The Fundamental Theorem of Calculus (Examples and Theory)

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

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

00:00
[Music] what are the most beautiful theorems in all of calculus one
in this video i briefly review the intermediate value theorem
the extreme value theorem and then the mean value theorem
then i explain the first and second fundamental theorems of calculus
and how they are important so let’s turn some coffee
into theorems hi everyone welcome back i’m dave
this is the series calculus one explore discover
learn series and in this video we’re going to talk about
some of the most famous theorems in all of calculus

00:01
and we’re going to work out some exercises at the end so stick around
and so let’s get started okay so calculus theorems and we’re going to
start off with the intermediate value theorem here
and in this theorem is uh something that we covered in a
previous uh episode um so i just want to mention that
the uh series playlist link is in the description below
so you can find out all the videos in the series in particular the one on
continuous functions is where we talked about the intermediate value theorem
so um let’s just quickly review the intermediate value theorem
so we need a continuous function here and we need a interval a b so let’s draw
something like that so let’s say we have some function like this right here

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and let’s say this is a and this is b then we can come up here and find the f
of a right here and right here we have the f of b so
let’s say that’s about right here f of b and
what this is saying the intermediate value theorem says
if you choose any number m between f of a and f
b so if i choose any number here so let’s say i choose
this number right here call that m then i should be able to go to the graph
and come down and find a c guaranteed always you choose any number here m
in between the f of a and f of b choose any number in between
if this is continuous function you’re guaranteed to be able to go over
the graph and come back down and find the c there’s always at least one c

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so that if i plug it in i get out of the m no matter which m you choose
so that’s intermediate value theorem um and we’re going to
use that here in a couple minutes here so let’s talk about the extreme value
theorem also real quick so the extreme value theorem
if we have a continuous function then the function
must have a maximum minimum and a minimum value on that interval a b so
i like to review this one here with a uh example of you know
an example how to use it [Music] so we’re going to look at this inequality here
0 is less than or equal to 0 to 1 x to the fifth over cube root of 1 plus

00:04
x to the fourth dx is less than or equal to 1 6. now
in the last video that we in the last episode that we did
we talked about how to do a definite integral and we defined it
using a limit of a riemann sum now we did lots of examples last time but
it didn’t always work in other words we used summation formulas we
you know we did special functions it may not always be workable
in particular limit may not exist or it just may be difficult to calculate
so in some cases you may not want to be able to calculate the integral
or you may not need to calculate the integral and it may be sufficient to
bound the integral between two numbers in which case that may suffice for your
application depending upon how you’re trying to apply calculus to the real world
but in this example i’m just trying to find out okay

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i don’t know the value of this integral so can i bound it between two numbers
so i know at least something about this integral here
and we’re going to use the extreme value theorem to do that
so first off we’re going to uh use a function so i’m going to say what my
function is so x to the fifth over cube root of 1 plus x to the fourth
so there’s my function and i’m going to say notice f is continuous
on my interval here 0 to 1. now it is continuous right there right i mean
this is cube root so this is always going to be defined
the only place where it may not be continuous is outside of its domain but
you know the domain here is going to be all real numbers

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but we’re just interested in on zero one so we’re continuous on zero one so
by the intermediate value theorem by this theorem up here
so i’ll just say by intermediate value theorem f must have absolute max
absolute max and min and so let’s recall how to find them
well we need the derivative first of all so we’re going to go and work on the
derivative and then come back here and write it okay so after simplification
we’re gonna get x to the fourth 19 x to the fourth plus 15
all over three and then we’re going to get x to the fourth plus one to the two
thirds okay so taking the derivative of this using quotient rule then getting a

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common denominator simplifying it all in the end you get this right here
so you know the only critical number is when we hit 0 right here
so x equals the only critical number is 0. the only critical number
is zero where the numerator here is zero because you know the denominator is
always defined here x to the fourth plus one so that would never be zero there
so this is the only critical number so now to
look at the extreme values for this function
we’re going to say the little m for the minimum is going to be f of 0
so that comes from looking at the left endpoint so i’m going to plug in 0 here
and we get out 0 [Music] and when i look at the right endpoint here

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we’re going to say this is f of 1 and then you know substituting that
value in here we’re going to get 1 over cube root of 2 which is about 0.79
okay so we have bounded this integral between 0
and this number here or we’ve bounded the function
um but you know actually we can do better so observe
that 1 plus x to the fourth is greater than or equal to one
for x in this interval for x between zero and one this is
greater than or equal to one and then you know taking reciprocal taking cube

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root so we can come up with this right here inequality right here
and this will be cube root of one plus x to the fourth less than or equal to x
to the fifth here so taking reciprocal taking cube root
cube root is a increasing function so and then we can multiply by x to the
fifth on both sides x is in this interval here so good
so now we can integrate we can say zero is equal to the integral
of the left-hand side here zero dx that’s zero and that’s going to be equal
to the integral of the middle one here which is what we want so i’ll just say
f of x here and then that’ll be less than equal to the integral over here
and so when we integrate this right here we’re gonna get one over six and so

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that’s how we bound this right here between zero and less than or equal to
one over six right here [Music] so the question is now how did we get to that um
so what we’re looking at here is two different ways of finding this right here
this integral right here either a limit of a riemann sum
or using a fundamental theorem here so we can write out the limit of riemann sum
and calculate this and get one to five here but
there’s going to be a much faster way so let’s remember this right here this
integral right here and we’ll come back and figure out why this is 1 6 here
but in any case here’s a function it’s continuous extreme value theorem holds
once we know that holds we just need to check the

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boundary and any critical numbers and we can find the
maximum and minimum values and that can allow us to one of the things that can
allow us to do is to write out some inequalities
okay so enough of that review for extreme value theorem
um now so here here here are they all are
all three of them the intermediate value theorem
the extreme value theorem and let’s not forget about the mean value theorem so
when we have this mean value theorem down here
um again we did a whole video or a whole episode
over the mean value theorem there so i’m not going to
talk too much about that one but we are going to talk about
next the mean value theorem for integrals so let me just say here though that
notice that all three theorems here talk about f

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is continuous on a closed interval and then intermediate value theorem
talks about choosing a number in the range
and coming up with at least one c that hits that number in the range
an extreme value theorem tells us that we have to have an absolute max we have
to have an absolute minimum and then mean value
theorem says one additional thing and we’re differentiable so then the
slope of the tangent line remember the left side represents the
slope of the tangent line is equal to the slope of the line
through the endpoints so there’s those three theorems and we’re
we’re going to use them here in a minute
but now we’re going to go talk about the mean value theorem for integrals
[Music] okay so we’re going to start off here with f is a continuous function

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again so we’re coming up with a new theorem we have
intermediate value theorem we have extreme value theorem we have mean value
theorem now we’re going to have mean value theorem for integrals
and for all of these theorems we’re starting off with f is a continuous function
and we have a closed interval here so we’re going to say what are the average
values that f takes on over an interval so to answer this about
answer this question we’re going to take a sample
of some numbers from the interval so say we sample n of them
so we’re going to sample and we’re going to take
and divide it into in sub intervals of equal with so

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for example if we’re looking here at say 2 8 and we’re going to do
eight sub intervals or let’s do say 12 sub intervals um
and they’re equal width so what will be the width of them
to be eight minus two so delta x will be eight minus two over in the n
here i’m using 12 so 12 and so this will be 6 over 12 so that’ll be one half
so we’re going to go here to 2 and 8 and i’m going to divide them up into 12
sub intervals here and each one is going to have a half so
what about if we chopped it in half and so

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what are we getting here we’re getting five right so three this way three this
way and then we can come up here and a half again one two three four
um but it’s just better just to since we know what this is right
so this is two plus a half then this will be three
three plus a half and this would be four
and then so on four plus a half and then five
and then five plus a half and then six and then six plus a half and then seven
seven plus a half and then eight so we got one two three four
five six seven eight 9 10 11 12 of them and so we can you know take our
continuous function that’s somewhere on this interval two to eight and we can

