll say plus six right because the bottom is four more

than the top so i’m just going to add four there and now how far are we going

so we’re going to go to 21 right because if i stop at 21

that’ll give me 21 plus 2 which will be our 23

and then this will be 21 plus 6 which will be the 27.

so here’s my conjecture for the sigma notation let’s try

so we’re gonna get when i is one we’re going to get three over seven

when i is two we get four over eight when i is three we get five

over nine and then the last one we’re going to get 23 over the 27.

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so the point is is that when you take a sum there’s lots of ways

to write in sigma notation and i don’t just mean the i and the k that’s a really

superfluous change but the function here

is different um you know i plus two over

i plus six even if you wanted to use a k it’s still a different thing

so either way we get the same sum out so if you’re given a sum there could be

multiple ways of writing it in sigma notation here are two ways of

writing this sum is sigma notation in fact i could think of many many more

okay so next example all right how about this one now this

one you might think looks overly complicated but actually it looks more like

what we’re about to do upcoming anyways and so here we go we’re going to write

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in sigma notation and what i notice is the summation right here here

and here and here oops i have an extra additional summation in there just

disregard that should be only one of those pluses there but any case

i have one minus something right they all have one minuses

and they’re all being divided by four and they all have a squared on it and so

it looks like the numerator is the only thing changing

so i’m going to put an i over here and i’m going to put

parentheses here now i need parentheses here

squared so i’ll go ahead and put the brackets and what are we summing from we’re

summing from the numerator there starts at one so i’m gonna say

i starts at one and then i goes to four and so there we have there’s the

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sigma notation we expand this out we get this up here

so this could be one way of writing this in sigma notation

sometimes people ask you to say start with zero

how could we start with zero if we if we wanted to

for some reason or another we wanted to start at zero so i’ll say j equals zero

and then i’ll just go to three and then i’ll say one minus

now to get to the one because because we our first term is one over four

so i’m going to say j plus one over four and then i’m gonna square that

and then i’m gonna have up close my brackets

so these will be the same right so if we if we go zero here then we get out of

one first and then the last one we go to is we get the four

over the four so this either one of these will work

as sig as a sigma notation for this problem here

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okay so there’s some practice on uh oops

we got one more all right let’s let’s do one more here so

this time um all right so there’s my sum right there

summation uh summation you know summation so i’m looking at how those

are separating things all right and so the first thing i notice is that this is

one over n and they all have one over n so it’s not changing

they all have square roots that’s not changing they all have one minus ones in

them that’s not changing so all that stuff is staying the same so

i’m gonna use an i down here but i have one over n and i

have a square root and i have one minus and none of that’s changing what’s

changing though the zero and then the one in the numerator

and then the n minus one in the numerator

so it looks like we’re starting at zero so i think i’m going to start at zero

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here and put an i over n squared and so that will give me a zero over n

and then a one over an n and then a two over an n

and so on and what would be the last one will be n minus one over n

so i’m going to go to n minus 1. that’ll be my last one here

so here we go there’s the sigma notation and we can test it out

if we want we’re going to start at 0 for the i

so we’re going to get 1 over n square root 1 minus

and we’re going to get 0 over n squared plus and now we’re going to get the 1

over n again square root 1 minus and now i is going to be 1 over n

with a squared and now the next term one over n square root one minus

and now we’re going to get a two over n squared

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and then all the way to the last term what would the last term be

with n minus 1 coming in here so we’re going to have 1 over n

square root 1 minus and now the last one is going to be the n minus 1 over n

with a squared and you can see that’s our given uh sum that we are given is this

expanded form over here and so this does the trick so what i did

was i just kept everything the same except what was changing only that

numerator was changing so i started where i needed with the

zero and i went to the n squared and that’s all we need there

are very good so before we move on we need to practice using some summation uh

rules here that allow us to add up things

um very nicely and so i want to just go over a couple properties here

so this first property here is called the constant term rule

so i’m going to give you just a quick example here

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nothing fancy just k equals 1 to let’s say our n is 10 and then our c is 5.

so what will this sum here be now according to this rule is just you know 50.

okay but what if you didn’t know that so what does this say to do

now there is no k here so each time i plug in a new k

all i’m getting out is as a constant 5. so when i plug in 1 i get 5

when i plug in 2 i get 5 when i plug in 3 i get 5

and then when i go to the very last one and then i plug into 10

but it’s a constant so i just get a five out so how many fives do we have

right we started at one we went to ten so we have ten fives there’s ten of them

here so it’s ten times five and so that’s fifty

so we can do the same thing with the original here

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when we sum from 1 to n of a c it’s just going to be c plus c plus c

and then the last one will be for the n and so we’ll have n times

we’ll just add up c’s but n times and so that’s just you know n times c

so that’s called the constant term rule and that’s occasionally useful

um so the next rule is the sum rule so let’s look at the sum rule here just

very briefly so when we’re summing on the left hand side we’re summing up a sum

and the way to do that is to sum up the sums okay so if for example

what if we have something like 10 plus 20 plus 30 plus 3 plus 5 plus 21 plus 78

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so how would you add those up you know first you might want to add these up and

then you might want to add these up and so you know that’s kind of

an idea here is that we can sum up a bunch of sums by summing them

individually and then adding those up and so you know if i have some from

[Music] one to ten of of a of a sum here i can just add these up here

from one to ten and then add these up here from

one to ten it’s really nothing too fancy here

let’s look at the next one we have the scalar multiple rule

and so this one says that if i have a constant c in here i can

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pull that constant out of the summation so think of that like this

what if we have 10 plus 20 plus 30 plus 40 plus 50

plus 60. what if we’re trying to add up all those numbers

well one way to do it is to say okay they all have a 10 in them

so this would be 10 plus 2 plus 3 plus 4 plus 5 plus 6.

so factoring out a 10 here and you know that’s pretty much what this

right here says because this is summation from say i equals you know

so we have a 10 so let’s say 1 and then 10 10 i and we’re going from

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one to six there’s a six there so this will be ten times one ten times two

ten times three ten times four ten times five ten times six

so the last one here will be six ten times six right and so

we can expand it out now you may want to add all those up but

you can just add these up instead ten times

and then add those up those are smaller numbers maybe it’s faster to add them up

right so we have a four and a six we have a five and a five that’s just 21 right

so 21 times 10 so we can write it as summation adding all these

up here i equals 1 to 6 of just i in other words we can take the 10 out of

the sum right there so we can do that in the general form also

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so summation k equals 1 to n of c of a i or a k and so this will be ca1

plus ca2 i plug in 1 i get ca1 i plug into 2 i get ca2

all the way out to the very last one we plug in an n so we get c

a n and i just factor c out of all of them a1 plus a2

plus a n and so this will be c of ak it goes from 1 to n so we can pull the c

out because c does not have a k in it c c is a constant all right so next one

this one is really a combination of the previous two

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so if we’re summing from one to n of c a sub k plus another constant d well

this is a sum of a sum so by the sum rule i can sum over this sum here

and so i can break this up into two pieces here

i can apply this sum to the first one and apply this sum to this one here

that’s by the sum rule and now we can apply the previous rule which was the

scalar multiply rule and so i can pull out the c

and so this is exactly what the linear linearity rule says

it’s really just a combination of the previous two here

i’ll just throw in some parentheses there

okay so i just use the sum rule for this equals and then i use the scalar

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multiple rule for this equals here all right so what about the subtotal rule

so sometimes i call this a take a break rule

so i have this m here is in between the 1 and the n and us [Music]

think about how we want to do this here something like 10 plus 20 plus 30

plus 78 31 plus one two three nine how would you want to add those up so

first off you know that this is just 60 here right that’s just 60.

so you know you could go to your calculator and you could say

this plus that plus 60. right so because i added these up first

and so that’s kind of the idea behind these is that if you’re summing from one

to n you may first want to sum from one to m

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so i may first want to sum these up and then i want to sum these

up here separately so this goes to one to m

and then i continue on and then i would sum up the other ones maybe using a

different technique or whatever you know here i just used my

head here maybe i used a calculator oh i didn’t use the calculator here uh

what about if it’s one one three two five all right so eventually you can get

to a biggest enough big enough number so that any human

would need to use a calculator or some kind of machine but the point is is that

you may want to break your sum up and deal with it differently

in different parts of it so that’s the idea behind the subtotal rule

okay so what about the dominance rule so the dominance rule is an inequality

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and what it’s saying basically is that if if the b’s are all greater than the

a’s then the sum of the b’s will be greater than the sum of the a’s

so for example i have a sum let’s say the a i’s are given by 10 20 30 45 52

and here the bis are given by 11 and then 21 and then 35

and here i have 48 and here i have 53. so if i add up all of these numbers

that’ll be less than or equal to if i add up all these numbers here

so this is just an it’s just called dominant the bis are

dominating the ais they’re just greater than

all the way through and so and that’s what we mean by for all k

right all the way through this one’s greater than that one so we can add up

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all these we can add up all those but whatever you get

when you add up all those it’ll certainly be less than or equal to when

you add up all these b’s that’s the idea behind the dominance rule

okay so one more thing here about summation before we get to the area

is some summation formulas so these summation formulas are going to allow us

to find some values of the sum even though we have a lot of numbers um

so when we look at this formula over here

um i’ll just write this out here this is sum from k equals 1 to n of k to the i

so i here is one or two or three or four or five or six

whatever you want the i to be so let’s choose say eyes one first

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and so this will just be one plus two plus three plus all the way out to [Music]

n and then if i sum [Music] where the i is a 2

then this will just be 1 squared plus 2 squared plus 3 squared

all the way out to the very last one which will just be n squared

and we what we want are summation formulas for the

for these so if i go to n equals if i go to i is 3 now what if the power

on here is three so this will be one to the third plus two to the third

plus three to the third all the way out to n to the third so i want to give you

some formulas to help us find these summations here so let’s look at an example

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for example what if this sum here is going to one hundred

so maybe that’ll be too many numbers for you to add up

would be nice to have a nice quick easy way to do that

if that’s not enough numbers for you then

what about if you have 100 to the fourth power

then this would be 1 plus 2 plus 3 all the way out to the last term would be

100 to the fourth power right so that would be way too many for you to

add up whether using a machine or calculator whatever

if four is not enough just go to 40. my point is is that it would be nice to

have a [Music] formula to add all these up now here’s a quick

nice uh way of thinking about things we have one in a hundred if we make that

pair that’s 101 and what’s the one that comes right before

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the 100 right that’s 99 isn’t it so the idea here is that 2 and 99

also make 100 and what’s the one that comes right before 99

that’s 98. so if i write them if i write enough of them out maybe i’ll

see a pattern right so the 1 and the 100 make 101.

the 2 and the 99 make 101 the 3 the 98 together make 101

so how many hundred ones are we going to have so you might think that this is a

101 times and there’s going to be half of them 50 because each one gets a pair

and so you would be right this is 50 times 101 and so it’s much much quicker to

multiply 50 times 101 [Music] rather than to add all these numbers up

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so this is what i mean by formula or some kind of way

of calculating the form the summation without actually going through and brute

force calculating some so here’s the first one

and you can kind of see how it’s working the 50 is the what 100 over 2.

