What Are Indefinite Integrals (What Is an Antiderivative?)

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

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

00:00
[Music] what is an anti-derivative what is an
indefinite integral maybe you’ve heard of them
their reputation does often precede them well can you answer this question can
you find those functions whose derivative is equal to a given function
in this video you’ll also learn how anti-derivatives
lead to finding areas hi everyone i’m dave welcome back to the calculus one
explore discover learn series in this video i’m going to cover anti-derivatives

00:01
first and then i’m going to talk about indefinite integrals and then we’re
going to see some applications of these ideas
and then stick around to the end where i go over some
exercises let’s get started what is an anti-derivative so we’re going to go over
anti-differentiation by defining an anti-derivative function
and working out some examples on anti-derivatives
so we’re also going to concentrate on the following problem
if capital f is an antiderivative so i’m going to use capital f often to denote
an antiderivative of a continuous function
so notice there’s a uh the case matters here
i’m going to use um lowercase f for a given function

00:02
and a capital f for an antiderivative now any other antiderivative of
of a given function f must have the form capital f of x plus c where c
is some constant and we’re going to show that if we have an
antiderivative function then if you add a constant to that
we’re also going to have an anti-derivative
and we’re going to use the mean value theorem here
to show that all anti-derivatives must have this form
that is an antiderivative plus a constant so these are the four main goals
of this video here and we’re going to go through them here
and and plus a lot of examples lots lots of examples so that
all of this here given to you up front right here makes sense that’s the point
of this video is to understand all that now in order

00:03
to really get the full value out of this video
is you need to have uh watch the mean value theorem video
so that one is sort of a linchpin and the idea here that antiderivatives
is a family of functions and and why exactly is that true
so um let’s get started we’re going to start off by first of all
what is an antiderivative so a function is called an
antiderivative of a given function okay so you’re given a function and that
function that you’re given is i’m going to call it lowercase f so
we’re given a function and an interval and then we’re going to say oh
i have an antiderivative of that function so this is similar to
when we’re talking about derivatives right so in order to know what a derivative
is first you’re given a function right and then you can find the derivative of

00:04
that function so we have an anti-derivative
capital f if you take the derivative of the anti-derivative
you get back to the original function f so that’s what that
first sentence is saying right there so you know you have an anti-derivative
if you take the derivative of that anti-derivative and you get back to your
starting place your given function f and so we’re going to say here
um you know let’s try to get comfortable with this definition here before you
before we move on here so let’s say our original function that we’re given here
is an x squared [Music] all right so we’re given a function how
do we find the antiderivative how do we find antiderivative
and i’m going to use the uppercase f what is the antiderivative equal to

00:05
what is it and you know to answer this question we need to
know that you know whatever we say this is we have to be able to take the
derivative of it and get back to the original function
so we’re asking the question what function do we take the derivative of
to get an x square so f of x you might think of something like an x to the third
is that is that an anti-derivative if i take the derivative of this will i
get an x squared let’s try and see what is the derivative of an x to the third
it’s three x squared right so th these are not the same
so this is not an anti-derivative it’s close
but we got the three in the way so what about if we divide by three then
how about this is this an anti-derivative well the question is let’s take the
derivative so if we take the derivative of this we get

00:06
the 3 over 3 and we get an x squared and so yes this is an antiderivative
because if we take the derivative of it we get back to our original function
so here we have found an antiderivative of x squared
what about if we have something like how about um something like
just an x what is going to be the antiderivative or what is going to be
an antiderivative of just an x so we’re asking what is the function we
take the derivative of to get an x so you might guess something like
x squared is this an antiderivative well let’s see if i take the derivative
we get 2x that’s not the back to the original

00:07
there’s a 2 in the way so how about we divide by
2 or or let’s say multiply by a half so is this an antiderivative
and let’s see what’s the derivative now the 2’s cancel
we get back to an x right so this is an anti-derivative this function
right here is an anti-derivative because
if we take the derivative we get back to x
now notice that what about if i add a 5 here
what is this also an anti-derivative the answer is yes because
if we take the derivative of this we still get an x
because the derivative of the five is zero right
so in fact that’s why we always you know we’re always careful and say
here’s a function and here i found one anti-derivative
in fact this is also another antiderivative or we could say minus three here’s

00:08
another antiderivative because this has the same
derivative as the original function so finding an anti-derivative is sort of
like reversing the question um you know and we we study derivatives
for so long in calculus one that we know it
here’s the function find the derivative now
we’re asking reverse here’s a function and now we’re asking find a function
whose derivative is equal to the original function here
all right so hopefully that gets you a little bit
better understanding of what an antiderivative is
and the the word the crucial word here is not only antiderivative but
in anti-derivative so we’re looking for an anti-derivative so because you could
have lots of anti-derivatives here’s the function here’s one of the
anti-derivatives is this an anti-derivative well what’s the derivative

00:09
2x minus 3 do i get back to the same no so this one is not an antiderivative
but if i just left the 3 here now the derivative of the minus three zero
yes this is an antiderivative right here [Music] okay so if
we have this um important theorem here now which says that
if you have an f antiderivative of a function then every antiderivative
must have this form here which is an antiderivative
plus a constant so in other words uh this example that we’re looking at
here a second ago when i have x squared here and i said oh i found an
antiderivative which was um you know one half
x right um what was an antiderivative of of of f here is this one