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um use these uh values right here so we’re going to say this is
the average of the n sampled values is and so we’re going to take
these c’s in here and we’re going to evaluate them into the function
and then divide by the total number of them so it’s just like taking an average
you’re just adding up and then you’re dividing by how many of
them you have so c1 c2 c3 we have we have uh n of them so we can write this with
uh sigma notation it’s so it’s just one over n
right it’s dividing by n and this is sigma notation so we start at k equals
one to n so to be f of c one f c 2 all the way and then add up f c n

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now here we’re making the substitution right here so the
delta x here is just simply b minus a over n right so
what if we divide both sides by b over a so b
over a divided by and then what is this one over n
so 1 over n is just delta x over b minus a so when i come back here and i
and i’m looking at that 1 over n there i’m going to replace it with
delta x divided by b minus a there and now the whole idea is we can
bring that delta x inside of the summation it’s a constant
so we can bring it inside the summation if we want
in the examples that we worked out last episode we often took the delta
x out of the summation because it didn’t depend upon k
for the examples that we did and so now we are going to integrate both sides or

00:18
take the limit so we’re past the limit and what that
means on the left-hand side is that we’re going to add up
all of the values not just a finite number of them so here we are
thinking about 12 sub intervals here but what if you do 20 and then 30 and then
10 000 and then you know past the limit go to infinity
so now your sub intervals are just at every point you have okay so you know
past the limit the limit on the left hands uh the right hand side though
is called the average value assuming it exists
and so we say if f is integrable that means the limit of the riemann sum exists
and then we have the one over b minus a um as a as a as a as a multiple

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so this is what we’re calling the average value of f on a continuous closed
on a on a closed interval here for a continuous function
okay so i hope that gives you some kind of idea
about why we’re calling this the average value
and it really is taking into account and
averaging the idea behind an average you add up all the
the values and then you divide by the number of them
okay so let’s look at an example find the average value of this function
over the interval and what i’m going to do here is we’re
going to find the average value so the average value of
f um on this interval right here 0 1 will be 1 over b minus a right so the

00:20
a 0 and the b is 1 so this will be b minus a and then integral 0 to
that’s our interval right here and then we have 3x squared minus 3 dx
so this will be equal to the average value right here
the average value of this function um over this interval right here now at
this point we don’t know how to calculate this yet um without
using a limit of a riemann sum and by the way this number right here in
front one over b minus a in this example here
just comes out to be a one so this is just three x squared minus three dx
and so we’re going to take the time here to do one more example of evaluating a
definite integral using a limit so this will be the limit

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as delta x approaches 0 and we have f of and so our function here
is uh 3x squared minus 3 so we have f of
and we’re going to use the left endpoint 0 so we can say 0 plus and then k in
uh sorry k um k over n and then times one over n okay
so this is using the uh a is zero b is one and delta x is b minus a
over n which in this case is just one over n so left endpoint plus the change
and then times the delta x here so what do we get for this limit here
so we can write this as n goes to infinity now
so i need to put k over n into our function
um oops i forgot of course one of the most important parts here

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which is obviously the sum limit of a riemann sum
okay so this would be sum k equals 1 to n now i’m going to substitute k over n
into the function so this would be three and i ran into
space there so let’s just jump down here real quick
so limit as n goes to infinity riemann sum here we got a summation so
we’re going to go k over n into here so 3 k over n squared minus 3
and we still have one over n out here okay so we’re summing up all of this
right here and it’s just because we took the k over n into the function
so i plugged in k over n into the function up here
and so now we can write this as well we can pull the 1 over n out and so we have
summation and this will be um let’s write it as k squared n squared

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right so let’s say here we have three over n squared summation k equals one to n
and then we have a k squared left and then minus and then three and then we have
summation k equals one to n it looks weird without something in
there so i’m gonna leave the three in there
so k equals one to n and then we have a three
okay so i’ve just taken this out of the summation right there
and applied summation to each of these terms here apply summation here and
applied summation here now when i apply summation here i pull out the three
and the n squared and i leave the k squared inside here because the index
here is k and then i have summation here to the three
so here we go next step we have limit one over n now we have a formula for

00:24
this so let’s write it as three over n squared and this is what i
was talking about uh earlier you know sometimes when we’re integrating functions
we have summation formulas we’re able to work with this and then we’re able to
find the limit but generally speaking if you write a function here
it can be very difficult to find this limit of a sum
in any case for k squared for the summation from one to n of k squared
this is going to be one over n n n plus one times two n plus one and then
the last one here we have a minus three in
all right so we have the limit one over n times all of that
okay there we go so far so good now we have the limit as n goes to infinity

00:25
and if we look at our denominator here we have um we have here the
six n squared here and then we have one over n still
so we’re going to get the three n n plus one and then times the two n plus one
minus three and then i’m gonna multiply by six n squared here
in other words this is over one so i’m multiplying by
six n squared over six n squared so i have a common denominator in any case
when you look at all the powers and you either apply
the habitat’s rule or divide by the highest power
or use some theorem whatever you’re doing you’re going to get -2 out of all this

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so you can kind of see that by looking at the minus 12 on top here over the six
look at the coefficient of the highest power
keeping in mind there’s dividing it in out over here
okay so this is the average value now we can come back and say okay this
is the average value here the average of value of f over this interval is -2
and the way we did that is we just simply found this integral here
and calculated it now let’s think about this a little bit more though because
if it’s true that the average value is a typical value you know quote unquote a
typical value of the function then the number
average value of f should have the same integral as

00:27
f does and that’s what that’s we’re claiming right here
that the integral of the average value should be equal to the average value
and so let’s go see that this actually works so let’s recall that the average
value of f you know is just because we just saw a moment ago y but
it’s equal to this integral here and i’m going to give that a constant k here
i’m going to say that’s some constant k the average value of f is a number in
the previous example we found that number to be
minus two but generally speaking the average value of f i’m just going to
call it equal to some k okay so here we go
so the left-hand side the left integral is a to b of the average value of

00:28
f which i’m calling a k now remember that k is a constant so it
can come out of the integral that was one of the properties of the
integral we found last time now if you have nothing in here you can
either just put a dx or if you want to you can put a 1
um but you don’t have to put the one but i want to put the one and the reason
why is because what does this represent right here so if i look from
a to b and my function is just a constant one height of one here
so what’s this area right here represent or or what is this area
this integral right here is just the area and what is the area it’s the width
times the height right so what is the width it’s b minus
a and the height is one so the area here is just b minus a so the area

00:29
is b minus a so that means this integral right here which is
that area is b minus a so this is k times b minus a
yeah so this is just from a to b so this width right here is all of this b take
away the a and that’s what we have here b minus a times the height
okay so this integral right here is b minus a now
what is k and that’s a capital k i should make it look better that’s a capital k
now capital k is equal to this integral right here
and then so we have capital k which is all of that
so b minus a times all of that so let’s write b minus a down again
and then times the k and k is all this so 1 over b minus a and then integral

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from a to b of the function and as we can see right here the b minus a’s cancel
and we’re just going to get the integral of f
and so we can see that the integral of the average value
is equal actually to the integral of the f so yes this
uh average value of f has the same integral as f does
okay so that’s just something that you would think would happen
and in fact it does okay so now we’re ready for a new theorem the mean value
theorem for integrals and for the mean value theorem of integrals here
for this theorem we’re going to start off with well no big surprise here
a continuous function on a close interval and what we’re going to say is that
there exists a number c such that the average value the