and the reason why we have 50 pairs here is because

we’re summing from 1 to 100 but we’re putting them in pairs

that’s where the 50 came from and then this is

n plus 1 right here so this is a hundred plus one

so you can see that formula right there is holding true it’s n

over two times n plus one so by using this formula so let’s use

this formula real quick let’s say here k is one and this time let’s go to 350

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350 k so we have one plus two plus three all the way out to the last one is 350.

so that’s a lot of them and i don’t want to add them all up

so what the formula over here says is don’t add them all up just simply go to

the n is 350 so 350 over two that’s what our formula says over here

n over two and then times n plus one so times three hundred fifty one

so to find the sum all you do is multiply this out 350 over 2 times 351

and that’s much much faster to find that sum right there that’s just easy

calculator or scratch work whatever so number two here is

what if we’re adding up a bunch of squares

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so let’s see an example of two suppose we’re summing from one to let’s say

a thousand and we’re summing up a bunch of squares

so what we’re summing what this sum represents the sigma notation represents is

one square plus two squared plus three squared

and then the very last term after you write them all out

the very last one is a thousand squared right so there’s there’s a sum there in

expanded form and as you can see it’s got a thousand terms on it that

number is going to be large lots of squares in there and so how do

we find the sum using this formula right here so

the formula says one over six times the n

the n is one thousand so so one over six times a thousand

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times a thousand and one times two thousand plus one so two thousand and one

so here i just identified what my n is it’s one thousand

so i have a n times n plus one times two n plus one and now i could

just calculate that number and it’s just multiplying and then dividing by six

and that’ll be this whole sum right here so these summation formulas are pretty

cool huh we’re going to use them in some calculus

in a minute together with some calculus but first

there’s the one for the third powers and so i’ll illustrate that when we

start working out an example so again here n can be um

a thousand two thousand ten million three

um and there that’s how you would sum a bunch of cubes

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you could use that formula there okay so let’s look at what’s next

so we have find the value of the sum here now

so let’s use some of these formulas and some of the properties that we

talked about and combine them together and so first off when i’m summing from 1

to 100 so we could write this out in expanded form

i’m going to do that real quickly just to convince you that we don’t want to do

that so we have the first term would be one times one squared plus one and then

two times two squared plus one two squared plus one and then

the last term would be one hundred so one hundred times

one hundred square plus one all right so that’s

got a hundred terms in it and we don’t want to expand that out

and try to calculate all that up we could but we don’t want to so how can we do

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this better using our properties and our formulas so

i’m going to sum from 1 to 100 and i’m going to take this

here and distribute it across so this is going to be

i times i squared so that’ll be i to the third

and then i times the one so that’ll be plus i here

all right now remember we can sum a sum so this summation can get applied to the

first one here i to the third and then i can apply the sum to the i here

all right very good so now when i um apply this right here i have i to the

third here so now i’m summing up a bunch of third powers

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and we know a formula for summing up a bunch of third powers here

and so i’m going to use this right here and if we look back

we had this formula right here and it was

the sum from i equals 1 to n of i to the third and that was 1 4 n

squared over n plus 1 squared so i’m going to use this formula right

here right in here except i’m not going to use it with an n

i’m going to use it with a 100 100 is my value right here for this problem

so i have a 100 here so i have 1 4 and then i have my n squared so i have

100 squared and then i have my m plus 1 squared so that’ll be 101 squared

and all that was for this part right here so now we have plus

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and now what was the formula for from 1 to 100 of just an i

so that formula is i equals 1 to n of i and this formula right here was just

one half and then the n and then times the m plus one

so using this formula right here now what do we have

our n is one hundred so i have one hundred so i have one hundred here and

one hundred here so plus comes from here and then i and

for this sum i have one half and then 100 times 101.

so there we go we’ve written out what this using the formula for this

for this summation right here is all right here and then plus

plus and then we wrote out the summation formula

with this part right here and now we’re just going to simply

add these sums up together multiply this out here i’m going to get 2 5 five zero

two five zero zero and here i’m going to get 50 50

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and so if we combine all those together we just get two five

five zero seven five five zero and so that’s the value of that sum right there

without having to actually um you know expand it out and add up 100 terms

we just did some substitution and using our formulas

and using our prop uh propositions for summation our properties for summation

okay let’s look at the next example this one is a lot shorter

we’re summing from uh one to four only so this one

we can just you know add it up it’s short it’s just one to four so

i’m gonna go from one to four of two i plus i squared

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and this will be two to the first plus one squared and now i is two

so this would be two squared plus two squared

and now i is three so two to the third plus three squared

and now i is four so to the fourth plus four squared and so sorry um

wrong button here here we go sorry so i uh just did a one to four of this

expression right here and then i used i equals one so i used a

one here and one here and i got this and then plus now i use i equals two so

two here and two here plus and now use the three so three here and three here

and then plus and now i use the last one so

four here and four here and then when we add all this up we’re gonna get

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3 plus 8 plus 17 plus 32 and all that combined together is 60. so

i found this sum right here it’s you know a small sum

it’s just one to four so i just crank it out plus i didn’t have a formula for

two to the i so all right next example [Music] okay so on this one right here

[Music] everything is um looking good here so we’re gonna go i equals one to n

and we’re gonna have i squared times i squared minus one plus one

okay so i’m going to distribute the i squared through here

that’s an i so summation i equals 1 to n i squared times

00:44

i squared so that’ll be i to the fourth and then i squared times minus i so

minus i to the third and then i squared times 1 so this will be i squared here

so i broke i expanded this out and then i have a sum of the expanded here

and now we’re going to write the sum to each of these terms here so the sum from

1 to n of i to the fourth and then minus and then the sum from

i equals 1 to n of i to the third and then plus the sum so i’m applying

the sum to each of these terms here so sum from i equals 1 to n of the i

squared here now the reason why we did that is

because we have a formula for each of these here

and the formula for this and this and this

now i didn’t write down the formula for this one a minute ago

00:45

but here’s what it is so let’s see if we can write it in red

if my red marker’s working or let’s try orange so this summation right here

is written as 1 over 30. and then n times n plus one times two n plus one times

three n squared plus three n minus one so all that is the form of the four

adding up a bunch of fourth powers from one to n

adding up the fourth powers you get this part right here

then i have a minus sign let’s go back here

so i now have a minus sign now we saw the formula for i to the third what was it

it was 1 4 n squared times n plus 1 squared so that was the

00:46

formula for this part right here and i have a minus in front and then now plus

now we have a formula for the i squared it was 1 6 n times n plus 1

times 2 n plus 1. so this right here is the formula for the i squared right here

now in this problem here we don’t have an actual value of n

so we’re not summing from one to ten or one to twenty or one to one thousand

were were in is unknown so in other words we’re trying to come up with our

own formula for this right here basing our results on the formulas that

we know for fourth powers third powers and squares here’s the formula for

fourth powers here’s the formula for cubes here’s the formula for squares can we

combine all this together to come up with our own formula

yes but you have to multiply it out i have to multiply all this out and to

multiply all this out multiply all this out and when you

expand it and then you bring it back together and simplify it

00:47

well you know actually you want to be a little bit uh

more delicate than that let’s use that word all right so we have an

n and we have an m plus one we have an n and an n plus one we have an n

and an n plus one can we factor out two n plus one

here’s a two s plus one here’s a two n plus one

but this one doesn’t have a two n plus one

and so we can factor some of that out to make it a lot easier to work with

in the end here’s what you get 1 over 60 and then

n which we can factor out all of them and an n plus 1

which we can factor out all of them so after you factor out the n

and the n plus one you’re going to get a bunch of stuff left that you have to

expand and multiply and then simplify and this is what you

get here you get 12 n to the third plus 3n squared plus seven n plus

eight and so there’s a nice beautiful formula for this sum right here

00:48

it’s equal to all of this so now if i wanted to go plug in

n into this one i could go find it by plugging in

into this one here now this is obviously something that we’re not going to

memorize but you probably want to come up with some kind of way of remembering

at least the at least remember the third power and the squares

and the single powers there all right so now let’s look at area [Music]