00:10
well what’s the derivative of this right so the derivative is so that’s not an
antiderivative right so how about one third x to the third
is this an antiderivative of x squared so we take the derivative and we get x
squared so yes this is an antiderivative now
we also said a minute ago that if we add a constant
we still get the same derivative so this is also an antiderivative of x squared
this this function right here is also an antiderivative
well how do we find all the antiderivatives
what if i have a 50 right so in other words we can add a constant to the
to an antiderivative and still have an anti-derivative
but what this theorem is saying is more impactful
it says that yes you can add a constant to an antiderivative
and still get an antiderivative but it’s also saying that

00:11
if you take any antiderivative it has to look like an anti-derivative
plus a constant so let’s see why this is true so we’re going to do this
theorem here in two parts so i’m gonna call this one here part one
part one and what we’re gonna assume for this first part here
is that if we have a function here g of x and it looks like
the anti-derivative plus c [Music] for all x on an interval and if we know that

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capital f is an antiderivative for all x and c rolex and i and c is any constant
see it’s any constant that’s the word constant here so
if we have all these three things here if we have an antiderivative capital f
is an antiderivative then if we add any constant to it
then that will still be an antiderivative is that true
let’s see so the derivative of g to find out if this is an antiderivative of not
or not to find out if g is an antiderivative i’m going to take the
derivative of g so the derivative of g is just the derivative
of f plus c and the antiderivative that we know of
and we add a constant to it is this still an antiderivative
well how’s the derivative work right it’s the derivative it’s f prime
and then plus the derivative of a constant is zero

00:13
we know that this is an anti-derivative so the derivative
of the antiderivative is back to the original function here
so what we just showed is that if you have an antiderivative
and you add a constant to it you still get an antiderivative
capital g is an antiderivative because the derivative of
g is back to the original function here f and this happens for all x and i
hence g is also an antiderivative an anti-derivative and anti-derivative of
f so what this is saying is that if you take any anti-derivative of f

00:14
and you add a constant to it then that and then then that expression there
right there is also an anti-derivative so that’s pretty nice once you found one
anti-derivative you can add a constant to it and you’ll
still have another anti-derivative so that is the intuitive part of this proof
and now we have to have the second part which says
that no matter what antiderivative you have it has to look like another
antiderivative plus a constant so let’s see how to prove that so part two
call this one here part two it’s the same proof but i want to divide it into
separate parts so that you can get the feeling for how the proof should flow

00:15
all right so now we’re going to assume that g is any antiderivative is any
anti derivative of f on i now i’m going to make up this function here capital h
which is just going to be g minus f [Music] and that’s you know for all x on the
interval so in other words we know of one anti-derivative
and this function right here capital g is
we’re assuming is is any antiderivative is any other antiderivative
and so we’re looking at this function here h now for all a and b

00:16
in the interval i for all a and b and i h is continuous
on the interval i uh interval a b so we’re randomly choosing we’re just
arbitrarily choosing we’re not putting any conditions on the
a and b other than we’re choosing it in i and once we do that we know that h
is continuous on a b closed a b and it’s differentiable [Music]
is differentiable on a b so we’re going to use the mean value theorem
so by the mean value theorem we know that the derivative of h
is equal to the difference of h on the boundary over the difference of a and b

00:17
for some c and a and b for some c in the interval a b
this is exactly what the mean value theorem gives us
because we know it’s continuous on the closed and differentiable on the open
so we can use the mean value theorem here to get some c
in this interval here where where the derivative of
h is equal to this expression right here okay good so now let’s notice
notice that the derivative at c or in fact the you know the derivative at c is
is what is the derivative of g at c minus the derivative of f at c
um but these are equal to f of c minus f c which is all in fact just zero
because remember these are anti-derivatives so the derivative of

00:18
g is f and the derivative of f right because these are
anti-derivatives so their derivatives is equal to the original function so if
i take the derivative of capital g and i plug in a c
that’s the same thing as plugging in c into f and the same thing here f is
f is our known anti-derivative and g is just an arbitrary anti-derivative
but any case they’re both anti-derivatives so they’re both equal to f of c
and so we’re going to subtract we get 0 here so what this tells us is that
in fact this right here has to be 0 right here this is 0 and
so these are this is not zero so this has to be zero right here so

00:19
h of a equals h of b so that says h of b which means h is constant
h is constant on the interval here a b or in fact i because
remember we chose a and b here for all a and b and i no matter what we
chose in i a and b that these values are the same
so h has to be constant on the whole interval so g of x is equal to
or let’s say here uh right so h is constant right so
so what is h h is just simply g minus f right so g minus f
is equal to some constant right and so thus g of x is equal to f
of x plus c where c is some constant and that’s exactly what we needed there

00:20
[Music] so if we take an arbitrary anti-derivative
it’s going to be equal to our known antiderivative plus a constant
so any antiderivative must have this form a known antiderivative plus a constant
all right so there’s that theorem there there’s the proof of it it relies upon
the mean value theorem here and that’s sweet okay so let’s go on part two
[Music] okay understanding the anti-derivative uh now much better
now we’re ready to talk about indefinite inter integral
so when we look at this expression right here
the right hand side we just talked about capital f of x

00:21
plus c where c is a constant now the left-hand side is the new
notation that we that we’re using right here
so this notation right here the integral of f of x
is the anti-derivative here so this is the anti-derivative
and this is indefinite integral this right here is the integrand
so here we’re given a function this is given to us
and we want to find an indefinite integral and it’s equal to the antiderivative
plus c [Music] and this right here is the the
differential right here that’s important this says what we’re taking the uh
anti-derivative with respect to with respect to x and so it’s important to