00:31
the the integral on the right hand side the or the whole right hand side that’s
the average value right so there exists a c that is equal to the average
value so that’s one way to write it in words
if f is continuous on interval close interval a b
then there exists a number in the closed interval such that
the function value at c is equal to the average value now in the example
that we did um we found the average value of this function right here
3x squared um sorry we found the average value of this
function right here 3x squared minus 3 over the interval and we worked it out
we found the integral using a limit of riemann sum and we found
-2 to be the average value and so what that means is

00:32
this theorem holds because this is a continuous function right so f is
continuous if it’s continuous on zero one it’s just a polynomial is continuous
everywhere three x squared minus three and so this theorem holds
and what it’s saying is that there exists so so by the mean value theorem for
integrals there exists a c we don’t necessarily know what it is there exists a
c n zero one somewhere where if you plug in c
into the function you’re gonna get to the average
value minus two you have to get to the average value we found the right-hand
side we found we found that integral we found the
average value on the right-hand side it came out to be minus two so you have
to be able to input some c into the function and get minus two let’s put in a c

00:33
and see if we can actually find it so three
and then c squared minus three and that has to be equal to minus two
so i plugged in c into my function that’s f of c right here
and that has to be equal to the average value right here
now whenever you plug in a c into a function it doesn’t always look so simple it
looks simple in this case because our function is simple it’s just
three x squared minus three so this is three c squared minus three
but generally speaking plugging in a number into your function
this could be very complicated over here and so you may not be able
to actually go and find the c so the importance of this theorem
is that it tells you that there exists a c
even if you can’t find it it does exist there is some c where the function
evaluated at c is equal to the average value now in
this example we can actually go and solve for c

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can’t we we can move the three over [Music]
and get what one so c squared is divide by three
take square root so it’s going to be one over square root of three
or if you like writing threes square root of three over three
now this number right here is definitely in this interval here
oops that’s a c take square root right so we don’t need the negative so because
we’re in this interval here so this is the c right here and if you
plug in this c right here into the function over here then
we’ll get the average value and so that’s the mean value theorem for integrals
it’s a very useful and powerful theorem um now
we’re going to use it when we start trying to work out
the fundamental theorem here and so we’ll see
how we can use that to our advantage here in just a moment
so for this example right here we found the c

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but generally speaking you may not be able to find the c
knowing that it exists is very powerful and allows you to accomplish
uh improve other theorems or work with integrals that are difficult to work with
now keep in mind that this right here is very different
from the discrete case so what do i mean by that so
for example suppose you have a classroom full of students
say you have 56 students and you have an exam
and then you average up all the grades on the exam and you realize
oh the exam average is a 76 so you have 56 students and you
add up all the exam scores and then you divide by 56
and you get your average you get your average is 76.

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now just because the average is 76 doesn’t mean that even one student made a 76
maybe no one made a 76 just because the average is a 76
in fact it may happen that no one even made a c on the exam
but the average turned out to be a c a 76 or if you want to simplify things even
further suppose just two people took a test one of them made 100 and
another one made a zero the average would be 50
but no one made a 50 on the exam right so this is very different this is for
continuous function right when you only have 56 students you
know you don’t have a continuous function going on
to find the average but for the mean value theorem for integrals
of course you have a continuous function and we’re looking at the

00:37
average value over a closed interval okay so let’s go on now
and look at the proof so how does the proof work here for the
mean value theorem for integrals so let’s see here um this one good so
f is continuous on on uh a b and what we’re going to use is the
extreme value theorem we know the extreme value theorem holds
so by the extreme value theorem which i’m going to abbreviate as evt by
extreme value theorem so we know f must attain its absolute max and min

00:38
we know f must attain its absolute max and absolute min in a b
okay so we know there’s an absolute maximum
capital m and lowercase m so let’s write that down like that
so i’m going to use the lowercase m for the absolute minimum and upper for the
absolute maximum and this x here is for any x in this interval here for all x in
the closed interval here okay so you know what we can do now is we can go
to the average uh the average holds so the average holds what does that mean
so the average holds means that if i do the b minus a here and

00:39
um well we we have this right here so um think of it like this
you know this holds for for an x it holds for another x another x it holds
for all those x’s and you know what if we average them up right
well it’s going to be between m a lowercase m and uppercase m so
we have here m is less than or equal to the average value here
so you may want to think about like this so for example you know
going back to the exam analogy which isn’t all that great because it’s
discrete but i think you’ll get the idea if all the exam grades or the exam
scores are between 0 and 100 well then of course the
average of all the exams is going to be between 0 and 100

00:40
so if all of the output values are between these two right here then
the average value which is this right here is between m the min and the max now
what i’m going to do is i’m going to multiply through by b minus a
and then i’m just going to get the integral here
so that’s an a b minus a so multiply by b minus a
multiply by b minus a they cancel and multiply by b minus a here and so
now what we can do is we can use the intermediate value theorem here
so the intermediate value theorem here so by intermediate value theorem
uh so let’s come over here maybe now let’s go here

00:41
so by intermediate value theorem there exists a c there exists a c in a b
such that if we plug in c so f of c is equal to the average value here
and that’s what we’re looking for right there f of c is equal to the average
value right there so we have here the boundary right here and
because it’s think of this as the um c coming from the
intermediate value theorem here there has to be some c

00:42
that hits this value right here which is between the outputs
and so that’s how we prove the mean value theorem for integrals there is
by starting off with the extreme value theorem and bounding it
using the max and mins that it has to have and so we
have this intermediate value theorem right here
because we have it bounded right there and so yeah let’s uh
see what’s next first fundamental theorem of calculus so it’s time to go
let’s do it [Music] all right so the first fundamental theorem of calculus here

00:43
and this theorem is simply amazing and what it does is it makes a
connection between the uh derivative the anti-derivative and the derivative
and so what we’re going to do is we’re going to build this function here
knowing that f is a continuous function the starting place of all of our
theorems in today’s talk and we’re going to define another function
which i’m going to denote by capital f and so
we’re going to claim the theorem claims that this function is
differentiable and that it’s derivative the derivative of f capital f
is just f of x and what that means is that the derivative

00:44
of the integral from a to x is just found by
substituting the x into the function and that’s all you
all you need to do this is an amazing theorem so let’s
look at this function right here first because
um you know when you first see this you’re wondering
you might be scratching your head wait what that’s not a function that’s an
integral so you know let’s take a continuous function right
let’s look at something like two x squared minus five and let’s look on
two three so 2x squared minus 5. this that’s a continuous function

00:45
and that is a close interval so what is capital f capital f
is this function here it’s going to be the integral from 2 2 is our a
so 2 to x and then it’s going to be f of t so notice the function the integrand
says f of t so f of t now what is f of t well i can just go put a t in here
so 2 t squared minus 5 and then dt and so this is just um a definite integral
going from 2 to x and here’s the function right here 2x squared minus 5 dt
now we know how to work this definite integral here we can write it out as a
limit of a riemann sum and we can find the

00:46
integral here we could do that we could calculate it out
and when we’re done though it’ll have some expression of x in here
now you know you don’t need to go do that to understand
to understand this what happens if we plug in say um 2.5
right so x is bounded between two and three right so it says this function is
defined for x between 2 and 3 so i can plug in 2.5 so this will be the
2 to 2.5 of 2 t squared minus 5 dt so how do you calculate inputs and outputs
of this function you input something here which means you put it here
you substitute it in here and now you have to go calculate that limit of the
riemann sum and you get out a number so you get out number

00:47
whatever that number is that’s the output for 2.5
so that’s interesting it gives you it’s a way of
defining a function using an integral so i can calculate say point eight
that’ll be the definite integral from two to two point eight
of two t squared minus five dt and again we go calculate that out
and the way we know how to calculate this out is using a limit of a riemann sum
so i could say limit riemann sum i could find my delta x’s and my a and my
my a and my b and i could go and calculate this limit out and we’ll get a number
so that’s how you find the output for the input 2.8
and you can keep doing this for any input between two and three
this in fact is a function because when you input something you’re going
you’re only going to get one output here so