okay so we’re going to begin talking about area

by actually not talking about area we we have so far talked about

summations right here and of course this is a calculus course let’s bring some

00:49

calculus into this problem now let’s take a limit so first i’m going to

find the sum and then i’m going to take the limit of it as n goes to infinity

so we’ll talk about area in a moment and right now we don’t know why we’re

taking the limit of this sum but it’s going to have something to do with area

so let’s see how to do this so the limit as n goes to positive infinity

of the summation k equals 1 to n of k over n squared so like i said first we’re

going to find the summation work on that and then we’ll take the limit so each

one of these steps we already know how to do

i just want to pull it all together so that we can we can see

now first of all what is the limit uh sorry what is the index here what’s the

variable in here so the this is k so this k right here

00:50

but the n squared here is a constant so the first step is i’m going to bring

the n squared out so the n squared is down here so i’m

going to bring out one over n square okay so the n squared doesn’t have a k

in it there’s no k in this so i can pull this out of some

out over here now i’m summing from one to n

now remember we have a formula for this right here

i’ll write it over here from one to n of just the k this was one half n times

n plus one so that was our formula that we

talked about earlier one half n times n plus one

so we can use this formula right here we know what all this right here

is we have a formula for it so let’s go over here

00:51

and say limit as n goes to infinity one over n squared

and now for this part right here i’m going to use our formula

so all this right here becomes one half and then n times n plus one

so that’s our formula right there so i just kept everything

exactly the same limit one over n squared but instead of writing the

summation out in expanded form i just wrote out the formula for it

so what we can do here is simplify this a little bit this will be

limit as n goes to infinity of n times n plus one in fact let’s just

multiply that out that’s n times n and then n times one so this will be n

squared plus n and what we get down here will get a 2n squared here

00:52

so what do we do next well we have indeterminate form here the

numerator is going to infinity the denominator is going to infinity so

i’m going to write here indeterminate form infinity over infinity

now if you didn’t understand what i just said then

go back and watch the episode over indeterminate forms and l’hopital’s rule

the link to all the previous videos is in the description below so

i’m going to take some derivatives here i’m going to get 2n plus 1 over

and then this will be 4n now this still goes to infinity so

infinity over infinity so we still have indeterminate form

infinity over infinity and so this limit right here now

00:53

by taking derivatives will be two over four or one half

so the derivative of the numerator is two the derivative of the denominator is

four so this is just one half so this thing

that we started out trying to find right here

the limit of the sum we worked on the sum

we found the sum right here and then we took the limit of it

using the habitat rule a couple times and we said one half

so the sum here didn’t have an n in it but what the end was doing was going to

infinity and so this is going to one half

all right so there’s our first one let’s look at another

and again we have a limit of a sum so we’re going to work on the sum first

and all right so here we go the limit as n goes to infinity

00:54

summation k equals 1 to n and we have to square all of this so 1 plus

2 k over n squared and then times the 2 over n

now when i’m looking at this thing right here i’m asking myself

what is the variable here it’s a k so i have a k here

so i got to leave all of this right here alone

but the 2 over n that does not have a k in it so i can pull the 2 over n outside

of the sum now i can’t pull the 2 over n outside of the limit because the limit

is n going to infinity so that’s going to matter it has an n in it

but i can certainly pull it out of the summation because the variable summation

is k so this could be our first step here limit as n goes to infinity

2 over n that’s coming outside of my summation so i have summation of

00:55

k equals 1 to n and then i have 1 plus 2k over n with a squared

so i just pulled the 2 over n outside the sum

so there it is now this summation right here

this has a k in it and i have to square all that and then sum it up right so

let’s see how to do that so i limit as n goes to infinity and then i have a

2 over n and now i’m going to try to work on the summation right here

in other words i’m going to square all this out so what happens when you square

something right so a plus b squared is just simply a square plus two a b

plus b squared right so just that familiar formula there

so i’m gonna get one squared one plus and now this times that times a two so

00:56

that’ll be four k over n and then plus the last one squared so that’ll be 4k

squared over n squared all right so now i expanded this i squared this out so i

just multiplied it all out okay very good now we have to break the sum

to each of these three terms here so here we go limit as n goes to infinity

2 over n now i’m going to put brackets here

because i need summation for the first one here a equals 1 to n of just one

and then sum from 1 to n of this one now when i’m summing of this one right

here i notice the 4 over n there’s no k in that part so i can bring

that out so i’m going to say 4 over n and then sum k equals 1 to n

00:57

and what’s left here is a k and then now this right here i can pull out the four

over n squared so i’m going to sum this third term and

i’m going to pull out the four over n squared here

all right so very good so we apply the sum

to each three of these when i apply it to this one i just have the sum from one

to n i apply the sum to the second one i notice i can pull the 4 over

n out of the of the sum and here i can pull out the 4 over n

squared but i’m still summing from k squared here

so now the idea is that we have a formula for each of these three here

we have a formula for each of those so i’m going to plug it in so

n goes to infinity 2 over n and then bracket now what’s the first one here we’re

00:58

going from one to n of one so think of this just like this like one

plus one plus one plus one plus one plus one how many times are we doing that

from one to n so we’re doing it n times so this right here is just an

n now this one will be four over n times now what’s the formula from one to

n of a k is one half n times n plus one and then close parenthesis here so

basically i took this this part right here and i wrote down the formula for it

now plus 4 over n squared i’m not changing that and now i

have a formula for all of this right here

and so we’re going to write that formula down now

00:59

so to sum from 1 to k from 1 to n of k squared that was the um one-sixth [Music]