00:22
have this terminology downright and this is a family of functions right here
right uh capital f of x is a is an antiderivative
and then we add a constant to it and that’s called the indefinite integral of
the given function f of x with respect to x and it satisfies the condition right
that the derivative of the antiderivative
is back to the original function for all x in the domain so it’s
important to remember that this represents a family of functions because c
is an arbitrary constant c could be zero c could be two c could be negative
square root of two c is an arbitrary constant so the process of finding all
anti-derivatives is called anti-differentiation or
indefinite integration so there’s two types of
integration that we’re going to talk about in calculus one
indefinite integration which we’re talking about first

00:23
and then we’ll talk about definite integration later
so in this video we’re talking about indefinite integration
and then in the next episode we’re going to talk about
uh substitution with indefinite integration
and then we’ll put it all together when we talk about definite integration
okay so here’s a first example here find um the family of anti-derivatives
of this function right here so the way that we find the family of
anti-derivatives is we find one of the anti-derivatives and then we find
and then we add the constants to it because that theorem says that all
anti-derivatives must look like one of the anti-derivatives plus an arbitrary
constant so we just need to know one of the antiderivatives

00:24
so what’s an antiderivative of cosine so an antiderivative of cosine you might
remember is right so what’s the derivative what do you take the derivative of
to find sine right so f of x will be equal to minus cosine x
and let’s check that what is the derivative of the answer
of an antiderivative right so we have a minus sign and the derivative of cosine
is minus sine so we’re going to have a negative times a negative
which is equal to sine x so this says right here
so this is an n is an antiderivative is an antiderivative of f
minus cosine is an antiderivative so how do we know my
minus cosine is an antiderivative because if i take the derivative
of it if i take the derivative of minus cosine i’m going to get minus and the

00:25
derivative of cosine is minus sine and that all comes out to be sine which
is our f of x so the family of anti-derivatives the family of anti-derivatives
of f is the minus cosine plus c now how do we write that right minus cosine
plus c we can write that using indefinite integral notation so the indef
the the indefinite integral of the the sine function the one we’re giving dx
is equal to the family of anti-derivatives minus cosine
plus c so this is the notation right here so this is saying the indefinite

00:26
integral of sine x is minus cosine x plus c what that means is if we take the
derivative of this right here we get back to sine
all right very good let’s go on and try our next one
how about x to the minus three so let’s try again
so first off we need to find an antiderivative not going to be minus cosine
and let’s try and see what happens so what is the
function that we need to take the derivative of to get back
to x to the minus 3. so we need to try to reverse
taking the derivative so normally when we take the derivative
we’re going to do what right what do we do when we take the
derivative of x to the minus three we we bring the power down

00:27
and then we subtract one so minus four so how do we reverse this process
right this is derivative anti-derivative is reversing the process
so instead of bringing the power down and then subtracting one
first i’m going to add one to the power and then divide instead of multiply i’m
going to divide so we’re going to do that process there
so if i add 1 to x to the minus 3 it’s going to be minus 3 plus 1 so it’s
going to be x to the minus 2 and then i’m going to divide by minus 2.
why is that why does that work let’s see if that works what is the derivative
what is the derivative here so minus 2 comes down and cancels
and then i do x to the and i subtract 1.
so the derivative x x to the minus three and that’s exactly what we needed that
is f of x so this is an anti-derivative right here because if i

00:28
take the derivative of it i get back to my original function here
so here’s an antiderivative so what is the family of all antiderivatives
is just find one of them and then add constants
now what’s a better way of writing x to the minus two over minus two
let’s just say it’s minus one half and then so let’s say it’s minus one over
two x squared and then plus c to find all of them
we just simply add the constant to them so let’s write that with indefinite
integral notation so the indefinite integral of 1 over x to the third
right that’s x to the minus 3 is equal to don’t forget your differential
is equal to minus 1 over 2x squared plus c so here’s an equation right here

00:29
in indefinite integral notation the indefinite integral of x to the minus 3
is equal to minus 1 over 2x squared plus c now how do we know this is right
if we take the derivative of this we get 1 over x to the minus three and you can
check that because because you know how to take derivatives
if you take the derivative of this you get back to your starting place here
okay so there’s our second example there um how about if we just do an arbitrary
n instead of looking at say minus three what about if we look at an
n for example what if your function is x to the fifth or your function is x to
the three halves any n you want except minus one
so how do we find an antiderivative of this
so i’m going to do the reverse of the derivative process
i’m going to find an anti-derivative so instead of

00:30
bringing the n down and multiplying and then subtracting one
i’m going to reverse all those so i’m first going to
add one and then i’m going to divide by it
and here’s why we can’t use n minus one here because that will make that zero
there down there so this is an antiderivative how do we
know it’s an antiderivative let’s find the derivative of the anti-derivative
so if i bring the n plus one down they’ll cancel and then i say in
x to the m plus one minus one which is just x to the n and that’s back
to our original function so this is an anti-derivative so is an anti-derivative
again how do you know something is an anti-derivative you take the derivative
of it and you get back to the original function
so this is an anti-derivative so now we can find the family of anti-derivatives

00:31
so the integral the indefinite integral of x to the n is equal to
an anti-derivative plus a constant so this is the family of anti-derivatives
is one way of saying it or this is the indefinite integral of x to the n
is equal to the family of anti-derivatives where c is an arbitrary constant
arbitrary constant or you could just say where c is any
constant or you could just say where c is a constant they all mean the same
thing it sounds fancier to say where c is an arbitrary constant
but they all mean the same thing where c is any constant
or where c is a constant okay so there’s the in indefinite integral notation