00:48
i just want to make sure that everyone’s okay that this capital f here
is actually a function now actually it says more it says is
actually a differentiable function so capital f is a differentiable function
and when you look at this right here you might say okay well
that’s a differentiable function what does that mean to me
because how do i take the derivative of an integral
i mean we’ve never seen anything like that for example
if my function is sine over x squared plus one i know how to take
the derivative i can use the quotient rule i can use chain rule if i need to
maybe i’ll put a three in here you know i can use my derivative rules
and i can use if i don’t have any derivative rules
available then i will have to resort to definition of derivative and so i can
take the limit of the difference quotient

00:49
and try to find the derivative so we can do the same here
to find the derivative of this function right here
we can since we don’t have any derivative rules for such
a function looking like this we could resort to the
definition of derivative and we can find out the derivative of this
but if you do that and you work it out what you’ll get is this right here the
derivative of the function at capital f the derivative of this is let’s get rid
of all this the derivative of this is very simple in fact
the easiest derivative to take if your function is defined
using an integral then this theorem right here says the fundamental
theorem of calculus says the derivative of this function right
here whatever it is well actually it’s just plug in the x so
that’s what this right here is saying is plug in the x and that is equal to

00:50
your derivative so the derivative of f [Music] the derivative of f of x is just
plug in an x 2 x squared minus 5. easy peasy just plug in the x
there’s your derivative so this function we can do inputs and outputs and we
realize it’s a function and to calculate the inputs and outputs
we calculate the limit of a riemann sum so if i input 2.1 into my function i put
2.1 here and i go calculate a limit of riemann sum and i
get my output however to find the derivative we just simply substitute in an x
and that has to be the derivative so that’s actually very powerful functions
that are defined like this it’s very quick and easy to find their
derivatives because their derivatives are sitting right here

00:51
and this says something very profound about anti-derivatives and derivatives it
really gives meaning to the word anti-derivative in the sense that
derivative and anti-derivative undo each other okay so let’s look at
um some examples now so let’s look at these examples right here
so in part a here the function g is given as an integral
from minus one to uh from minus one to x
and we’re given this function right here t square root of t
squared plus one so g prime is so what do i check before applying the

00:52
fundamental theorem i check my lower limit of integration is a constant
and my upper is an x so to find the derivative i just simply
plug in the x this is going to be x square root of
x square plus one and this is going to be defined for x greater than
or equal to -1 there so part a here’s the function and there’s the derivative
now for part b though it’s a little bit different because
we don’t have a constant lower limit and an x in the upper limit so we do
have a constant lower limit but we don’t have an x here instead we have an
x squared so we have here an x squared sitting right here

00:53
so what we’re going to do is use the chain rule
so the derivative instead of having an x there we have a function sitting inside
of our function so we have a function sitting inside of that x there
composition of functions so the derivative will be [Music]
put in x in here so we have x sine x times the derivative
of the x squared which is 2x so in other words we just have the regular
x and then times the derivative of the x squared
and that comes from the chain rule and so another way of writing that is just
2x squared sine x so there’s part b there now for part c um

00:54
we have 2 to square root of x so for part c g prime will be sine of the x over x
[Music] then times the derivative of the square root of x
and so if we work that out we’re just going to get minus sine x
over the x to the third x and then we’ll get another thing here so

00:55
x squared here or x to the third okay so here we’re getting the derivative of
square root of x sorry this is sine x over x
and then we’re gonna get here so that’s what x to the one half right
so one over two square roots of x okay so for part d
so there’s 4c there now for part d we’re going to have a problem because
for part d we have x squared to x to the third so first i’m going to write d as
g of x i’m going to rewrite it in a form i we can use so i’m going to go from x
squared to a c natural log of t dt and then
i’m going to continue on and go from c to x to the third
and so this is for some c between x squared and x to the third

00:56
so this is a property of integrals that we’ve been working on here
and now this one here doesn’t match or this one here doesn’t have a fixed
lower limit so we need to change the order here
so let’s write this one first c to x to the third
natural log minus and now switch the order of integration
c to x squared and the natural log of t dt okay so
for this for this d here we can do this uh simplification here
um we’re not simplification we can break the integral up into two integrals here
and so for this first one here we’re gonna get natural log of x to the third

00:57
times three x squared and then plus natural log of x squared times a 2x
but with a minus sign here now [Music] you got to be careful when you’re doing
these right here this wasn’t 0 to x squared these should have been
x squares here so we’re going to treat the 0 to x squared
we’re going to treat the x squared as the x and then times the derivative
of the x squared so this will be 2 x to the third and then sine of x squared
and right so just simplifying that 2x to the third there
and this one here for c we’re going to treat that square root of x as an x

00:58
so we’re just going to do sine square root of x over square root of x
and then times the derivative of square root of x so these should have square
roots on them and then we can simplify this a little bit sine square root of x
over 2x so just sine squared of x times one and
then right square root of x times square root of x
is an x if you have a square root of x still up here so
square root of x all right so there we go there are some good media
examples there you gotta watch out if you don’t have an x there
zero to x square so i just treat it that x squared is a regular x like i would
just put that x in x squared in there for the t
and then sine t but because it’s not an x it’s an x squared i then do times the

00:59
derivative which is 2x and the same thing here for c
for c here i uh it’s 2 to square root of x so i put the square root of x in
wherever i see a t but the square root of x is not an x so
i do times the derivative right there all right and so here we have natural
log of x to the third times the derivative of x to the third
and then natural log of the x squared and then times the derivative of the x
squared okay so there’s some good media examples there for us
now let’s see if we can take a look at the
proof here and see why this theorem is true the fundamental theorem of calculus

01:00
[Music] so [Music] we’re going to start off here and fix an x [Music]
in the interval here a b [Music] now because we’re going to be using some
theorems coming up here they give the existence of
other numbers i just want to make it clear that i’m fixing this
because typically you let x vary and we’ll talk about that in the end
but right now let’s just fix the x and we’re going to suppose
that x plus a little bit of an h is in a b also
in other words h is really small so if you choose an x in here
i can add a little bit to it really small little bit if need to be

01:01
and i’m still in here but h is not zero [Music] okay so [Music]
so h’s are think of h is something very very small
small enough to keep x plus h in here okay so then what can we say
well the function here that we’re defining right here capital
f if we substitute in x plus h minus and then we substitute in x
we can write this out so this will be a 2 x plus h and then f of t dt
minus and then this will be the integral a to x f of t dt
right so that’s just minus the f of x and this is f of
x plus h so i just have an x plus h here

01:02
and an x here and the difference here is this right here
now we’re going to use the mean value theorem for integrals so by
mean value theorem for integrals so mvt i mean value theorem for integrals
there exists a number c remember we may not always be able to
find that c but we know it exists so between this x and x plus h
so think of x is fixed and we have x and x plus h
such that if we plug in the c we get the average value
so if we plug in the c we’re going to get the average value
over this interval here so it’s going to be x plus

01:03
h minus x and then the integral of um x to x plus h and then f of t dt
f of t dt so i’m taking the average value here
over this interval right here where x is fixed
and a b and x plus h is a little small change in x so we know there’s some c
no matter how small h is there’s some c the mean the mean value theorem for
integrals guarantees that c i’m equal to the average value right here so if i um
divide this right here by an h i’ll divide you know by h
then what happens so we get um f of let’s go here [Music]

01:04
so we’re fixing x in here and we got a little bit of a bump in
an x but to keep it in here and then if i plug in x plus h i here
and i plug in an x i’m here and now we have the mean value theorem right here
there exists some c which is equal to f of c is equal to the average value so
now what happens if i divide this right here
by h so we have all of this right here and divide by h
so it looks like we’re setting up the derivative right here right
this is the difference quotient for capital f
and if i divide this by h over here then we’re going to get 1 over h
and we’re going to get the integral from x to x plus h right because we