and then n and then m plus one and then times two n plus one then i

gotta close off this bracket here okay so we could put parentheses around

this right here like we did here so i’ll put this parenthesis here and here

although technically they’re not needed alright so now we have so when we’re

looking at this limit of the sum right here because of the expression that we’re

summing over has a squared in it and so we had to expand that out

right here and then we have three terms so we had to go three times we had to

use three formulas here now we need to combine all of this together

01:00

into something really nice including the two over n here when we do that

that’s going to require some some practice there

but we’re going to get to some n going to infinity of the 2 over n here and here

again when we get the common denominator and

expand all this out here we’re going to get 13 n squared plus 12 n

plus two over three n so that is after expanding and simplifying

expanding all this out and expanding all this out

the common denominators is six n squared so we need to get a common denominator

of six n squared everywhere so i would multiply by six n squared

over six n squared but the point is is that after you do all that you simplify

01:01

and you’ll be able to cancel some and you’ll get this right here

now we’ll be able to do uh the habitats rule if you notice we have an

n squared here and we have an n squared down here

and so the coefficients of the highest powers are the same

so this is going to be 26 over and then this will be three n square so

this will be over three so i’m also going to put equals dot dot equals here

so i have two dot dot dots and what that means is

we have to go fill in our previous knowledge and we know how to algebraically

simplify all of this so that’s why i put dot dot here

01:02

is to come up with this in algebraic simplification

then how do we calculate this limit here so i wrote dot dot dot when you

calculate this limit here you should get 26 over three now there’s

two ways you may want to calculate this limit

or perhaps three ways so the first way is to uh i have la hop

i have indeterminate form infinity over infinity so i’m going to use the

habitat’s rule and then i’m going to use the habitat’s rule again

you’re going to get 26 over 3. another way of doing that

is to divide everywhere by the highest power the highest power is n squared

so you can divide the numerator by the highest power in the denominator by the

highest power and you you can get 26 over 3 that way the third round

would be to say that you have some theorem that you know and you’re

applying the theorem and the theorem says that if you have the

01:03

highest powers matching right so here if you look at the numerator what’s the

highest power it’s n squared so what’s the number in front of n

square what’s the coefficient it’s 26 and here the in the denominator the

highest power is n squared and so the coefficient is the three now

since the highest powers match n squared over n squared i’m just going

to look at the coefficients which is 26 over three so you have one or two

three one of three ways you can fill in the dot dot dot there

different people require different things depending upon

what you’re trying to accomplish so the answer to this problem the limit as

it goes to infinity of that summation there all of that is just simply 26 over

three and there’s the idea behind it all is

to use our summation formulas and our properties of summations

01:04

and we got it all right so let’s look at how we can take all

this and combine it together with area so here’s our theorem over area

and what this is saying is that if f is continuous and positive throughout the

region f of x is greater than or equal to zero

and we have an interval a b then we can talk about the area

of the region under the curve over the interval

is that limit right there so let’s see why this is true let’s go here and

um kind of get a realization of why it’s true here

uh a visualization i mean so i’m gonna just draw this in the first quadrant here

01:05

and so we got this function here f and we got this interval here a b

and f is continuous and it’s above the x-axis or perhaps it’s touching but

the one i draw is not touching the x-axis um throughout the interval a b

so you know once you get past a b once you get past the b it could do anything

you could like that and then like this you know it’s it’s not continuous here

and maybe even has an isotope right here isotope here like that so

but on a b f is continuous so i’m only looking at f on a b

01:06

not f globally just a b here on a b f is continuous and it’s

greater than or equal to zero okay so here i am at b right here

and here’s right here at a and so here we go

now the area of the region under the curve over this interval so we have

this line coming down here x equals a x equals b and we’re looking at this area

here and the way to do that is to the way to find that area

exactly so notice this is an equal sign right here

there’s an equal sign right there and that means that this is

that the area a is equal to this limit this is not an approximation now

01:07

the way that we’re going to do this is this delta x here is b minus a over n

so what we’re going to do is we’re going to break so b minus a

is the length here right because it’s all of b

and then take away this part right here so this is the width

of the interval right here b minus a and then i’m going to divide it up into

n pieces so for example if n is 2 then i would divide it into two pieces i

would have this part and this part maybe n is four then i would divide it

into four pieces one two three four um if say

we have ten then i would divide it into 10 pieces

so i’m going to divide it into n pieces and

n can be whatever we want it to be so i got it

in pieces here so i’ll just draw a bunch of them

01:08

like that so just imagine in pieces now so the delta x here

is just the width of each one here so that that there’s a width there and

there’s a width there and there’s a width there and so this is the width

of the n sub interval right the the width of the nth sub interval

and they’re all the same width here so this delta x that we have going on

right here in the in this limit that’s the width of the sub interval here now

when i do a plus k times delta x so here’s a and now k

k is our variable for our summation so k starts out at one and then it goes to

case two case three and so on in fact hey let’s just write that

let’s just write that limit out in expanded form

01:09

just so we can see that this is a equals the limit as delta x goes to 0

of the summation so what is the summation here so when k is 1 we have what f of

a plus delta x times delta x now when k is 2 now we have 2 plus

now we have two plus uh delta x times delta x and then

dot dot dot and then look we can do one more maybe f of a plus three delta x

times delta x and then what would be the last one plus

da da da and then the last one would be right so the last one here is the sum

goes from one to n so from one to n so the last one here will be n

01:10

a plus two or no not two n in in delta x times delta x

so we’re taking the limit of this summation right here and what do these uh

you know what do these ref reflect so we know that this right here is the width

of the sub intervals here so what is a plus a delta x so here’s a

and then plus the delta x so this is a right endpoint here

of this sub interval right here and f of that means go plug it in right so

this right here is this number right here and this right here is width

times the height and so this expression right here is the area

01:11

of this rectangle right here similarly when we go on to the next one here

we’re going to go um to the next one here so now i’m doing two delta x’s

so i’m starting at a and i’m going to 2 so here’s

a and then i go to 1 and then 2 so now i go

here and i again i plug that into the function to get the height

so now i’m coming up here and i’m getting another height right here

and when you multiply with times height you’re getting the area so i’m getting

these two areas here the first one here so the first one here

is the area of the first rectangle this is the area the second rectangle

and then now we can go on to the last one

here where we’re going to get the right endpoint here

and we’re going to get the height we’re going to calculate the height

so we’re going to get width times height again so the width

times the height will give us the area when you multiply them together

01:12

the width times the height you’ll get the area of this right here

and so i’m getting all these areas here how many did i do i did n of them

and so what we’re doing here is making these rectangles

underneath here to get the to get the approximate area here

so now i’m going to take the limit as the delta x here approaches zero

now notice that as delta x approaches zero what’s happening to

n as delta x approaches zero what is n doing right so this is approaching

zero right here so in n is going to infinity so the number of

uh rectangles here is going to infinity so we started off by thinking

01:13

about maybe n is 2 or 4 or 10 but n goes to infinity the delta x is

going to zero so this area here when you pass the limit when you don’t

have the limit on there when you just look at the

summation you have an approximate area but when you take the limit you have an

exact area okay so let’s look at an example to make sure all this makes sense

to get uh some intuition as to why it makes sense

find the exact area under the curve so here we have x squared plus four

and we’re looking on two to ten so we’re going to use our area formula that

we just learned so the limit as delta x goes to zero

01:14

of the summation from one k equals one to n of f of a plus k delta x

over times delta x now we have to identify what everything is

so here we’re using a is two and b is 10 and what’s our delta x

it’s b minus a over n so b minus a over n so

you know using all that information in here let’s write it out

again this will be the limit as um delta x goes to zero

n goes to infinity right because because delta x is eight over n right so

delta x goes zero and goes to infinity so we’re not gonna have a delta x in our

problem we’re gonna have n’s we’re gonna have k equals 1 to n

01:15

f of and here our a is 2 so 2 plus and we have k

and we have a delta x is 8 over n so i have eight k

over n and our delta x is eight over n [Music] okay so we transform the formula

into a specific example here this is the general formula here

and we’re using a is 2 because we’re going 2 to 10

we’re going delta x is b minus a over n so using these values right here in a

specific example this general formula becomes this and

now we need to calculate this this is why we’re practicing finding

summations first and then finding limits of summations

so let’s see if we can find this limit here

so i need to plug in though don’t forget that this is

01:16

this is not multiplication this is not f times this

this is plug in 2 plus 8 k over n plugging all that into the function here

so here we go we have limit as n goes to infinity summation 1 to n

now i’m going to plug in 2 plus 8 k over n i’m going to plug in all that

into my function my function takes whatever you give it and

squares it and then adds four so i’m gonna do two plus

eight k over n i’m gonna square that and then i’m gonna add a four and then

all that is still times eight over n so there we go we did this first part

right here so this part right here it’s substituted for all this right here

i found that actually that is one of the most common places to make a mistake

01:17

is that part right there okay so now let’s go and square this out and add four

and and pull this out of the summation and so here i’m hoping that you feel you

feel a little comfortable because we’ve done limits of sums let’s try

so we have summation here but i’m going to pull this eight over n

out because summation has a k here going to pull this out here is

eight over n and so just gives us a little bit less to worry about

we cannot pull the four out though because the four is inside of the

summation right here okay so here we go we’re going to sum from

one to n and we’re going to square this out

and then add four to it in fact what are we going to get here we’re going to get

2 squared but we already have plus 4 so i’m going to say 4 plus 4 that’s

8 and then now this times this times 2 so that would be 32k over

01:18

32k over n right so it’s two times eight

times another two because we’re squaring it

and then the last one here is a squared k squared over n squared and i already

added that 4 to the 2 squared here and so there’s my sum right there i need to

take the sum of this so my next step is to use formulas and factor

so we have eight over n and i’m going to sum the first one here

which is just an eight so the all right i’ll just apply this to each

one i don’t want to lose anybody here so i’m gonna apply this to each term here

this will be 32 k over n and then summation k equals one to n

01:19

so i’m just applying the sum to each of these three

okay so i apply the sum to the eight apply the sum to the next term

apply the sum to the last one now for each of these

i can pull some things out of the summation before i use the formula

so let’s do that so this will still be the sum from this will be

eight in why because we’re adding up eight

eight plus eight plus eight we’re doing it in time so that’s eight

in now this one here is 32 over n and then what’s the formula for the

summation of k so that was one half and then n and then n plus one

plus and now we can pull out the 64 over n squared that’s a constant

01:20

and now what was the formula for summation for k squared so that was what uh 1

6 n n plus one times two n plus one all right so good so we we factor out

the 64 over n squared that’s a constant here

and then we got the summation of the k square so there’s the summation of the k

squared using the summation formulas that we learned earlier in this video

now dot dot because we have to use lots of algebra here

what’s our common denominator six n squared so i need to multiply this one dot

top and bottom by three n and i need to multiply this one by six n squared over

six n squared and then i’ll get a common denominator of six n squared everywhere

and then you can expand the numerator out so after some work what you’ll get is

01:21

eight over n times um here i’m going to put dot dot for

algebra and then l’hopital’s rule and then your final result here will be 1088

over three [Music] 1088 over three okay so i talked here about that in the

last example about how um you need to do the algebra

then you need to compute the limit right both of those things you

should know how to do but you also need to practice

so don’t wave your hand here if you’ve practiced this before

make sure that you can fill in those algebra steps and make sure you can fill

01:22

in those limit steps there now i want to show you something exciting though

this is a sneak peek at the next video i’m going to look at this x squared

um plus four here and i’m going to go two to ten and x squared plus four dx

and this is 1088 over three this is no no i’m not going to do it i’m going

to have to wait till the next video okay so but

we will do one more example here though all right so let’s look at this one now

01:23

now when we look at this one right here we have an x to the third we have an x

squared now that’s going to set this up again here

so the area here says find the exact area

so i’m going to use the limit as delta x goes to 0 of the summation

of f of a plus k delta x times our delta x so this is the right endpoint we

evaluated at the function to get the height

so this is width times height and we’re doing a bunch of rectangles how many of

them we’re doing we’re doing n of them and then we’re

letting that number of them go to go to infinity and so this will be the limit

as n goes to infinity and so what is our delta x what is our a

a is zero b is one and so delta x is b minus a over n

01:24

so this goes to zero that’s going to infinity now this will be

f of so that’s zero so zero plus oh forget my sum

sum k equals one to n f of zero plus k times delta x is 1 over n so k over n

times our delta x which is 1 over n and so our exact area is equal to this

specific limit here for this problem here now

this is 0 plus k over n in other words it’s just k over n

i need to substitute k over n into my function

that’s the first thing we need to do here of course i could take my one over n

out of my summation because it does the summation depends upon k

01:25

so here we go limit as n goes to infinity 1 over n and then summation

from 1 to n and that happens when i substitute k

over n into the function so i’m going to put here four k over n to the third

plus three times my input k over n squared that’s k over n so there we go so

i take my input but my inputs k over n so my input

cubed my input squared here i’ve got my 4 by 3

and so this is the limit i’m trying to find now

so let’s write this out this will be equal to the limit

as it goes to infinity of one over n and now i have summation for each of

01:26

these terms here so i’ll just do that summation k equals one to n

and then we have a four k to the third over into the third

plus summation equals one to n of a three k squared

over n squared so i put summation to each of these terms here

and i cubed it and i squared it notice the force not q right if you go

back to the original problem up here it’s four and only the x is cubed so

this is my x and this is cubed this is four and then k cubed over n cubed

and three k squared over n squared okay good so now we can pull

up pull them out this is one over n and then our first summation

01:27

um the summation index here is k so it would be four over n to the third

summation k equals one to n of k to the third plus and then on this summation

on this summation i can pull out the three over n squared

okay so pulling out the three over n squared here we got this right here

so now i can use the formula that i know for the powers

the third powers and the squares um so this is n is going to infinity here

one over n and then four into the third now the formula for right here for the

01:28

thirds is one fourth n squared n plus one squared

so it’s four over n to the third times that so four over n to the third times

the formula plus three over n squared times the formula for the k

squared which was one over six n times n plus one times two n plus one

all right so we substituted in this formula right here and we got this part

right there and we substituted in this formula right

here and we got this part right here but don’t forget to multiply here

multiply here but in the end it’s the addition that’s separating them

okay now after some algebra we’re going to get the limit as

n goes to infinity of three plus seven n plus four n squared

all over two n squared so this is after a lot of simplification here

01:29

the common denominator here is what um 12 n to the third

and so expanding all that out and simplifying it you get this

and then now using whatever method you use here

to find your limit l’hopital’s rule highest power or theorem you find the

exact area to be a two it’s the exact area here

okay so now it’s time to start talking about the riemann sums [Music]

okay so riemann sums riemann sums is where we start to understand

01:30

uh area even better and we start to um and we start to

define the definite integral okay so let’s start off here uh with what a riemann

sum is now when we’re looking at a riemann sum here

we need a function and we need a an interval a b and we can talk about

the summation right here and we have all of these x i’s

and we have these x i asterisks and you know let me show you how to do that so

i have this function here and i have this a and the b

and i’m going to make a partition and so i’m going to say this is x0

01:31

and this is xn and i’m going to make a partition here so i’m going to

put down some x’s this is x0 so this is going to be x1 x2

and then all the way to the very last one b i’ll say b is the x n so this was

x 0 here so a is the first x and then b is the last one

and then i’ve chosen a bunch of them in here

and so this is a partition here so i’m going to use this as a script

p to be x0 x1 all the way through xn where it starts at

a and it ends at b and this is my partition here now i’m going to be able to

pick an interval sub interval here so let’s pick this one right here i’ll

01:32

put it in orange right here now when we look at that sub interval right there

the left endpoint here is going to be called x i

minus one so that’s the left endpoint and when i look at the right once here’s

the x i right here uh minus one and then i look at the next

the right endpoint here and that’ll be the x i so that’s the right endpoint

[Music] and in between the x i minus 1 and the x i

i’m going to choose something in here so i’m going to choose i’ll put that in

blue i’m going to make a choice and that’s my choice is going to be x i star

and so i do that choice there i put a star or an asterisk um

01:33

and i do that to each one of these so i make a choice a whole bunch of tick

marks any choice you want and then i go through each one of them

there’s a finite number of them i go through each one of them

and i make a choice now your choice could be left endpoint

right endpoint midpoint or anything in here that you want

is your choice but these are called sub interval representatives sub interval

representatives sub interval representatives now the delta x i here

the delta x i this is the um x i minus x i minus one

so this is the width of each of one of these

now i have to put an i here so in the last example we were using something

01:34

like delta x was b minus a over n now in that case

i was making them equal with sub intervals but here i’m not here i’m just saying

lay down a bunch of tick marks you can use whatever you want

now if you use whatever you want then each one of them could have a different

width sub interval every every sub interval can have a different width

so i’m keeping track of all my widths delta x i this x i the right endpoint

minus the left endpoint right that’s how you get the width right

it’s the right endpoint take away the left endpoint and what’s left over

is the width there now when you when you just lay down a

whole bunch of tick marks just a whole bunch of points here one of them

one of the sub intervals is going to be the largest because you have a finite

number of them one of them is going to be the largest

but in fact they all might even be equal to the largest for example if

you made them all equal with then all of them are equal to the

largest but since we have a finite number of them we can go through

01:35

and we can figure out you know what is the largest

with and i’m going to denote that by the nor i’m going to call that the norm

of p that’s the script p so this is the width of the largest

sub interval and you know there there may be some ties but they may all be the

same with but in general speaking um if i just randomly choose these

x’s you can go through there and find which one is the largest

and so we come back here now we can talk about this riemann sum here

so this riemann sum here is when i’m going to go to my representatives here

the xi’s and for each one so i chose an x i i chose another x i

01:36

i chose another one i chose another one i chose this one

and this one this one and this one and this one now for each of these

representatives that i chose so now i’m writing them in red for each of these

representatives that i chose in each of these sub intervals

i go and i substitute that into my function and i calculate a height and i get a

height for that sub interval and then for this sub interval here

i now go up to the function and compute a height

and that height will represent that that sub interval right there

so what we’re getting here is the riemann sum is going to be width times height

and it’s going to be the sum of them so when i sum up the f of the x i stars

times the delta x i’s these are the widths of the sub intervals and these

01:37

are the height of the that represents the sub interval

and so this right here represents area this part right here represents the area

of the first one and then we’re going to sum and we’re

going to add up the next area and add up the next area and so on

and so what we’re ending up with is a bunch of rectangles

and we’re finding those areas so what’s different about the riemann

sum is that these are sub-interval representatives

they don’t have to be the right endpoints like we did in our previous

examples so we can choose the partition on our own

and we can choose our sub-interval representatives on our own

and so now let’s see lots of examples illustrating this procedure here

okay so just to make clear we’re clear on the terminology

01:38

the delta x i’s are the widths of each sub interval the sub intervals

are indexed by i and we have our partition that’s just the

all the tick points or all the x values collected together

it’s called a partition now out of all those sub intervals one of

them has the or not one of them but there’s a largest uh with and that’s

called the norm and we have the sub interval representatives they’re chosen

we can choose them whatever we want them to be so we can choose the partition

and we can choose the sub interval representatives

and so let’s see with an example how to do that [Music]