00:32
and there’s our family of anti-derivatives and again i know that’s right because
if i take the derivative of this i get back right here with x to the n
okay so let’s look at next one um now now that we’re kind of getting a
better feel for this process because we’ve taken so many derivatives
up to now in calculus one that you can go back and look at all of
the derivatives that you found and you can make new antiderivative
formulas for them so for example um what are some of the um derivative
theorems that we know of for example we know about the derivative
of the constant you can pull out the derivative of a constant
so you can do the same thing for anti-derivatives
we know how to do the sum rule how do you take the derivative of a sum
you take the sum of the derivatives and so we can do the same thing for
anti-derivatives the difference say same thing

00:33
and putting them all together we have the linear rule
if you take the anti-derivative or the indefinite integral of a linear
combination it’s going to be the linear combination of the antiderivatives
okay and now we have this um this power rule that we’ve been working on
and we have some very familiar um and indefinite integral formulas here
so what is the you know how do i know this one right here is true
that the indefinite integral of e to the x
is just e to the x plus c if i take the derivative of this
i get back to e to the x right what’s the derivative of e to the x
plus a constant well the derivative of the constant is zero
and the derivative of e to the x is e to the x

00:34
so the indefinite integral of e to the x is e to the x
plus c what about next one well we did this one right here in a
previous example right the indefinite integral of sine is a minus cosine u
plus c and you can check that what’s the derivative of minus cosine u plus c
about this one the antiderivative of cosine
we can check that that one is true check for
right so i’m going to look at sine u plus c and i’m going to take the derivative
what’s the derivative of sine it’s cosine
what’s the derivative of a constant it’s zero so plus zero so this is a check
this is a check that number four is correct i took the derivative of the
right hand side and i got the integrand back
what about number five so number five we know that the derivative

00:35
of tangent derivative of tangent plus c the derivative of tangent
is secant squared plus derivative of the constant derivative of secant is
secant tangent so number six works the derivative of the right hand side is
equal to cosecant cotangent the derivative of
minus cotangent is cosecant squared ah remember the inverse
formulas we had an episode on derivative of inverses
what is the derivative of sine inverse right so we can check that formula right
there and because we know what the derivative of
tangent inverse is and because we know the derivative of

00:36
secant inverse so those formulas come because we know what the derivative of the
inverse functions are okay so let’s look at some more examples
so we have some basic formulas that we know of
but integration is harder than just that so here’s an example
if we want to look at say this one right here the integral of
number ten right the integral of one plus x squared dx
by looking at number 10 here is just you know tangent x plus c

00:37
or tangent inverse x plus c but this example right here wants us to
look at the integral of the indefinite integral of
x squared over one plus x squared right and so we know what this one is
but we don’t know what this one is and when there’s an x squared here
that makes the problem very very different
what you cannot do is take the x squared out and have the integral of 1 plus
x squared and now we know what all this is it’s just that right so this does not
work that’s just a bunch of junk so how do we do this
we need to find the we need to find an antiderivative of this what do you take
the derivative of to get an x squared plus x squared plus one
and so that can be a little tricky so what we’re going to use is x square

00:38
plus one i’m sorry x squared over x square plus one i’m going to add a one
and subtract a one in order to make this work so
i’ll take the x square plus one right here with the first two
and then i have a minus and the reason why i do that is because
this is a one right here this is one minus so we just we’re trying to work with
this function right here x squared over x squared plus one and i
wrote it down like this and the reason why i wrote it down like this
is because i know how to work with these two right here so here we go
the indefinite integral of x squared over x square plus one

00:39
is equal to so instead of using this i’ll use this one right here so one minus x
squared over x squared plus one okay so this one looks harder to work
with this one looks a little bit easier to work with
and the reason why is because we have our difference rule so i can just work on
this one right here and then minus and then i can work on this one right here
sorry this is a one here that should be a one here right all of this is equal to
the integral of all of that just integrating both sides
so this right here will be the integral of one over x squared plus one
so integrating this finding the indefinite integral of this one
i broke it down into two simpler ones this one is straightforward it’s just x

00:40
and then it’s actually plus a constant now how do we know this is right well
what’s the derivative of this derivative of x is one derivative of constant zero
right so this is the antiderivative right here
minus and then this one right here remember this one was
tangent inverse and then we have a constant
now we have to be a little bit careful about the constants
because this one is an arbitrary constant coming from this
family of anti-derivatives so we could put a c1 here
and this is a constant coming from this family of anti-derivatives right here
so we could put a two here in fact we have a minus here so that
should be minus for the whole thing or if you want uh parentheses
right because we have this one this family of antiderivatives
minus this family of anti-derivatives okay
so you could write it like this where c1 and c2 are

00:41
arbitrary constants or you could just write it like this where c
is an arbitrary constant where c is a constant so c 1 minus c 2
where c 1 and c2 are arbitrary is the same thing as writing an arbitrary
constant over here so this is really unnecessary right here
we could go straight from here to x and then
minus sign and then the and then this right here
is you know the derivative of tangent inverse is
1 over 1 plus x squared and then plus a constant
so i just basically saved up all my constants from both
and put him together as one arbitrary constant
okay so we found this right here and we did a little trickery