01:05
came up with these two right here um but actually that doesn’t have
uh what we need in here because what we need here is this is f of t dt and
[Music] so this is x2 x plus h and we didn’t write that out so actually
this can be written as now because it has this minus here so
let’s actually write this out here a little bit more and let’s go a to x
f of t dt and then plus x2 x plus h f of t dt

01:06
and then the minus the this one right here and so these actually add up to zero
don’t they and so we’re getting here is just equal to the x x plus
h and then f of t dt okay so if i go from this right here goes from
a to x plus h and this one right here goes from a to x
but that’s not really in the form that we need we need it in this form right
here because we’re looking at the average value right here this right here
on this interval right here from x to x plus h so we can um go here from a
to x plus h let’s go first a to x and then finish that off and go from
and then finish that off and go x to x plus h
so so this one right here is equal to the sum of these two

01:07
minus this one and now we see these can’t add up to zero
so we just get this one right here so f of x plus h minus f of x
without the a’s in it i get this right here so now if i divide both sides by one
over h or i’m sorry if i divide both sides by h i get this right here
and i get one over h here but this is exactly what this is so this is f of c
right here [Music] so this is f of c right here perfect now what is left is to
take the limit as h approaches zero now what’s happening as h approaches zero
so if i take the limit here as h approaches 0 then i get the derivative
and what do we get over here when we take the limit as

01:08
h approaches 0 of the f of c so we’re taking the limit as h
approaches zero over here now where is the c c exists because of the mean value
theorem for integral c exists and it’s between x and x plus h but h is
going to zero so this is getting really small and c is in here
c squished and as h gets smaller and smaller
c is getting closer and closer to x so when i take this limit right here
that’s f of c so i take the limit of both sides here
limit of this side is the derivative and the limit of this side right here
because the c is getting closer and closer to f
and f is continuous this is f of x right here so the derivative of of f
is just f of x and that’s it so the main lifting uh of this theorem right here
comes from the mean value theorem for integrals which
if we look back at the proof of that it was

01:09
we used the extreme value theorem and the intermediate value theorem
so now we have a mean value theorem for integrals and we have a
first fundamental theorem of calculus and so that right there
gives us our derivative you just simply substitute f uh x into the function
given all of these conditions here okay so let’s see what’s next
okay so the fundamental theorem of calculus usually comes
in two parts the first part is very exciting because we’re taking the
derivative of an integral and when we define a function that way
the second part here we also begin with continuous function

01:10
and to me this is the most exciting one right here because what this tells
us is that the um definite integral defined on the left hand side there
which up to now we’ve been using to find that value the limit of a riemann sum
and what this theorem is saying is that on the right hand side
you have an anti-derivative and we worked a great deal of time for two
episodes on finding anti-derivatives so if you find those anti-derivatives
all you got to do is plug in b plug in a and subtract
and that gives you the definite integral and now for a large number of functions
will be able to use the fact that continuous function is inequal
and we can find the actual value by looking at anti-derivatives

01:11
this is incredible so we’re going to be able to start integrating now without
finding limits of riemann sums and so let’s look at some examples so first
example is our first three examples are these three right here
and so let’s get to work on them so when i’m looking at number one here
i need to find my f of x is two x minus three
and i need to find an anti-derivative of two x minus three so we’ve already
studied how to solve this problem right here
we can find the antiderivative of two x minus three
and if you remember how to do that we’re going to do 2 x squared over 2 minus 3x
plus c and so there’s our f of x there’s an anti-derivative
in fact there’s all anti-derivatives because c is an arbitrary constant

01:12
so i can put any constant i want because it says here in this theorem where f is
any anti-derivative now if i can choose what anti-derivative i want
and that’s what the theorem says then i can choose my constant to actually be
zero so this is an anti-derivative right here
and i’m going to use this one right here to find
the definite integral right here so this was an indefinite integral
and you know this simplifies it’s just x squared minus 3x that’s my
anti-derivative so here we go we can solve number one
using this anti-derivative right here so the definite integral from -2 to 0
of 2x minus 3 will be a limit of a riemann sum but

01:13
we don’t need that in this example we’re going to use the fundamental theorem
and so what i need is a antiderivative which we found right here
and i need to plug in the b right the b is the
zero so i need to plug in zero here and i need to plug in minus two now we
have a special notation for that so i’m going to write down my f of x
which we found down here using indefinite integration
and i’m going to use this notation right here so this notation means
plug in zero first and then minus and then plug in the lower limit
of integration so this will be zero squared minus three times zero [Music]
so i plugged in zero here and here now all of that let’s put uh parenthesis here

01:14
minus and then plug in two everywhere so minus two squared minus three times
minus two now when we calculate all this we get zero here
so zero minus and now we’re going to get four and then plus six
so we’re going to get minus 10. wow and that was so much easier than
doing a limit of the riemann sum so we found the anti-derivative of this
right here here it is using indefinite integration and you
don’t need to go do this all down here we’ve practiced that a lot we can just
do it right here so 2x the antiderivative is x squared of that part
the antiderivative of minus 3 is minus 3x and so we substitute in 0 everywhere
here minus for the lower limit and then we
substitute in the lower limit everywhere when we calculate all those numbers out

01:15
we get minus 10. so number one here is minus 10.
so let’s look at number two now so number two is zero to two
we have two minus four u plus u squared du
okay so the way we’re going to do this is we’re going to need an anti-derivative
so i’m going to integrate this right here it’s going to be 2u minus 4
and then we have u squared over 2 and then plus
u to the third over three so i found the i found an anti-derivative of this and
now i’m going to evaluate it so we have two to zero
okay so this means plug in two everywhere and that gives me my f of b and then

01:16
minus and then i plug in my a everywhere so let’s do that
so actually may simplify this first well it’s not much of a simplification
so here we go two times two minus and that’s already a two
and then i put in the two to get four plus eight thirds so all of that
minus and now i plug in zero everywhere so two times zero
minus two times zero squared plus zero over three
and if we calculate all these numbers up we get four thirds
so there’s number two there we find the antiderivative we plug in the lower
i mean sorry we plug in the upper limit and then
minus and then we plug in the lower limit there this is all zero here and we can

01:17
calculate that that up that’s just four thirds
okay so let’s try our third example now this is much more fun than doing limits
of riemann sums perhaps so number three here and
so to work on number three i’m going to simplify here a little bit first
i’m going to say 3x to the fourth and then minus 2x plus 1 i’m looking at those
powers there and i’m going to do some dividing first
so i’m going to divide each of these by 2x squared
so i’m going to get 3 over 2 x squared so that’d be the first term there
and then we’re going to get minus two x squared over two x squared so that’s
minus one and then the last one is one over and so i’ll just write that as a

01:18
one half x to the minus two so that could be your first step there
is just to simply divide it out and now i can use the integral for each
of these terms here so i’m going to integrate the 3x squared
and then i have an x squared here so x to the third over 3
minus x and then now i’m going to add one divide
so x to the minus one over minus one so i’m gonna minus one half and then
it’s gonna be over x and this is all going to be evaluated from one to two
okay so [Music] yeah now so that uh one half is there and then
this is x to the minus one over minus one
that’s why that becomes a negative there okay so

01:19
if i substitute into two everywhere i’m going to get
three over two times eight thirds minus two minus a fourth all of that bracket
minus now plug in a one everywhere so when i plug in a one i’m getting
three over two times one third minus the one and then minus the one-half so
when you’re using a minus here make sure and use parentheses around all of this
because this is everything that you get from plugging in two
minus everything you get from plugging in one
and so you know when you start working all that out
you get 11 force when you simplify it but you want to make sure this
minus sign goes for everything and then you just work it out and you get
you get a number so this integral right here the