so here they uh told us exactly what partition to use

but they didn’t shoot they didn’t tell us our sub interval representatives

so we can choose our sub interval representatives as we want so

01:39

i’m going to divide this um not divide this i’m going to organize this into a

table so because we’re being indexed by i so i is going to go from 1 to two to

three to four because there’s going to be four sub intervals from zero to one

is the first sub interval from one half to three fourths

that’s our next sub interval so our sub interval here is zero to one half

and then one half to three fourths and then three fourths to

the 5 6 and then the last one is 5 6 to 1. now on this example here

they gave us the partition so we break the zero one

01:40

up into the sub intervals so here’s the zero one and

they gave us this partition here for whatever reason this is the partition

and so one half is right here and so zero to one half is the first sub interval

and then they gave us three fourths so that’s about right here

and so the next sub interval is one half to three fourths

and then they also gave us five sixths so here’s the first sub interval here’s

the next sub-interval and the next one and then the last one so you can see how

we have four sub-intervals and there’s the actual sub-intervals there

zero to one-half one-half to three-fourths three-fourths to five-sixths

and 5 6 to 1. so now we have the sub intervals now the next thing we need

01:41

to do is to make our choices so notice in this problem here it says find a

riemann sum so or or compute a riemann sum but

the the important word there is a in other words it doesn’t say

find the riemann sum so someone else can come along and make different choices

here so you just have to make choices but how do you make your choices

well it just has to be something in the sub interval

so i could just choose left endpoints as my representatives here

or i could choose right endpoints so i could choose one half

three fours five six one or i can choose any random number i want in the sub

interval i’m actually going to choose one third

and then i’m going to choose two thirds then i’m going to choose eight tenths

and i’m going to choose nine tenths now the fractions may look a little weird

but i just wanted to make sure that something was in this interval here

01:42

and so those are the numbers i chose so these are choices here you can choose

whatever you want but it has to be in the sub interval

here so there are a million ways in one to find a riemann sum here

but those are my choices now what do you do with the choices

well you put them into your function you calculate a height

so this is the height column here so i’m going to plug in one-third into my

function which means i need to cube it so what’s one-third

cubed one over twenty-seven what’s two-thirds cubed

and what’s eight-third uh sorry eight tenths but cubed it’ll be 64 over 125

if you write it out and then what’s nine tenths cubed would be 729

01:43

over a thousand all right so let’s start making our table a little bit

we got different columns in our table here

then i just like to put that row right there

right so to make a riemann sum so far is when we look at the partition we

realize what our sub intervals are and that tells us what our index is and

then now i go and choose some representatives

so i choose x i star so in other words x one star is one third

x two right the i is two here x two star is two thirds x three star is eight

fifths and x four star is nine ten so just to write one of them out

x four star right that’s the fourth one there

is nine over ten so that’s my fourth sub interval representative is nine over

ten okay so now to build our riemann sum remember we need to do

01:44

with times height so now i’m going to calculate my width

so my delta x i that’s the width so what is the width of this sub

interval it’s one half minus zero so it’s one half

and now i have three fourths minus a half that’s one fourth and now i have five

six minus three fourths so that’s one twelfth

and now i have one minus five six which is one

sixth and now i’m ready to build my riemann sum

for my choices i made here so the riemann sum looks like this

it’s a summation from i equals one to four in this example here one to four

and i have my f of my sub interval representatives

times my widths so i use my sub interval representatives to calculate my height

and i do high times width and so i’m adding up all the areas

01:45

so what are we going to get here well let’s just expand this out so you can

see it you don’t really need to but i’m gonna go ahead and do it just so you

can see one of them expanded out it’s only got four

our next one will have more so i won’t expand it out on our next one but for

this one this will be f of x one star times delta x one plus f of x two star

and i say star but it’s really an asterisk delta x

two and then the last and then the all right and then f

of x three asterisk or star and then delta x three and then plus our last one

uh i is four now so x four star and then delta x4 and so we can go

and find all of these numbers we have all these numbers in our table here

01:46

so this is x once a stricken delta x1 so i put those right next to each other

so we’re multiplying these multiplying these

multiplying these and multiplying these and we’re just going to add them up

so here we go we get 1 over 27 times 1 half plus 8 over 27 times 1 4

plus 64 over 125 times the one one twelfth plus

and then the last two you can see right here seven two nine

over one 1000 times the 1 6. and now we can go

and calculate all this up and then when you do that

you calculate that times that plus that times that plus that times

that plus that times that when you calculate all that up you get two seven

01:47

seven three ten eight hundred so that’s the riemann sum there and

that represents the approximate area so we can come over here

maybe and see if we can build something of a sketch here

we have y equals x to the third something that looks like that and we’re

bounded on zero to one we chopped it into one half and then three

fourths and then 5 6. and each time i chose a representative i chose one

third first to calculate my height one third right there

and then on this interval right here i chose two-thirds

so i chose two-thirds in here and i calculated my height

01:48

and then this interval three-fourths to five-sixths i chose

eight-tenths somewhere in here and i went up and i calculated my height

and then five six to one i randomly chose nine tens

and i went over here and calculated my height and so what we’re getting is an

approximate area under the graph and i’ll put that in black here

as you can see it’s an approximation sometimes it goes over sometimes it goes

under now when i’m making these choices here i just randomly chose them

and that’s because the function i’m working with here is just x to the third

so it’s nothing too fancy that’s just x to the third

but in real life you’ll want to look at your function here

and you want to decide do i want to over or underestimate the value of of this

01:49

and so you can strategically make these choices here so that either your sum

is easy to find or you’re getting some kind of approximation

that you’re interested in for example an over estimation or an underestimation

so sometimes you go over the area and then sometimes you go a little bit

under the area it just depends upon your function

and your choices here okay so there’s our first riemann sum there here’s a

riemann sum width times height and we got the sub-interval representatives

and we make our computation all right so very good so let’s look at another one

so we’re gonna um given this function right here

and our close bounded intervals zero to five

and they give us another partition here in this in this case

and they say compute a riemann sum so this time how many sub intervals do

01:50

we have what’s our index so that says sub interval and we have

zero to one one to two two to three three to four and then four to five

so these are our sub intervals so we have one two three four five of them

and now i’m going to make our make make some choices [Music]

so i’m going to make my sub interval representative choices

and i’m always going to choose in this case

i’m just going to choose left endpoints all the way down i want to be lazy i

just want to choose them now sometimes it’s to your advantage to

you to choose left endpoints it’s really not a matter of being lazy

or not it’s just if you look at that function there you

might have some information about it that might lead you to this strategy

01:51

here or someone may specifically tell you to use left endpoints

in any case i choose those for this example so now i go and substitute my

representatives into my function so now i got to plug in 0 into my function

and i get out 2 i plug in one into my function

so i have to do two plus six times one minus five times one plus one to the

third and for all that i’m going to get out of four

so i have to calculate all that out i get a 4. so now i have to plug a 2 into my

function when i plug a 2 into my function i get out of 2

when i plug a 3 into the function i also get 2 out again

when i plug a 4 into the function i get a 10 out

01:52

and now i’m going to calculate my sub interval widths what are the widths

now in this example they gave us a partition that is

regular in the sense that we broke up zero five

into equal widths so this is a constant all the way down here they’re all ones

so this is an example something called a regular partition

where our sub enter for widths are all the same

now i can build the riemann sum just by multiplying and adding

so the riemann sum here the sum from i equals 1 to 5 of f

of f of x i star times delta x i star this is my riemann sum

it’s width times the heights and i’m summing them up

and i’m going from one to five so this is going to be this column times

01:53

this column and i’m going to add them all up so i’m going to do two times one

plus four times one plus two times one plus two times another one and then the

last one ten times one so you can see why you might want equal

width sub-intervals here if these are all ones

then it really makes this easier to calculate

even if they’re not all ones but they’re all equal with like if they were all

like 1 8 then that would still make this sum here pretty easy to calculate

in the end this is just 20. so if i look at this function here

on the interval 0 to 5 if i use that partition

i can find an approximate area by using the left endpoints and i found it to be

01:54

- so here’s what that would look like so here is our um function right here

two plus six x minus five x squared plus x to the third

and i’m looking at it on zero one and we chose left endpoints

to find our height so my first left endpoint was zero

and so my height here was a two so my height here was a two

when i chose my left endpoint so that right here is height of two

it’s going all the way across on the first sub interval

and then i use my left endpoint to get the height again

it’s right now i get this rectangle right here so on the first sub interval we

underestimated the area didn’t we we have this air of this part gray here

that underestimates it on the second sub interval we overestimated

the limit uh sorry we overestimated the area

01:55

now from two to three we choose the left in point

we calculated the height so here it looks like we overestimated the area

then on three to four we definitely underestimated the area

and then on four to five we underestimated the area also so by

looking at the function if you had a graph of the function

you could strategically choose which endpoints or which representatives

you want to use so that you always underestimate or perhaps you

over estimate is your goal so you can have some strategy in those choices that

you make okay so um let’s do one with eight sub intervals now

so let’s get started on this one so we have six x squared plus two x plus four

01:56

and we’re looking on one to three and we’re going to have

eight sub intervals now on this one they don’t give us a partition so

all they tell us is use eight so in other words be reasonably accurate if

you’re going to just use two sub intervals you know

depending upon the function obviously but you’re probably not going to have a

lot of accuracy so eight sub intervals and then it’s only a quadratic so the

behavior isn’t too wild it’s just just a nice smooth parabola there

in any case for my partition i get to choose they don’t tell us

so i’m going to partition one to three one one three is our sub interval here so

i’m going to choose one three here as my yeah one three here okay so here we go

from our partition i’m gonna it has to start at one and end at three

01:57

but you know how am i gonna break it up after that

well i’m gonna look at five fourths three halves seven fourths two

all right and then i’m going to go between two and three i’m gonna break it up

nine fourths five halves and eleven fourths and three

so there’s that’s gonna be my partition that i chose

for no particular reason okay so now i’m gonna make my

table i’m gonna organize my data into a table

to make it easier to find my riemann sum so what are my sub intervals [Music]

and what’s my index so i’m first going to have 1 to 5 4.