00:42
to it so integrating is much harder than finding derivatives
all right let’s look at the next example find the general anti-derivative or in
other words find the indefinite integral and
if i wanted to say find the derivative of this function
we could use a product rule or we could expand it all out and then take
the derivative of each term finding the antiderivative though there
is no product rule we’re going to find the anti-derivative
and the way to do that is to expand it out so we’re going to have
f of x we’re going to rewrite it so i’m going to say one times each of those two
so that’ll be 1 minus 4 over x squared and i’m going to say 1 over x times each
of those two so 1 over x and then minus 4 over x to the third

00:43
so we can write f of x like this and in fact we may want to write it down as
1 and then plus 1 over x and then minus 4 over x squared
and then minus four over x to the third maybe you’d like to write it like that
in fact maybe you would like to write f of x like this
one plus x to the minus one minus four times x to the minus 2 and
then minus 4 times x to the minus 3. in any case however you want to write f
we have to go find the antiderivative of it
all of them so the antiderivative or the indefinite integral of f will be
all of this one plus x to the minus one minus four x to the minus two minus four
x to the minus three okay good now i can go and look at them

00:44
separately one individually one by one and we could write all that out
in fact you should see it at least once so i’m going to write it out and then
i’m going to erase it because i don’t think that you should write it out
but i think that you should visualize it so how can you visualize it if no one
has ever done it for you at least once so this is one dx
plus x to the minus one dx plus minus four x to the minus two dx and then plus
and then the last one here minus four x to the minus three dx
and then maybe put parentheses here all right so all of that
so we’re gonna go and enter and find the indefinite integral of
each one of these four here so what is this one right here

00:45
it’s just x plus c plus some constant i’m going to save up all my constants to
the end so x plus c1 okay so i’m just gonna find the
indefinite integral of the one and then this one and then this one and this one
so here we have now this is a minus one here so this is going to be
the natural log of the absolute value of x plus a c2
and then this one right here is going to be i’m going to add 1
to the power and then divide so i’m going to get minus 1 so i’m going to say
minus 4 x to the minus 1 and then divide plus another constant and then plus
and then this last one right here is going to be minus 4 and then
x to the minus 3 plus 1 so minus 2 over minus 2 and then plus a constant

00:46
so what are we getting for all this we’re getting a natural log of the
absolute value of x plus plus x and then what’s this right here plus
4 over x and then this one right here all becomes um 2 over x to the third
or squared and then plus c [Music] okay so this is our general antiderivative
or this is the indefinite integral of this function here f
and you could you could you could test that by taking the derivative of all
this right here taking the derivative of all this right here and you’ll get this
function right here f now to be honest you don’t really need to do each of
these steps sometimes you do sometimes you don’t
just depending upon what they’re asking you but if this is your 200th

00:47
example then you can move here much faster
the first time you see it you could see all these steps here c1 c2 c3 c4 they’re
all constants and then you just put them all together
into one constant and then you just say where c is a constant c is a constant
okay so can you go from right here to right here how many examples
how many exercises will you need so that you compute the one it’s just the x
the x to the minus one is just the natural log of the absolute value
and then the you know use the power rule there
or indefinite integrals to get each of those okay anyways next example
now in this example here we’re going to this is called an initial value boundary

00:48
problem and it’s basically we’re going to reverse the process of differentiation
and we’re going to find the constants so to work on this one right here
we’re given this second derivative and we need to find the function f
so you’re used to doing the opposite here’s the function f
find the derivative find the second derivative
now we’re given the second derivative and we’re asked to find the function
so we’re going to find f of x f prime by integrating so this will be
x plus the square root of x dx and what is this so
when we integrate the x we’re going to get x squared over 2
and when we integrate the square root of x right so this will be x to the
one-half power but i’m just looking at this part right here

00:49
the square root of x so how do we do that so we’re gonna
add one to the power so one half plus one that’s three halves and then i’m going
to divide by three halves plus c right in other words this will be
two x to the three halves over three two thirds so let’s write that up here so
when i integrate the first one the x is just x squared over two
right the power once we add one to the power and didn’t divide
we do the same thing once one half so it’ll be x to the three halves
over three halves and then it’ll be easier just to write it like that
two thirds so this will be two thirds x to the three halves and then plus a
constant all right so so far so good so we know what the second derivative is
they give that to us and now we know what the first derivative

00:50
is well we don’t know what the first derivative is entirely
because c is an arbitrary constant so what they do is they give us a boundary
condition this is called the boundary condition
we know that f prime at one is two so if i plug in
1 into the derivative i should get out of 2.
so let’s do that if i plug in a 1 into the derivative
i should get out of 2. well what happens when we plug in a 1 into the derivative
this is the derivative so what happens when we plug in a 1 so
we’re going to get one half plus this will be two thirds
plus a c so solving this right here for c so we have to move the one half over
move the two thirds over and we get so c is 5 6 when you solve that
right just do two minus a half minus two thirds

00:51
you know just put all that over six multiply by
you know six top and bottom you get twelve
and then minus three and then minus two so right you get five six
in any case our first derivative we have all the way pinned down now
the first derivative is all of this plus the c we just found
using this condition right here the derivative at one is two
we’re able to use the derivative at one is two by plugging in one
we found the c so the first derivative is uh let’s say it say it’s one half x
squared plus two thirds x to the three halves power
plus our constant which is five sixths so it asks us to find the original
function instead we found the derivative well we can integrate one more time