01:20
the integral right here number 3 is just 11 over 4.
all right let’s do three more okay now number four here
we’re going to expand this out so i’m going to have
square root of x times now let’s do x plus 1
times x minus 2. so it’ll be x squared and then it’ll be a minus 2x plus a x
so that’ll be a minus an x and it’ll be one times minus two which is minus two
so that might be your first step and then i’m gonna write it like this
this will be x to the two plus one half which is better known as five halves
minus x to the one plus a half which is three halves minus two and then

01:21
x to the one one-half and then now dx so i would write it like that
so that we’re ready to use the power rule so i just distribute square root of x
times each of those and we’re going to get these fractional
powers but that’s okay so when i integrate this first one here
it’s going to be x to the five halves plus one so
seven halves and then i’m gonna divide by seven halves
which is the same thing as multiplying by two sevenths
minus and now x to the three halves plus one so five halves
and then divide by five halves which is the same thing as multiplying by two
fifths minus two times and then x to the one half plus one so three halves
and then divide by three halves which is the same thing as multiplying by

01:22
two thirds and then now i’m going to evaluate from two to zero
so when we’re coming in here with zero everywhere we’re going to get zero and
zero and zero so all that’s zero then minus and you can put a zero here to
help you keep track of that so plugging in zero everywhere we get zero
and then minus and then now we’re going to plug in a two everywhere
so we’re going to get two sevenths times two to the seven halves
all right let’s write it like that two to the seven halves
minus two fifths and then two to the five halves
and then here we get four thirds and two to the three halves
and so we’re gonna get zero minus all that

01:23
so you really gotta be careful with your minus signs there
now when we simplify all that we’re gonna get 208 square root of two over

  1. and you should really check that to make sure that you can do that
    just because you’re in calculus doesn’t mean
    you can ignore the arithmetic make sure you can do that that’s four thirds okay
    so this integral here and sorry this integral here number four
    we found it right here all right so now let’s do the next one number five
    here we go number five so we’re integrating from zero to pi

01:24
of sine two x cosine x dx so what can we do we don’t have a way of finding an
antiderivative formula for just sine 2x cosine x
in other words we don’t know what this right here is off the top of our head
but we know substitution don’t we so should we try this to be the
u that’s just a guess let’s try this and see if that is anything useful
so this will be d u will be what 2 cosine 2x dx
right but we don’t have a cosine 2x we need a
so the derivative of sine will be cosine of 2x
and then times the derivative of 2x which is 2
but now we’re not worried about this 2 we can put in the 2 and divide
but it’s the cosine 2x that we’re having a problem with

01:25
so let’s try to think of something else if we try u to be cosine x
then what will the d u be minus sine x dx right but we have a sine 2x
we don’t have a sine x so we’re going to need some something else
here how about let’s use a trig identity sine two x equals two sine x cosine x
let’s see if this helps this is a trig identity
sine two x equals two sine x cosine x so let’s use that trig identity right here
so this will be 2 sine x cosine x times another cosine x so that’ll be squared
um and so i don’t know maybe now let’s try this derivative here

01:26
because now i’ll have a u squared and i have a sine x dx we have a sine x dx
the only thing is to adjust the constants so i’m going to
pull the two out and put a minus sign in and write it like this
minus two and then integral of cosine squared and then with a minus sign
x times a dx here so this will be a negative 2 times
a negative those combined together to give us the positive 2
and this is still cosine squared and this is still sine x
dx so the way i wrote it like this is because there’s our du sitting right there
looking pretty so this will be minus 2 the integral of
this is the u so u squared u so now we can integrate this will be minus two

01:27
u to the third over three plus c and now we can come back in with the cosine
so this will be minus two thirds and this will be u to the third so this
will be cosine to the third x plus c okay so that’s just kind of a review
of what we went over in the indefinite integrals by substitution integration by
substitution episode so we can integrate that
using a u substitution and we get this right here
now to solve our original problem we need an anti-derivative of this
and we found one that’s sitting right here here’s an antiderivative of it
we don’t need the plus c we just need an antiderivative
that’s one of the beautiful things about this theorem is that
f is any anti-derivative and so i’ll choose this one right here

01:28
so this will be equal to minus two-thirds
and then what was it cosine to the third right cosine to the third x
now let’s write it better cosine to the third
and then so there’s our antiderivative and then zero to pi so
now we can continue on by substituting in here a pi
cosine to the third pi minus and now substitute in the zero
so minus two thirds and then cosine to the third of zero
so that’s minus one so that’s a positive so that’s two thirds
and that’s a one um actually it’s a minus and i’m having a lower limit so that

01:29
should have been plus and then so that’s a one there and one to the third
cosine zero is one one to the third here’s another two thirds
so the answer is four thirds now when you write it like this it’s kind of
a big jump and we erased our scratch word down here
so i can show you this up here but that’s kind of a big jump to go from
here to here i found an anti-derivative and i had to go look at
all the scratch work over here i’ll show you a much better way of doing this
right here without having to go and work in two different places
it’s kind of awkward when you have to work in two different places and
piece together your work so let me show you how to do this
without doing that although some people prefer to
work it out in two different places and to be honest depending upon your
integral it may be that your scratch work is 90 of the work
and that could take a great deal of time and energy and space

01:30
and so you know having it in two different places
is sometimes appropriate but for this example here i think it’d be easier
to just work it all out together without doing it in two places
so let me show you how to do that now so let’s work on number five again
but this time i’m just gonna put it all in one place so here we go
so first step was to make a trig substitution right sine 2x is 2 sine x cosine x
dx so that was our first step there now we’re going to make a u substitution
with u is cosine x and so d u is minus sine x dx
so that’s still our substitution so we still want to substitute
an integral in all x’s into an integral of all u’s
now we have this 2 here and so we have a minus here so we still need to

01:31
adjust the constants so what i’m going to do is
i’m going to put a minus 2 here first and then integral
0 to pi and then i’m going to say um actually that should be squared there
right so just to make sure we’re clear on that sine 2x is 2 sine cosine
there we go so this is sine 2x is 2 and sine times cosine so i’m going to
have two cosines that’s the cosine squared there
sorry i just forgot to put the square now with this 2
i’m pulling it out and i’m putting a minus in front of it
and i’m writing cosine squared and minus sine x dx
okay so i still have minus two times the minus sign
so basically the only difference that we’ve doing so far is we’re just still

01:32
placing the zero to pi on there and we’re not changing anything
about the definite integral there we’re just zero to pi
zero to pi this is the same work we did here and here
but now the next step is important so i’m not going to put my
limits here yet i’m going to go from all x’s to all u’s so this is u squared
and this is all d u right here that’s d u but now this is no longer zero to pi
so this is x this is x equals zero to x equals pi but let’s
transform them to u’s so for example when x is 0 when x is 0 what is the u
so what is cosine of 0 so i’m trying to transform the lower one here
when x is 0 cosine of 0 is 1 so this is going to be a 1 here now what

01:33
about the upper limit here so this is when x is pi so when x is pi
what is cosine of pi that’s minus 1 cosine pi is minus one so this will be
minus one here so now we’ve gone from all x’s to all u’s
and we’ve changed our limits here so now this integral is
much easier to work with this will be minus two times u to the third over three
[Music] and then i’m going to go from one to minus one
now when i substitute in my minus one i’m going to get what minus one third
and then minus and now substitute in my lower limit
and so we’re going to get two times what minus two-thirds or so differently
four-thirds which of course is what we got

01:34
before so this doesn’t necessarily save you any steps
it just makes it look cleaner to me in the sense that here
we’re using equal signs throughout here and so i can i can start right here and
say this integral is equal to four thirds and so it’s just
a little bit of a cleaner argument instead of having to work in two places um
you know so some problems you might want to take this route
especially if your steps are really short if you only need two steps to
solve your problem it’s just easier to do it one location
okay so that takes care of number five two different ways
now let’s look at number six okay number six here lean board