01:58

and then five-fourths to three-halves and then three-halves to seven-fourths

and then seven-fourths to two and i’m going to break this into two tables

so i’ll have another i and sub intervals over here i’ll just put my

i don’t think i’m going to have enough room here

so i’ll go over here and do another i [Music] now just keep going 7 4 and then

2 to um nine fourths and then nine fourths to five over two

and then five over two to eleven fourths five over two [Music]

01:59

and then i’m missing my last one here which is eleven fourths to three so

11 over 4 to 3. all right that’s good enough

all right so how many eyes do we have we have one two three four five

six seven and eight now of course they told us we had to have eight

so there’s our choice there’s my choice for my partition

now for each one of these i need to choose a representative

a sub interval representative [Music] all right so for my sub interval

representatives i need to choose something in each one of these intervals here

i’m going to choose left endpoints by force three halves seven fourths two

02:00

nine fourths five halves and eleven fourths so there’s our

my sub interval representatives i just chose left endpoints all the way through

now i need to go calculate my heights [Music]

so i’m going to substitute 1 into this function over here on the left

6x squared plus two x plus four i’m gonna plug in a one into that

so i’m gonna get six plus two plus four so i get a twelve

and i’m gonna skip ahead a little bit here i’ll let you check my answers

i’m gonna plug in five-fourths into my function and i get out 1 over 27

127 over 8. i’m going to plug in 3 halves into the function and get out

41 over 2 and then i get 207 over eight and here i get 32 nine fourths i get

02:01

three eleven over eight for five halves i get ninety three over two

for eleven fourths i get 439 over eight perfect now for my

widths for my widths all right so i calculate all my heights

so that is certainly the probably the hardest

one to fill in but it’s just plugging in numbers

just plug in 11 force into that function there you go there’s your number there

now what is the width of each one of these right

because i chose them all different this first one here the width is one-fourth

[Music] one-fourth one-fourth fourth one fourth one fourth

02:02

so as you can see i made a regular partition here now the last column here is to

compute width times height width times height so this is the width

times the height so i’m going to be multiplying across here

to get with times the height so just multiplying it i get um 127 over 32 here

and then 41 over 8. and then 207 over 32 and then 8 and then 311 over 32

93 over eight and then for the last one here i get 439 over 32.

okay and so then the last thing is to add up the last column

02:03

just add them all up three plus right so the riemann sum would be some

adding up all the uh heights of the representatives times the width

so if we add them all up three plus this one plus this one plus this one plus

this one we add them all up we get four 493 over eight [Music]

and so that will be our approximation for using eight sub intervals

and i used a regular partition left endpoints

so there’s nothing very complicated about it

other than understanding the overall procedure

i choose the partition because they didn’t give me one

so i chose a regular partition and by that i mean

02:04

all of these are equal with and you know what else i chose the left in points so

there’s all of our work there and then we went all the way down to eight

all right that looks sweet all right so this is um

something of a sketch here if i could do that for us here

let’s see if i can get some kind of sketch going on here i’ll just erase

this part right here and you know x squared plus four right

or sorry uh six x squared plus two x plus four

so if i look on this interval here one two three

so we’re gonna look something so from one to three

and it’s just going to be increasing and now because i always chose left end

02:05

points 1 and then 5 4 so i i went from 1 to 3 right so i did halfway

let’s let’s let’s divide this up first so i went halfway

half ways again and then halfways i got my eight [Music]

now because i always chose left endpoints and it’s increasing

so it’s important to know that it’s increasing and i always chose left endpoints

so did i under value or overvalue the limit that means that i over value or

undervalue the area so here so here is the approximate area that we

found in blue 4093 over eight as you can see here we undervalued the

02:06

area we didn’t count all those little trying

all those little areas in there that look like triangles

they’re not triangles but we underestimated the limit [Music]

that’s not the exact area we underestimated the limit the area okay so

now we’re ready to start talking about the definite integral [Music]

okay so that was a two hours so far that was a two hour and

getting ready to understand what the definite integral is so we have now

to say what the definite integral is and we’re going to start off with of course

02:07

a function we have a closed interval we’re going to partition that’s we just

did in our last couple of examples we’re going to partition the interval

we’re going to do it into n sub intervals by choosing points

and so by partition i’m going to say that here

that they’re strictly increasing so i just put them in order

and there’s n of them so we did some examples where we had four of them

and then our last example we had eight of them

but here i’m just going to say that we have n of them

and i’m going to call that partition p and i have the i sub interval

a delta x sub i i’m missing my x on that that should be delta x i

as we had the force delta x i is just x i minus x i minus one and the

largest of these with is called the norm if it’s a regular partition then they

all have the same width but in general you don’t

02:08

have to choose them all the same width so some of them will be larger than

others perhaps so i make sure that the largest one

is always going to zero when i take my limit so i’m going to choose sub interval

representatives these are the x i stars or asterisks

and i choose each one from the sub interval

and then i form the sum so the riemann sum here

associated with the function and the partition

and the chosen sub intervals there so if any of those change then your riemann

sum changes it’s not just your riemann sum of your

function but it’s your riemann sum of your function your partition that you

chose and these sub-interval representatives so that’s not anything new

that’s just what we talked about with riemann sums we just did some examples of

02:09

of all that stuff so now what’s new well we’re going to take a limit so the

definite integral is defined as the limit of the riemann sum here

notice the left-hand side here is new notation that we haven’t seen before

and it looks a lot like the anti-derivative symbol that we’ve used before

and we’ll certainly explain why that’s true

but we’re saying that this limit of this riemann sum

now of course the disclaimer maybe the limit doesn’t

exist right you you’re very familiar with limits

you know that sometimes you take the limit of something and you realize oh

that limit doesn’t exist so if the limit exists then we call it

the definite integral of f over the interval a b

and we call the function f integrable over a b and

02:10

you know we’re using riemann sums to define what a definite integral is

okay now of course as i’ve been mentioning all along

for special cases we can define particular types of riemann sums

in other words if your intervals all have the same width then i don’t

need to use the i subscript anymore so in other words your delta x i is just x

i minus x i minus one but if they’re all the same

then we can just drop the i from here because they’re all the same which means

which means we’re all going to choose them to be exactly the same here

okay so we can form a riemann sum using regular partition

which means we don’t need to use the i on the delta x

i’s and this is called a regular partition so we did two examples so far with

02:11

regular partitions all right so for regular partition the norm goes to zero

if and only if n goes to zero and so the definition

is sometimes written like this so this is the definite integral

the limit as n goes to infinity and then we have the limit of the riemann sum

now the same disclaimer goes that assuming the limit exists if the limit exists

then the function is integrable over that interval okay so here’s a theorem now

and the theorem says if your function is continuous then the limit will exist

so this is sweet what that’s saying is that this is a continuous function and

you know what was the function that we did before

this was the function we we looked at in our last example

02:12

well this is continuous it’s continuous on and in the last example we had 1 3.