00:52
so to find f of x i’m going to integrate the first derivative
so two-thirds x to the three-halves plus five-sixths so i’m going to
integrate all of that to find the original function here
so here we go one-half x-squared so what do we get
so i’m going to increase the power by 1 and divide so x to the third
over 3 but i already have a one-half here so this is going to be 1
6 x to the third less now this is going to be right so we’re
going to add one so to be it’ll be uh you know three halves plus one
so that’ll be five halves over five halves but we already have a
two thirds in the front of here right so so what is all that when i try
to when i try to integrate this piece i’m using the power rule
so three halves plus one and then divide right so this will be two thirds

00:53
times so if i divide by five half that’s the same thing as multiplying by two
fifths and then x to the five halves so this is what four fifteenths
x to the five halves so sometimes you might need to do some scratch work over
here but if you’re doing the integral of two thirds
x to the three halves all of that right there
is this right here so i’ll put it over here four fifteenths x to the five halves
and then now the last term so we integrated the first one we got that
we integrated the second one we got that now we’re going to integrate the 5
6 and that’ll be 5 6 plus a constant so 5 6 plus a constant 5 6 x
right the derivative of 5 6 x is at 5 six plus a constant now i’m gonna have

00:54
another constant here and i want to confuse it with my previous constant
my previous constant was c so i’m gonna use a constant k capital k there
so in order to find out the function the final answer
i need to find out this capital k here now they tell us that if i plug in one i
get out one so let’s do that if we plug in one here
we get out one now if i plug in a one here i plug in a one everywhere here and i
should get out one so let’s plug in one plug in one here i get one
six plug in one here i get four fifteenths i plug in one here i get five six
and then if you add all that up you get right you get 1 minus 1 6

00:55
minus 4 15 minus 5 over 6 so when you find that k it’ll be minus 4 over 15.
so we have our final function now all of this plus our k which we just
found so it would be 1 6 x to the third plus 4 15 x to the five halves
plus five six x plus a constant which we found
by this initial condition here f of one equals one
and that was minus four over fifteen so there’s the function there
we know this works we can check it if you find the second derivative of this
function right here it will be x plus square root of x and
if you plug in one into this function you should get out one
and if you plug in one into the derivative of this function

00:56
you’ll get out two so this is our function right here so
that’s pretty sweet there’s the work there we start by
integrating to find the first derivative and then we integrate to find the
function we have to find the missing information here the constant but
in the end we get the final function there
okay so let’s move on to the next part applications of the indefinite integral
okay so our first application is going to be finding area
so this is an application to mathematics itself of the process of antiderivative
so let’s read this theorem here and try to make sense of it
it says that if we have a continuous function

00:57
and that function is positive or zero on a closed interval a b then the area
bounded by the curve the x-axis and the vertical lines
so all that bounces certain region and to find that area of the region
you use the antiderivative of f and so let’s look at a
just a sketch of what that would look like so we have here um
a continuous function it’s positive it may touch
down to zero but let’s just draw one here like that f that’s the graph of f
and we got an a and a b and i’ll just draw it in the first quadrant here

00:58
just to illustrate it so the area bounded by the curve which we have right here
the x-axis right here and the vertical lines x equals a and x equals b
we’re bounded by those vertical lines there so here’s the region that we’re
bounding right there and now we have x equals t so let’s move this
to a t here instead of a to b we’re looking at well let’s put a b
here because we want to view this as a function of t
so i’ll erase some of this and say here’s the x equals t
vertical line and so let’s choose a t somewhere in here
and so here’s the area here and we’re going to denote the area function by

00:59
a of t [Music] is the area of the bounded region
we know that it’s bounded because f is continuous and we have these two
vertical lines and we know that the function f of x stays above the x-axis
and so we’re also bounding it below by the x-axis
so here’s an arbitrary t between a and b so as you move t back and forth you’re
going to get different area and so this is a function of t here
when t is actually equal to a well then your area is zero
because you don’t have any width to your region here
and if t comes out to be equal to b then you have the whole area of the
region so we’re thinking about t being the variable and a of t is the
area the bounded region and what we’re saying is that this theorem is claiming

01:00
is that the area is an anti-derivative so this is an anti-derivative of f
so here’s the function f and this area here is the anti-derivative of f
so well when i say anti-derivative i mean with respect to t
on a b so we’re given f of x and so we just use t as the
variable and then we have the anti-derivative
so let me show you an example a concrete example here
so find the area under the parabola over the interval
0 1 and to do that we’re going to use this
theorem here finding an anti-derivative so here we have y equals x squared

01:01
and we’re on zero one [Music] and so we can go back here and say um
look at this theorem here with this example right here we have x squared and
we’re on zero one so what’s going on here so we have
y equals x squared looks like that we touch zero right there
but we’re continuous and let’s say one is about right there
so we have this right here and then we have zero one
so we have x equals zero vertical line and we have the x-axis
so we’re looking at this area in here and let’s pick a t between zero and one
let’s call this right here t so how do we find this area here a of t

01:02
is an antiderivative so what is an antiderivative of x squared
right so we know after having done some practice
that the indefinite integral of x square dx
is equal to x to the third over three plus a constant right so we have
here a of t is we can do this with t now t squared dt
and this will be one third t to the third plus c
and how can we find c well we know that a of zero is zero in other words when t
is zero we’re back here we have no width then the area is zero
so when we plug in zero when we plug in zero here we should get out zero