01:35
number six so zero to pi and we got absolute value here
so the absolute value here is what makes this fun
now let’s actually integrate it without the absolute value
and just see what we can do with that so what is an antiderivative for cosine
it’s a minus sign is that true no it’s sine
if you take the derivative of sine you get cosine
so zero to pi so i substitute in pi first minus now i substitute in zero
now sine of pi right so i substitute in pi first
and then minus and now i substitute in zero that’s our fundamental theorem
now sine of pi is zero and sine of zero is zero

01:36
so this is all 0 and i think that’s interesting because if i look at cosine
from 0 to pi what do we what do we look like right so here’s what um
here’s zero um and then pi over two and then pi right here we’re already at
minus one and so when we look at zero to pi
think of think of it as like this area right here so why is the area here zero
well this this part right here and this part right here
undo each other this is positive area because we’re above the x-axis
i like to sometimes refer to this as negative area
just intuitively because it’s just area that’s below the x-axis
so this is negative this is positive they’re equal in the amount so they

01:37
come out to be zero here this integral is zero if we integrate from zero to pi
over two then we’ll just have that area there
so if we integrate 0 to pi over 2 of cosine x well it’s just still sine x
but now 0 to pi over 2 and so pi over two comes in what’s sine of pi over two
minus sine of zero sine of pi over two is one
that area right there is one exactly one and this area right here is exactly
minus one so no wonder this integral right here from zero to pi
comes out to be zero because they’re positive and negative area there
so what about absolute value of cosine how does that change so let’s get rid of

01:38
all this fun stuff now when i’m looking at cosine from 0 to pi
i get this part right here above and i get this part below but what is
absolute value doing absolute values taking the negatives
and turning them positive so now i’ll get this part right here
and it becomes positive all these all these negative y’s all these negative
outputs now become positive outputs here and so
now i’ll put the tick mark down here pi right but these areas are still the same
but now they’re both positive so this integral is just going to be 2
without doing any work now i mean i just use geometry and and the
integration from zero to pi over two but you know let’s put some work in here
so i’m going to integrate from zero to pi over two and just say cosine plus

01:39
and now i want to go from pi over two to pi
in fact i’m still going to just leave perhaps the value on here i want to take
this one step at a time so i’m going to i’m going to i’m just gonna break up
zero to pi zero to pi over two and then pi over two to pi
so i didn’t do anything to the function here
i just said this is zero to pi over two and this is pi over two to pi
and the reason why i picked on pi over two is because that’s where it hits
zero and to the left it’s positive for cosine and to the right it’s negative
but for absolute value it’s always positive
so this will be zero to pi over two and now here on this part
i can remove the parenthesis uh the absolute value
because from zero to pi over two cosine is already positive

01:40
so i don’t need the absolute value here on e
on this part right here though from pi over two to pi
that’s when we’re down here so here i’m going to put a minus cosine x
you might say oh that’s changing negative well no
because cosine x on this domain the outputs here are negative
and so to be a negative times a negative which will make it a positive
so now we can pull out the negative here or we can just integrate each one of
these we could just do the antiderivatives right away so this will be
minus sine x or sorry sine x from 0 to pi over 2 plus
minus sine x from pi over two to pi so we can do the

01:41
sine of pi over two which is the one minus the zero plus now we can do the
pi here so minus sine of pi now minus and now use the pi over 2 here
got to be careful for all these minus signs
so for for this integral so i have a sum of two integrals
this is the first integral here so i use pi over 2 i get a 1
and then minus the lower one when i plug in zero i get zero
so these this right here is from the first integral and then plus and now i
have the second integral so first i plug in the pi so it’s a pi so minus sine pi
minus and now plug in the pi over two and so now let’s check all of the this
is a one plus let’s see we get for all this sine of pi that’s a zero

01:42
so plus a zero minus a that’s um sine of pi over two is a one
this is some negative negative so that’ll be plus one
all right so we get out the same two there so you sorry using
whenever you have an absolute value in your integrand
a good idea is to look at the function very closely
and break it up according to the domain according to where it might be positive
and where it might be negative and then integrate multiple integrals
because absolute value of cosine x is really a piecewise
function you want to think of it as a piecewise function
on this part right here it looks like cosine on this part right here it looks
like minus cosine and so on so breaking it into two integrals
and then just integrating each one of those
but to be honest this problem was a lot more fun thinking about
the geometry of the problem so number six was a lot of fun

01:43
i hope you enjoyed it all right let’s look at two more here seven and eight
so what can we do with seven so we’re integrating um [Music]
one over x squared plus sixteen we’re going to go from four
square root of three and we have one over x squared plus 16
and then dx so here we need a formula here from
we have one over x squared plus one dx i believe that’s arc tan of x plus c
one over x squared plus one is tangent inverse plus c so we have a 16 there

01:44
so what can we do we we need a one here so let’s try to get a one here and see
what happens if we do that so have zero square root of three here
and i’m going to um say you know divide this by 16 and multiply it by 16.
so let’s say we have x squared plus 16 and then i’ll do over 16.
so let’s divide the top by 16 too all right so i divide the top and bottom
by 16. and you know to be honest i want to try to help as many people as
i can so let me just try to practice the algebra here first or
or not practice but show you the algebra here so that we can focus more on the

01:45
calculus but i wouldn’t i want to look make it look like this
but i have the 16 in the way so like i said if we divide top and bottom by 16
what does that do is that a good idea well this is 1 over 16
this will be x squared over 16 and this will be plus 1.
and so writing it like this 1 over 16 and then this will be x over four
with a squared and then plus one so that’s think of that as x over four
and then all of it squared and then plus one and so let’s write it like this
1 over 16 out here and then 1 over x over 4 squared plus 1.
so when i write it like this this looks like something i would want to integrate

01:46
because i can make a u substitution right here for this
and then we can get it to look like that so the u
will be x over four okay so i want to do this
trick here and this is the first time you’re seeing it there’s really no magic
involved it’s just simply factor out a 16 is probably another way to think about
it so factor out a 16 here all right so let’s do this calculus part
over here now again we’re trying to use this form over here x squared plus 1
to get the tangent inverse here so let’s say we’re going to factor out a 16
and then i’m going to put it out here so i’m going to say 1 over 16
0 4 square root of 3 and then here we’re going to get x over 4 squared plus 1 dx
so you don’t necessarily want to show every tiny little bit
of algebra in your calculus problem to check that then i factored out 16

01:47
right just simply multiply it through back
so the 16 times the 1 of course that’s the 16
and then 16 times this will give me the x squared over
16 but times 16 those will cancel and i just get the x square back
okay so there’s the good step there and then i showed you all the little tiny
bits of pieces of whatever you want to see but now i’m going to say u is
x over 4 and so d u is 1 4 dx so this will be my u and i need a 1
4 now i can borrow part of that 1 16 and put a 1 4 in here so now let’s write
it as 1 4 integral 4 square root of 3 1 over x over 4 squared
plus 1 and then a 1 4 and a dx so i just take the 4 and the 4 gives us a 16
but i put the 4 over here so now i have my d u

01:48
i have my 1 over u squared plus 1 so now i’m ready to go to all u’s
so going to all use i’m going to wait on my limits here
make sure i can get all u’s this is to be 1 over u squared plus 1.
and then this right here is my du and then now this is this is our this is our
x’s so x equals zero and x equals four square roots of three now i
need to change them into u’s so what will the u be so whenever x is zero
whenever x is zero x is zero so u will be zero
so the lower for x is zero becomes u is zero now for the top when x is four
square roots of three when x is this one
then the force cancel and i get a square root of three
so perfect we went from this integral here which seems to look pretty simple but