now in the last example we worked on one three

and we did a regular partition and we broke it up into eight

sub intervals so what this right here represents

one two three of of this function right here six x squared plus two x plus four

this is equal to the you know the definition says

that this is equal to the limit as n goes to infinity

of the riemann sum so k goes from one to n and then now i have f of um

the a plus k delta x times delta x now if i’m using a regular partition so

02:13

i can find this and this definite integral and this is definition so this is

definition so we can find this integral right here

by looking at the limit of the riemann sum here

now this is from one to three so what is our delta x from one to three and what

is the a so the a is 1 the b is 3 and so delta x is 3 minus 1 so 2 over n

so delta x is 2 over n right and so this is one

right so we’re on one to three so this is one and our delta x is two over n

so this is two k over n and then this is two over n so on one hand

in our last example we used eight sub intervals and we found the area to be

02:14

approximately 493 over eight and we used equal with uh sub intervals of

with one fourth now so we know how to approximate these area

using riemann sums and partition but we also know that if we take the limit

we you know we’ve practiced uh solving problems like this where we have

a summation and we take the limit so we know how to work this out also and

this will give us the exact area whereas this is the approximate area

so this will be approximate area this will be

exact area here so we can find the exact area

under the graph of this function right here using a limit of riemann sum

as we did in a in previous examples where we worked out

02:15

examples of the theorem or we can use a riemann sum to approximate it

now if you don’t have limit here then this is an approximation right here

because this is just a riemann sum right here um

using um a a regular partition here so think of it like this there’s the

approx approximate area and then when we take the limit as n goes to

infinity when we complete the whole process then we get the exact area and then

so we can find better than this number right here

because remember we we did this number right here this was an

underestimation of the area and this will give us the exact area

and so that’s why it was important to talk about summations

and talk about limits of summations because it gives you the definite integral

this is the definite integral right here and this is the definition of the

02:16

definite integral right here using all this right here

okay so let’s look at some examples um now we did lots of them

with limits of sums um but what if you have functions and

you’re not going to be able to work with summation formulas

and you know finding the limit of a riemann sum is very difficult in general

so let’s look at some using some geometry

so what if we try to express this limit here of the summation of this

um expression right here square root of four minus

right it’s p is a partition and we want to write this as a definite integral

and find its value using geometry in other words look at the graph what

does this graph say that the definite integral should be

02:17

so if we look at the shaded region here so what i’m thinking about is this

is a circle right part of the circle and right here’s the center and we’re

looking at the so this is radius two right here

right so we’re looking at x squared plus y squared equals four um but we’re just

looking at this part of the circle so if we solve for y here we’re going to

get square root of 4 minus x squared and we’re just looking at this

area right here this tick mark here’s a 2 and this is a 2 up here

and so that’s that’s what we’re getting over here and

so the function this is the function that i’m thinking about here

is square root of 4 minus x squared the c k is a representative and the delta x

02:18

k is the partition here so how would we find this area here

here from 0 to 2 of 4 minus x squared dx and this will be equal to the limit

as the length of the width of the partition go to zero

or if you want to put here n goes to infinity and then summation k equals 1 to

n um now we have this function right here we have our representatives and our

delta xk here so it’s all represented right here or

you can write it out right here like this using the definition and

you know because we have an in here we have it in here but generally speaking

02:19

um know what is the area here of the shaded region here right so

recall the area of a circle with radius 2 is 4 pi the area of the circle is 4 pi

now of course this is only a quarter of it so

this whole thing right here is just equal to pi

so this this expression right here if you want to write it out as a definite

integral or if you want to leave it as a limit of riemann sum

this is four and the reason why i mention all this is because

if you look back at our examples where we did limits of summations well we use

summation formulas right but this is a square root on it so we

didn’t have any summation formulas for that

and generally speaking it can be very difficult to find

that type of expression there we really kind of need a different way

of finding definite integrals rather than appealing to limit of riemann sum

and this example is saying hey maybe try to use some geometry

02:20

if that is at your fingertips so here’s another example

in this case though we’re going to just be asked to find an approximation so

in this example here we’re going to say that our function is tangent of x and

we’re going to uh you know look at this graph over here

um and that says from zero to pi over four right we’re looking from zero

to pi over four so when i look from zero to pi over four

i’m looking at that area of the shaded region there and that

is written as this limit right here the norm goes to zero the length of the

largest and so this will right here will be one to n

02:21

of the function chosen with our representatives

sometimes people rhyme with ck and then delta x

k and then let’s put parentheses so we can keep the argument separate there

but you know this is just the limit that they give us

and we can write it as a definite integral and what this represents is the area

of the shaded region now how can we find the area

of the shaded region well it says to find an approximate value in this case

so when i look at the tangent right here i’m looking at something that goes like

that pi over four and i get a height here of one

so you know to approximate this area here you could like draw a triangle so

the approximate area would be the area of the triangle that would be an

02:22

approximation so what is the area of that triangle

that’s chopping right through here just a straight line going through here

we know this is not a straight line because it’s given by tangent

right but but if you just chop a line through zero through pi over four one

so what would the um you know area of that triangle be well

you know what don’t think of the area of the triangle yet just think of the area

of the rectangle what’s the area of you know your width is pi over four

and your height is one so it’ll be one times pi over four

so this right here is equal to um pi over four times one

um but then we’re going to chop that into a half

and right because because you know this is only the the triangle is only half of

the area of the rectangle and so this is going to be power eight

and this is just an approximation though so just by looking at what this

02:23

represents as in terms of area we can use some geometry to get a basic

approximation to this it’s nothing too fancy here okay so now um

let’s look at some basic properties of the definite integrals and so

the first one is called the order of integration

if you switch the order of integration this switches the sign

so for example if i’m looking at something like

the integral from 2 to 3 of the function and i want to integrate from 3 to 2 for

some reason well if i switch the these are called limits of integration

i just put a minus sign there the next one is called the zero with subinterval

02:24

so if my sub interval is from a to b okay but what about if my sub intervals

from a to a so when i look at the definition of the

uh riemann sum here i mean sorry if i if i look at the definition of the

definite integral it’s the limit riemann sum and i have f of

you know some representatives times some delta here

if if these widths are all zero here right these are all zero here if i

integrate from a to a right so the delta x i are the or the widths

and if they’re all zero right so if you choose a regular partition it’s

b minus a over n if a and b are the same then the widths are zero here

so if you just think in terms of say regular partitions here

um you know but anyways if you have all these is zero then i’m going to

02:25

have zero times the height zero times the height

zero times the height so we’re going to be adding up a bunch of zeros we’re

going to be taking the limit of a bunch of zeros

so in the end that this is just going to be equal to zero here

if your width is up zero here then the area under the graph is just zero

so for the next one we can pull constants out of our integral here

um one way to think about this is that if you

scale your function that you’re trying to find the area of

and you can find the area and then scale the area

that’s one way to think about that and we have sum and difference so

you know how would we think about why this one is true well here’s our um

let’s go from a to b here’s our function and it’s the limit of a riemann sum and

02:26

what about if we’re integrating a g same interval here this will be a limit

of a riemann sum also right so what happens if i want to add these

together well it’s the same as adding these together

now we know a limit rule for adding limits and we know a rule for adding

summations and so we can use those rules and we could write out something like

this this this integral right here f plus g

will be the limit of a riemann sum and this will be f plus g of x i star

and so when i apply the definition here of adding functions

and then i apply the sigma notation to each one of those

02:27

and then i apply limit to each one of those

what we’ll end up with is a sum of these two here so long story short

this sum and difference here work because that’s the same properties that

hold for limits and summations okay so additivity the way i usually like to

give the intuition behind this one is in terms of area

so you know this right here you can think about it as

area if your function is above the x-axis right so we can think about it as

something like area we have a to b we have this area right here

now someone wants to come in here at a c and

02:28

put a c in there and what we can say is that the area here

and the area here if you add those areas up you get the area of the original

so this will be the area from a to c that’s the area in blue plus the area in

the red c to b of g of x so if you add up this area plus the area

of the rest you get the area of the whole thing

so that’s the uh probably probably the way i think about the additivity there

and so we had um you know to think about in terms of limits

or summations it wouldn’t be too hard to write out a proof

but in any case here’s another interesting one the min max inequality

so the middle expression there um the integral represents the area think

02:29

about it in in terms of the area intuitively so here we have

a function and we have a to b in fact let’s make the function look a

little bit more interesting let’s say we have something like this

and it has a minimum value and a maximum value so here’s the maximum value

and so if i’m looking at and stopping at a

and b i’m going to stop right here at a so i don’t need to look at that anymore

and we’ll stop right here at b and so i don’t need to look at that anymore

now where’s the minimum maybe let’s make it a little bit more

clear i’ll make it come up right here so at b right there it’s not the minimum

so here’s the maximum and here’s the minimum right here

02:30

okay now the area in blue so so what is this part right here from a to b

of f of x right so this right here which is the middle uh in the inequality

that’s the area under the graph so that’s the area of all of this in

here all of it so now what is the expression on the left hand side it’s m

where m is the minimum so come over here and put an m

so when i do m times b minus a so b minus a is the width right so b

and then take away the a so here’s b minus a here’s the width and so if i draw a

dashed line across here this area here is less than so the part on the left here

that’s the area on the left that’s certainly less than

02:31

the area under the whole graph on the other hand

if i draw a dashed line over here at m the maximum value

and now i look at the end area from a to b and so now

that’s the whole thing and that is certainly greater than the area

um under the graph so that’s kind of the idea behind the max min inequality

and then now for the last one the dominance inequality there

and it’s basically saying if g wins throughout the interval

then the area under the graph of g is greater than the area

under f of g and so i guess i could draw a picture out real quick

02:32

of the so if f of x loses to g of x um for x in on an interval here

a b right so we have a g that’s winning throughout and an f who

knows what f is doing but it’s not winning so here’s a and here’s b and so if i

look at the area here the area under g the area under g

is greater than just the area under f so the integral here of g from a to b

that area is going to win this area is going to be greater than the area under f

so that’s all that one is saying okay so there’s uh seven properties of

02:33

the definite integral and so let’s look at an example using them now so

now these properties are very useful because remember the definite integral

is the limit of a riemann sum so the fewer times you have to calculate the

limit of a riemann sum you know maybe the better off you are in life

well maybe not you need to practice how to find the limit of room on sum

and make sure you’re good at that but let’s see

there are better ways to do things so how do we evaluate this limit right here

i mean sorry this integral right here this definite integral

it goes from zero to two and we’re calculating two

times f plus five times g minus six times h

they don’t even tell us what f g and h are but they tell us some

values of their integral so f of x we know the integral of f of x from zero

02:34

to two is three so here’s how we do this zero to two

of two times f of x plus five times g of x so we’re going to use these

properties that we just talked about so i can integrate two times f of x dx plus

zero to two of the next one five times g of x dx and then plus

the integral from zero to two of the minus 6 h of x

dx i’ll put a parenthesis around that minus 6.

okay so there’s there’s the third one so i can apply this integral to each of

these terms here using the sum and difference rule for integrals

so integrals of each one now for this one right here

02:35

i can pull the two out so 2 integral 0 to 2 of f here i can pull the 5 out

and these are basically working because these are the rules

that work for limits and summations and we put them all together

so they work for the limit for the for integrals and so this would be -6

0 to 2 of h of x so now we can go find this one right

here this is three so this would be two times

three plus five times this one right here is

minus one and then this right here is minus six

times this last one over here for h of x is a three and so when we calculate all

that together we’re going to get minus 17.

so perfect so this integral right here without using a limit of a riemann sum

using just properties and knowing the values of the other integrals

02:36

we’re able to calculate this integral here it’s a minus 17.

now in this last part it says now they tell us the integral value is zero

so this this one up here we didn’t know was -17 now we do

this one they tell us it’s zero but they say find s

so that it is a zero so let’s see how to do that so i’m going to integrate

0 to 2 5 f of x plus s g of x minus 7 h of x would integrate all of that dx

and i’m gonna skip and combine some properties together so this will be five

integral zero to two of f of x plus s times zero to two of g of x dx

02:37

and then minus seven integral zero to two of

h of x so i use the sub indifference rule and the scalar multiple rule all

together at once i integrated the first term and i pulled

out the five integrated the second term and pulled out the s

which we’re trying to find and i integrated the third but i pulled out a

minus seven so now we know values for each of these

this is five times and we know the value for that is three

minus s or plus s and we know the value for this one right here

it’s minus one and then minus seven times and we know a value for for the

eight here it’s three [Music] and so then when you calculate all these

numbers up we’re going to get actually you can’t there’s an s so we’re

going to get 15 minus 21 and then we’re going to get a minus

s here so we’re going to get minus s minus 6

02:38

and they tell us that that has to be equal to zero

they gave it to zero right here so this has to be zero

so therefore let’s put therefore s is minus six minus a minus

minus six or you know six s is minus six anyways so s is minus six so

those properties of integrals are very important

very useful okay so now for this example here we’re going to go back to

using some geometry to find an integral so we’re looking at the integral here

uh let’s go to this one here we’re looking at the integral here um from -1

to 5. let’s take that off for temporarily here

so i can see the full graph here it goes from minus one to five

02:39

now let’s look at this graph here a little bit um so that’s the graph of the

function f notice between minus one and one it’s constant at a two

between one and four it looks like the line three minus

x so it has a negative slope and between four and five it looks like

the line two x minus nine so between four and five it looks like a line that has

positive slope and so we’re going to try to find this this integral here without

looking at a limit of a riemann sum just using that definition of the function

and by looking at the graph of the function so let’s see how to do that

so first off i’m going to break the integral up into its pieces

the reason why i do that is because f is given to us in pieces

02:40

so this will be the from minus five minus one now it says from minus one to one

it just looks like a two so i’m going to just

write it like that and then where else are we broken

between one and four we’re the same and then the last piece we’re gonna go

from four to five so i broke this from minus one to five

now i could have broke it up at any point here i didn’t have to use a one

but why do i want to use a one and that’s because of the way the function

is defined it looks different to the left of one and to the right of one

so it looks different so i’m gonna break it up at one same thing with four i’m

gonna break it up this is how it looks to the left of four

and this is how it looks to the right of four

and so now let’s actually write those numbers in so

02:41

f of x between minus one and one the f of x is just a two

now the f of x on one to four what is the f of x between one and four

is the three minus x and now between four and five what does

the function look like looks like 2x minus 9.

so i just substitute in what is f of x on this interval

what is f of x on this interval what is f of x on this interval

now we could go to limit of riemann sums to find what this definite integral is

in terms of our definition however let’s interpret the indefinite

integral in terms of what it is uh in terms of area so when i look at this

graph over here when i look between minus one and one

i just get a constant two so i’m looking at this area here

02:42

of this rectangle between minus one and one and the height here is two

so i’m looking at that area of that rectangle right there

and that area so because the width is a two and the height is a two

this is just four so this value right here is just a four that was just four

now what about the next one now between one and four what’s going on

with our graph over here between one and four

so between one and four we have the so we have one and four here

and we have this line going through here so the height here is two

02:43

and so if i look at the try to find this area here

what’s the area of that this is one to three there’s the three right here

so this is with two and this is height two

but it’s the triangle so it’s only half so i’m gonna say plus a two here

but this is below the x-axis so this think of this is negative area

so what’s the area of this triangle well this is one and this goes to

minus one down here that’s the height so this is what

um one this is one by one so this whole area here is

one and i cut it in half so i’m going to say minus one half here so between

between one and four between one and four i had to look at it two different ways

02:44

because between one and four we’re above right here on this part right here

and then we’re below on this part right here so

when i’m above between one and three that that width is two

and the height is two so that area of that whole

rectangle will be four but i’m only interested in the triangle it’s only the

triangle that’s shaded so that’s the twos coming from this part right here

and the minus one half is coming from right here and it’s a minus because it’s

below the x-axis all right so now let’s do this last part here

that’s 2x minus 9 and what does that look like right here

so now i’m between 1 and 4 so i’m sorry now i’m between four and five

so now if i look between four and five let me take this off for a minute so

when i’m between four and five i have half of it below and half of it above

02:45

so they’re going to cancel out to each other so here i’m going to say

plus 0 or if you want minus 0 depending upon if you’re half full or half empty

with your glass but in any case this comes out to be

you add them all up it’s 11 over 2. so this integral right here from -1 to 5

by breaking up into its pieces i get 11 over 2.