01:03
so a of zero is one third zero squared plus c so
c is just zero so we found a of t a of t is c zero here one third t to the third
so that is the area so for example if t is one half
then we will have found the area under the curve
you know but only going to one half the midpoint now the problem asks us um
what is the area on zero one so that means t is just one so a of one
is just one third one to the third or in other words just one third
so one third unit squared or you know so that’s the area of the graph
of x squared on zero one if you went all the way
and t was equal to one you would just plug in 1 and you would get the area

01:04
so by finding the general anti-derivative here
with the c but being able to find the c so now i have the area function right
here and you know we can use any value of t between zero and one for example we
can find out one half we would just plug in t’s one half or whatever [Music]
okay so that illustrates that theorem there um next application
so let’s see here a ball is thrown upward with a speed uh 48 feet per second
and from the edge of a cliff 432 feet tall and so let’s see here find its height
above the ground two seconds later when does it reach its maximum height

01:05
when does it hit the ground so let’s try to answer all those questions there
so the motion is vertical and we choose the positive direction to be upward
at time t the distance above the ground is s of t
so let’s say s of t this distance um above the ground [Music]
and let’s use v of t is the velocity and a of t to be the acceleration
and um we’re going to use a of t which is the derivative of the velocity

01:06
is minus 32 because this is taking place on earth so two times
okay so how do we find the velocity then all right so we’re going to take the
derivat uh the integral of the derivative so the velocity will be [Music]
taking the indefinite integral of this right here and so this will be minus 32
dt and this will be minus 32t plus a constant
we can always check that taking the derivative we get back to the minus 32.
now can we determine c what is the velocity at they give us that the

01:07
initial speed is 48 so v of zero that’s a zero is 48 so
you know when we plug in a 0 here minus 32 times 0 plus a constant so [Music]
c is 48. in other words our velocity velocity function is minus 32t
plus our constant 48 so there’s our velocity
it’s you know find the height above the ground
when does it reach its maximum height so on so we can see that
the ball reaches the maximum height when the velocity is zero
when the velocity becomes zero that’s when it reaches its maximum height

01:08
so we’re going to look at that now and say v of t equals 0 which is
minus 32 minus 32 t plus a constant minus 32 t
plus the constant so we need this to be equal to zero
so when we do that t is going to be equal to move the 48 over divide by minus 32
and we’re going to get t is you know 48 over 32
or 1.5 and we’re going to go with seconds here
so that’s when it reaches its maximum height
it’s going at 48 feet per second so this is seconds
so that’s certainly the time there when it reaches its maximum height
so we’ll just say here max height at t equals 1.5 seconds

01:09
okay so now we want to find the s of t to answer some more questions
so the s of t is going to be the think of it as position right so it’s
going to be the um so v is the velocity there so
s of t is going to be the integral of the derivative here minus 32 t plus 48.
so we’re going to be able to find the position function
and we’re going to get here so add 1 and divide so we’re going to get
minus 16 t squared plus 48 t plus a constant now i already use c as a
constant down here so i don’t want to get it confused i’ll say k perhaps

01:10
and we’re giving some initial conditions here so s of um so
we’re given here 432 right so f of 0 is 432 so when i plug in 0 i plug i get a 0
here and actually that should be 48 t right plus a constant right
so again check derivative of this we get back to the original
so when i substitute in 0 here i get 0 0 and k so when i plug in 0
i get 0 plus 0 plus k so k is just 432 okay good so now we have our s of t

01:11
now it’s going to be minus 16 t square plus 48 t plus our constant
so given any point in time this is our position right here so
the height of the function is given by this right here right that’s the position
so what is the ball hit the ground so s of t
equals zero when the height is zero so we need to solve all of this
equal to zero so when we solve this equal to 0 right here
well we can use the quadratic formula here and we’re going to get t is equal to
3 plus 3 square roots of 13 over 2 which is approximately 6.9
seconds so 6.9 seconds later it’s going to hit the ground so we found the height

01:12
function right here to answer the question find its height above the
ground two seconds later right there and when does it hit the ground
when the height is zero and then so yeah using the quadratic formula there
and then we get uh about 6.9 seconds later
all right very good so there’s you know some elementary position velocity ball
type of problem now let’s look at something like some you know economics maybe
okay so now we have a manufacturer estimating and
they want us to find the domain demand function
so we’re given the marginal revenue given the marginal revenue so let’s find
the revenue first knowing the marginal revenue we’re going to find the revenue

01:13
so the uh so the revenue is going to be the integral of the marginal revenue
and so this is going to be the integral of 240 plus 0.1 x
dx so in order to find the revenue knowing the marginal revenue we’re going
to find the indefinite integral and we have 240x
plus so we’re going to increase and then divide so divide that by 2
so we’re going to have 0.05 x squared and then plus a constant so that gives
us the possible revenue um can we find the constant [Music]

01:14
so you know the revenue is zero when we have produced zero units right so
well if if we substitute in a zero we should get out zero here
so let’s substitute in zero so put a zero here and a zero here
we’re going to get zero plus zero plus c you know so
this is zero the revenue r of zero is equal to zero
so c is zero okay so the revenue function is just 240 x plus 0.05 x squared
you know plus zero so there’s the revenue function there
now the domain the demand the the sorry the demand function
is the revenue divided by the number of units
or think of it you know differently the revenue
is the price or the demand function times the number of units right so to
find the demand function you’re just going to divide by the x

