Definite Integrals and Riemann Sums (Using Limits To Find Area)

Video Series: Calculus 1 (Explore, Discover, Learn) Series

(D4M) — Here is the video transcript for this video.

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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

00:38
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

00:43
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

  1. 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

  1. 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
and i’ll see you next time if you like this video please press this button
and subscribe to my channel now i want to turn it over to you
math can be difficult because it requires time and energy to become
skills i want you to tell everyone what you do
to succeed in your studies either way let us know what you think in the comments

About The Author
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David A. Smith (Dave)

Mathematics Educator

David A. Smith is the CEO and founder of Dave4Math. His background is in mathematics (B.S. & M.S. in Mathematics), computer science, and undergraduate teaching (15+ years). With extensive experience in higher education and a passion for learning, his professional and academic careers revolve around advancing knowledge for himself and others. His work helps others learn about subjects that can help them in their personal and professional lives.

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