now this doesn’t work in general this works because

most of my graph as you can see right here most of my graph

is made up of rectangles and triangles so you know if you can do that then great

that could save you a lot of work because this was a lot less work than if

i had to try to find the limit of a riemann sum three different times

so three different limits of riemann sums and that would have been a lot of work

02:46

instead we just relied upon some geometry and how to find the area

of some simple geometry formulas okay so next one

okay so suppose f and h are integrable and we know these values here

so now let’s find some other values some other integrals how about a here

what happens if you do um the integral from one to nine of minus two f of x

so when i try to do this i have to make sure the bounds match

because these three given integrals the first one is from one to nine

and the other ones are not and so i’m going to look at how to find this

02:47

right here so -2 of f of x and i’m going to pull the minus 2 out

and say from 1 to 9 this will be -2 and the number they give

us for this integral from 1 to they give us a minus one so this will be

minus one here so this would be two so that’s for part a for part b

we’re going to look at the integral from 7 to 9 of f of x plus h of x times dx

so what would that integral be so i’m going to go from 7 to 9 of f of x plus

02:48

7 to 9 of h of x and do they give us the values for these for f and

h from seven to nine and so they do from seven to nine were five

and from seven to nine for h were four so this will be

five plus four which is nine okay for part c

we’re going to look at the integral from seven to nine

of two f of x minus three h of x okay so this is a simple matter of

pulling out the two pulling out the three

and then substituting in the values that we know so we know this one here is the

five minus three we know this integral right here

02:49

is a four and so when we calculate that we get minus two two okay for part d

we’re going to look at the integral from one to seven of f of x

they don’t give us from one to seven so what can we do

so to go from one to seven i’m gonna go from one to nine

because they give us that value and then i’m going to take away 7 to 9.

so to integrate from 1 to 7 i integrate all the way to

9 and then i take away the seven to nine so think about this in terms of

area this is the area from one to seven so i’m gonna find the area from one to

nine that’s too much area and then i’m gonna take away the area

that was in this region so they give us these two numbers here

and so this will be minus one minus five so it’s minus and then a five

02:50

and so this will be minus six that’s part d there

for part e we’re going to look at the integral from 9 to 1 of f of x

and so to do that i’m going to say minus the integral from 1 to 9 of f of x and

they give us one to nine it’s a minus one

so if i do a minus and then all this is a minus one so now i get a one

and then for the last one we have the integral from nine to seven of

h of x minus f of x all right so let’s see how to work that

the integral from nine to seven and so i’m going to integrate here

02:51

from seven to nine of f of x and the reason why is because

that has a negative on it so when i integrate that one i’m going

to change the order of integration and then minus and then i’m going to

integrate 7 to 9 of the h of x and i switched the order of integration

here because i know the integral from seven to nine of h and so

since this one’s positive h with seven with nine to seven i’m gonna use the

minus sign now and so this will be 5 minus 4 which is just 1. so there’s f

alright so there’s some fun with integrals now let’s look at some interesting

questions here um think about the this is to help you with your

um thinking about the definite integral in terms of area and try to have some

02:52

kind of intuition there so we’re trying to minimize the value of this integral

in this problem here and they give us the function here this is the graph

of the function x to the fourth minus 2 x squared so there’s the graph of it

now between a and b well we don’t know what a and b are

let’s imagine though that a is minus 2 and b is 2. the whole thing right

minus 2 to 2. what would that integral be well there are some parts of the graph

that are above the x-axis you can see the shaded region

and there’s some parts where the graph is below the x-axis

and you can see those shaded regions as well now because you have some above and

some below you’re going to have some positive area

and you’re going to have some negative area

02:53

so if you were to integrate from -2 to 2 it may not be as much as if you were to

integrate just from the the part that’s below the

x-axis so we really kind of need to figure out what that is

so i’m going to start by trying to solve that

i’m going to try to solve the x to the 4th minus 2x squared and

figure out where that is 0. and the reason why is because if we look right here

you know where is that zero at where is that zero at

and the reason why i want to know that is because if i integrate between this

point and this point all of the area will be negative

which means i’ll have the minimum value of this integral

because remember this integral represents area and i want the area to

be the below the x-axis in this example because i’m trying to minimize it so

let’s see if i can find those zeros so let’s go back here and see if we can

02:54

solve this so i’ll factor out an x squared and i get x squared minus two

and i get x equals zero and then i get x equals

plus or minus square root of two and so those are the places on the x-axis

where we’re getting zero there so you know what we can say is that

is that the function x to the fourth minus two x squared is less than or

equal to zero on the interval minus square root of two the square root of two

that’s we found the zeros and in between the graph is negative

so a is minus square root of two and b is square root of two and this

right here will be minimum this will be minimum for this function f

02:55

of x right here this will have minimum value minus squared as minimum value

all right next example um and this one we want to show that the integral

lies between these two numbers here so we’re looking at the integral uh

between zero and one so there’s the graph it just looks like

the square root of x plus eight um it doesn’t look too curvy on this

interval here between zero and one now what we’re going to notice is that f

of x is increasing on this interval right your function square root of x plus

eight is increasing on zero one and so you know why is that important well

if i want if i try to find the area here um

i’m trying to find the area here under this graph

02:56

you know square root of x plus h just coming through here

if i want to find the area here at one what’s our height height is three

so the area of the shaded region here between 0 and 1

is less than or equal to three is the maximum right here

and you know what’s the width zero times so this is zero times three so this is

less than three that’s three is the area of the rectangular region

and the area of this one right here it just slices through there

so this area this one overestimates we have some part in here

and what about uh underestimating so you know because it’s increasing

02:57

throughout um zero here’s the minimum and so

you know what’s happening at zero here if we come across at zero here

then what is it it’s zero it’s one times and then what’s the height what’s

the height here when we have f of zero so this is two square roots of two

so this is the minimum and this is the maximum so we have this

inequality here and it’s given to us just by looking at

the geometry of the problem this is the minimum at zero here

because it’s increasing throughout it’s not going to come back down anywhere

else it’s just increasing throughout so that gives us the

approximation so even if you didn’t know how to find what that is

equal to you can still bound it between this is about

uh sorry this is supposed to be two square roots all right square root of eight

02:58

when x is zero uh but this is about two point eight and three is about three

so you know even if we don’t know how to find this integral here

maybe it’s hard to find the limit of the riemann sum here

we know it’s between these two numbers here so that may be good enough for some

problems it’s just to find an approximation depending upon what you want to

accomplish in real life okay next thing so just to formalize

the intuition that we’ve been trying to develop suppose

that f is continuous and the graph is above the x-axis

then the area under the curve is given by the definite integral

so let’s practice it sorry wrong button okay sorry so let’s practice this here

02:59

no geometry no properties let’s just go back to the definition one more time

and make sure we can do this so we’re looking at the definite integral here

we’re going to go from 0 to b [Music] we’re leaving b unknown

so this is 3x squared so we could go from 0 to 1 0 to two zero

to three point four b is unknown and we’re going to find the

area under uh area of the region so just to give a quick here’s

the probably three x squared and over here’s our b

it comes up so we’re asking for this area right here exactly and this is a nice

curvy parabola so this is not a rectangle this is nice and curvy and so

03:00

what is the exact area well we’re going to use the definition

we’re going to use a limit of a riemann sum so we have the limit

as n goes to infinity a summation here and we’re going to use f of our cks our

representatives times our delta and we’re going to sum from 1 to n

and what are our delta x so delta x we’re going to use as a b minus a over

n or in this case b minus 0 over n so just b over n and what are we going to use

for the ck for the representatives we’re going to choose n times delta

03:01

n delta x which is just going to be the b’s so making those choices right there

we’ll be able to calculate this limit of the riemann sum here

so n goes to infinity summation k equals 1 to n and then this is f of rb times

b over n so using the representatives here b and then b over n here

um sorry this is the last sub interval representative so

what are representatives going to be so we’re just going to choose [Music]

03:02

regular partition so our ck will be a times delta n

we just choose them to be case not ends or b’s so ck would just be k

so sorry about that this is just k here now when i put plug in k to my function

i’m just going to get right so the function is

3k squared and i can pull the b over n out of this so my next step here will be

limit as n goes to infinity b over n and then we’re going to sum k equals 1 to n

of 3 k squared okay so good so we can pull this out and

then plug in our representative into the function

03:03

so we get 3k squared so we can write this out

let’s just go down here so limit as n goes to infinity

i can pull the 3 out and say 3b over n and then sum from 1 to n of k squared

right so now we’re going to use the formula for this right here

so now let’s come over here and write that so we have the limit as n goes to

infinity 3 b over n and then for this right here we’re going to get

1 6 in n plus 1 times 2 n plus one okay so we can simplify this down a

little bit right all we did was simplify um

substitute in this formula right here here it is

03:04

and now we can try to simplify all this here we’re going to get um so this um

it’s gonna be this b right here is so i think the problem is the ck

still so the ck here is um [Music] okay so let’s go back up here to this um

03:05

a plus k delta x times delta x and let’s say what the a is the a is zero

the b is a b and the delta x is b over n okay so

this is going to be the limit as n goes to infinity

summation k equals 1 to n and so this is 0 and this is k b over n

so i’ll just write that out so that’s k b over n for the delta x that’s kb over

n and then this is just b over n okay so that’s good

so this right here is summation 1 to n now i’m taking my kb over

n and i’m going to do substitute it into my function

03:06

so to be 3 times the input squared and then times my b over n still okay good

and so let’s see what constants we can pull out of all this we can pull a 3 out

we’re gonna get b to the third and n to the third

so i’m gonna say three b to the third over

n to the third we still have summation from one to n of the k squared

okay so that’s to the third okay so this is perfect

so now let’s just use the formula for that right there

so this will be limit as n goes to infinity

three b to the third over n to the third and then this right here is one sixth

n n plus one times a two n plus one there we go so integrating from zero to six

03:07

using the limit of the riemann sum and then

we identify a b in delta x we have it and then we substitute the

input into the function by cubing it and multiplying it by three

and then we pull the three b squared b out and the n to the third

out we’re left with just k squared and then we just use the formula for the sum

of the k squares and there it is and now we can simplify this here a little bit

we simplify this right here we’re going to get b to the third

over two times two plus three over n plus one over n squared

all right and then n is going to infinity so this limit right here will turn out

03:08

to be um that’s zero zero so it’ll just be b to the third

so there we go this integral right here from zero to b of three x squared dx

is just b to the third and so notice the connection there between this one

and this one and that means that we’re ready for

the next video but first let’s talk about some exercises

okay wow that was a long video and now it’s time to look at some exercises

so lots of exercises to do over this video here

03:09

now some of these exercises are for riemann sums some of them or for

definite integrals but the idea behind these videos is to give you a chance

to work on some problems and to get some feedback from me

so in the comments below let me know if you’re able to solve these problems

let me know which ones were easy which ones were hard which ones you’d like to

see me do in a video and there’s nine and ten and here’s some more 11 and 12

[Music] 13 and here’s 14 and 15 so some of these are just working with

summations and looking at riemann sums there’s 17 of them so far here’s 18

03:10

- you have to use a riemann sum and they tell you

how many sub intervals to choose they choose they tell you which endpoints to

use they say equal length sub intervals so you have to make riemann sums a very

specific way and then here using some properties of

some integrals here as we did in some examples and then 21 and 22

write the expression out as a definite integral

use some properties here of definite integrals

okay so i want to say thank you for watching and our next video is going to be

over the fundamental theorem of calculus

and that will put everything together so we’ve talked about anti-derivatives

and indefinite integrals and today we talked about

definite integrals and so we’re going to

put the connection between them together in the fundamental theorem of calculus

03:11

and that’ll be our next video and i just want to say thank you for

watching and if you like this video please subscribe and like below

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