01:49
the but the problem is the 16 you know doesn’t work so now it’s a one here so
now we can we know an antiderivative of this so i have my one fourth
and an antiderivative here is using this right here so
we have a u and a u and so we’ll have a u here plus c so tangent inverse of u
we don’t need the constant and then zero two square root of three
so now i’m going to use the square root of three in here
minus and then one-fourth and then tangent inverse of zero
okay so now we need to know the tangent inverse square root of three

01:50
and then we’re just going to get pi over 12 multiplying that by one fourth
right so this is just you know pi over three
and then multiply it and that and that’s zero of course
okay so we found the answer is pi over 12 and we did all this in one location um
in the sense that we wrote down a formula over here
i always find it practical to keep writing the formulas down
while you’re learning them as a student keep writing them down in your scratch
work to help you remember them but you know we did all this in one location here
just equals equals equals all the way through here we found this right here
we converted it to use that’s sweet we have the antiderivative formula and
then we’re able to find the number okay so there’s seven

01:51
now let’s look at eight let’s get a clean board here
now for number eight what should the substitution be
so we’re integrating from one to e and we know what the integral of e to
the x is is just e to the x plus c so if we could try to use
that the integral of e to the x is just e to the x plus c
so we’ll try to use the fact that the antiderivative of e to the x
is just e to the x so when i’m looking at this e to the natural log of x squared
i’m thinking there’s going to be a substitution there
so e to the natural log of x all that squared dx
so i’m going to let my u be all of this up here and see if that works
so my u is natural log of x and all of that is squared and my d u will be

01:52
2 natural log of x and then times the derivative of natural log of x
so and then dx okay so there’s my udu and let’s see if it works so here we go
equals so this this will be can i go from all x’s to all u’s already
this is e to the u um do i have my uh d u in here we have a natural log we
have an x we have a natural log and an x
we’re just missing a two so to put a two up here
i’ll uh divide by two so i’ll say one half
and you know after a while you’re going to start stop writing this step out here
but i’ll write it out here one more time this will be 2
natural log of x over x dx so i just put a two in here and to

01:53
balance it i wrote a one-half out here and i just put that over here so now i
see my d u so this will be one-half integral
of e to the u d u so now the integral is much simpler
this is e to the u and that’s all my d u now what are my limits though
so this is when x is one when x is one natural log of one is zero so that’s zero
this is when x is e when x is e natural log of e which is
one so this will be one squared this would be one here so this will be one half
and then this will be e to the u from zero to one
this will be one half so e to the one minus e to the zero so that’s e minus

01:54
one over two so there we go there’s number eight right there
so wow these are really fun these are so much more fun than say limit of a
riemann sum which is very tedious this is very fun find a u substitution
and we’re doing this with a definite integral all right so
one thing left to do which is to write a proof we have
uh said this theorem is true but we haven’t looked for a reason why yet
and so let’s do that now so i hope that those eight examples help you
understand this uh theorem here better um so we’re going to say

01:55
g is a function here we’re going to define it as a to x of f of t dt
right so f is continuous on a b and i’m defining this function here g
here’s g knowing that f is continuous this integral exists so this is a function
and we know that g is an antiderivative of t so g is an antiderivative [Music]
of f simply by the way that it’s defined so i’m going to let now
f be any antiderivative of f so let f be any antiderivative

01:56
of f that’s exactly what this theorem says here
above is any anti-derivative of f so knowing that both of these are
anti-derivatives of f we we know what f here looks like so by previous theorem
so this was the episode where we talked about anti-derivatives
and we had a theorem that said all anti-derivatives have the following
form f of x is g of x plus c so all anti-derivatives are equal to an
anti-derivative plus some constant now we’re going to
try to find this constant here so we’re going to do that by letting x
be equal to a right and so what happens um when we substitute in f of a

01:57
so we get g of a plus c right but what is g g of a so g of a is from
a to a so this is g of a here is zero so this is just c
g of a is zero right because if you plug in a you’re going to go from a to a
that’s zero so this is zero so f of a is just a c so what is f of b
right so let x be equal to b now so what is b so that’s g of b plus a c
and you know g of b is the integral from a to b of f of t right

01:58
g of b is 0 to b right g of b is 0 to b and that’s just the integral and then
plus our c constant c so putting this all together thus or therefore therefore
f of b minus f of a is the integral a to b f of t dt plus the c
right that’s f of b and then what was f of a it was just c
so minus c so the integral in other words is equal to the antiderivative f
whereas in any antiderivative of f f of b minus f of a
that’s all that it is it’s just equal to the integral
so there’s the proof right there um defining g would be this function right here

01:59
and then so using that you know fundamental theorem part one
and then taking any anti-derivative of f and using that theorem that all
anti-derivatives look like an anti-derivative
plus a constant then we it’s just a matter of figuring out what f of b minus
f of a is a is just a given integral there so that’s a nice proof
and now we’re ready for um some some more exercises some more examples let’s do
let’s do two more examples um make sure that we connect it back with
area that we have been working on here so for number one here
our function is x squared minus two x plus two and we’re looking on
minus one to two so we’re gonna find the area
now in the last episode we did this at great lengths

02:00
using limits of riemann sums and so now we’re going to say the area is minus one
to two of x squared minus two x plus two dx
and without using a limit of a riemann sum
this is just x to the third over three minus two x squared over two plus two x
and we’re gonna evaluate from minus one to two so we’re going to come in with
the two everywhere so we’re gonna get eight thirds minus four plus four
minus now plug in a minus one everywhere so we’re gonna get minus one third and
we’re going to get here minus one and then we’re going to get here minus two
so when we plug into two everywhere we’re going to get eight thirds
the twos cancel so we’re gonna get minus four plus 4
minus and now you use a minus 1 everywhere so minus the third

02:01
and then minus 1 and then minus 2. now when you calculate all those numbers
up you’re going to get 6. now what’s happening here is we have this
function right here x square minus two x plus two
and it’s coming through here like that at minus one
it has a height here and 2 here it has a height here
and we’re integrating from -1 to 2. and so what we’ve done is we found this
exact area here that that exact area there is six
and we found it very quickly you know doing some arithmetic
but that’s a lot easier than you know the limit of a riemann sum okay so
there’s number one and what’s the polynomial you just need to practice it

02:02
power rule okay for part two for part two now now we’re gonna say the
area is the integral from one to two of one over x squared
and that’s just one to two of x to the minus two
and that is just x to the minus two plus one and then divide and then one to two
or so differently this is what minus one over x from one to two
so this will be minus one half minus a minus one
so what is this uh minus one plus one so one half
so you know that’s just a minus two power then i add one and divide to integrate
it and we have our limits from one to two so from one to two then

02:03
we use a two first so minus one half here and then the
minus sign and then use the ones we get a minus one and then adding it all up
now over here what we have going on is from this is coming down through here
and so from one to two we’re getting this area in here
and that area is there is just the one half
so that’s just the function there one over x squared coming through here
between one and two and so that area there is one half
all right so there’s two examples there that connect it back to area
just to kind of convince you that hey finding area now is actually quite easy

02:04
all right all right so i have a couple of exercises here for us to look at
um and these exercises are for you to try
and to comment on below in the video and for these exercises
i want you to you know give them a try and see how you do let everyone know how
you do in particular let me know how you do and
um you know i can make a video for you if you if you want to see how i do them
and so we have two three more here exercise three four and five here and so
give those a try and let me know how it goes
and so now i want to say thank you for watching
if you’ve watched all the episodes wow great job i look forward to seeing you
in the next video if you like these videos please like and subscribe below make

02:05
sure and have fun with your studies and i’ll see you in the next one [Music]
if you like this video please press this button
and subscribe to my channel now i want to turn it over to you
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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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