01:15
all right and so we just found the revenue so we just need to divide it by
x so we’re going to get 240 plus 0.05 x so that’s the demand function there
all right very good um and so if you’re not like um clear on the domain function
and the revenue and all that stuff in our last video
i’m sorry last video was the hopital’s rule the previous
uh video on the applied optimization we went into that in more detail
this is kind of a follow-up on all that um
of course in that first initial video we just knew derivatives
now we’re applying anti-derivatives so given some information uh before
we were given the demand function and we’re asked to find something like the

01:16
marginal revenue which with it so this is kind of the reverse of the problems
we did before okay so let’s go to the next one here [Music]
okay so an excellent film with the very small
advertising budget must depend largely on word of mouth advertising so in this
case the rate of which weekly attendance might grow
is modeled by this function right here this is a derivative where t is
in the in the time in weeks since its release
and a is in the attendance in millions so we want to find the function that
describes weekly attendance in other words we want to find the function a

01:17
and then we want to use that function right there to make some
predictions so how can we find the a knowing the derivative of a
so we’re going to say that a of t is the indefinite integral of
d of a of t right dt so we need to find out what this is
and this is our a function our our d a right here is actually minus 100 over
c plus 10 and we have a square on all that and then plus 2000
all over t plus 10 to the third to the third and then so we need to find

01:18
the antiderivative of all this i’ll just write like that so the d a
right we have right here i just use the d a so now i want to find the
antiderivative of each of these here [Music]
okay so when i’m looking at this first part here um
i’m thinking about this like this i’m going to write it down here so it’s 100
sorry 100 and then what is this right here it’s t plus 10 to the minus 2 right
so i’m going to add 1 and divide so it’s minus 1 and i’m going to divide by it
so i’m going to get 100 and then t plus 10 right because it when i add
what’s gonna be minus one so it’s going to be in the denominator here
and then minus and then i’m thinking about this one right here is
2000 and then t plus 10 to the minus 3 right and so i’m going to add 1 and then

01:19
divide so i’m going to be chopping that 2 000
in half so i’m going to get minus 1 000 and then t plus 10
squared and then i get some arbitrary constant let’s call it c so
this is my a of t here is all of this now if you’re not sure about that take
the derivative of this you take the derivative of this right
here you get this right here and if you take the derivative of this
right here t plus 10 right there you take the derivative you get this
right here and then the derivative of the constant is zero
okay so we found our a of t he’s in time in weeks since its release and a is
in millions of of people so now when t equals zero then we can find the constant

01:20
so when the what’s the attendance when there’s no time that has passed right
so this will be 100 over 0 0 plus 10 minus 1000 over 10 squared plus
c and this is just c and so c is zero as it should be c is just zero
now of course when time is not passed your attendances should be zero right
so this is our attendance function here let’s just go and write it out real
quick it’s just 100 t plus 10 minus 1000 here over t plus 10 squared [Music]

01:21
and so number part b here is what’s what’s the
attendance when it’s in its 10th week so we’ll just plug in 10. [Music]
so we get 100 over 20 minus 1000 over 20 squared
and if you work that out you get 5.2 or two point five million people
all right so just plugins in get our number 2.5 2.5 million people there okay so
last example here [Music] here we go now we’re given the marginal

01:22
cost for a product and so sometimes people denote the
marginal cost with that bar over the mc and its fixed cost is 340
and we’re given the marginal revenue is eight adx and we’re looking for find
the profit or the loss from the production and sell of some
different units okay so let’s first find the cost function
knowing the marginal cost we should be able to find the cost function
now also knowing the marginal revenue we should be able to find the revenue
once you know the cost and the revenue you should be able to determine the
profit so let’s get started so i’m going to look at
c of x first the cost and it’s going to be the
indefinite integral of the marginal cost so 60 and then square root of the

01:23
x plus 1 dx so what will our cost be so i’m looking at the x plus 1 with a
square square root so let’s look at it like this x plus one to the one half
right so we’re gonna add one and divide so when i add one i get three halves
and then i’m gonna divide by three halves now we have a sixty of course
that’s constant so sixty and then x plus one to the three halves
and i’m gonna divide by three halves and of course we have a constant i’ll
say constant k um now when we substitute in zero we should have the fixed cost
340 there so cost at zero is 340 and what happens when we plug in
zero what do we get for all this right here
we plug in zero here that’ll be one to the three halves so that’s just one

01:24
so what’s 60 divided by three halves or in other words multiply by two thirds
plus the constant k so we need to find the k now multiply 60 times two thirds
and then subtract that from the 340 and k is 300. okay good so
we have the cost function here now it’s going to be the um
you know 40 one plus x to the three halves plus the constant we just found 300
[Music] um yeah so now we can find the marginal revenue
so the revenue function oh sorry now we can find the revenue

01:25
we know the marginal revenue so let’s integrate and so 80x
and so this will be x squared and then divide by two
so we’re going to get a 40. so 40x squared plus some constant let’s call it c
um now the revenue when we’ve made zero units should be zero
so if we plug in zero here the whole thing should be zero
so c is zero also or in this case just like the last example because r of
zero should be zero when we we have no units we should have no revenue
so if i plug in 0 here 0 here the whole thing should be 0 so c has to be 0 also
so our revenue function is just simply 40 x squared all right so now we’re ready

01:26
for our profit so remember that the profit is revenue minus cost
so we found these two let’s put it together so we have 40x squared
that’s the revenue minus the cost and here’s the cost right here so then we have
40 and then one plus x to the three halves and then this is minus
all of the cost so this should be minus all of the cost so minus this minus that
so minus 300 there so there’s the profit function right there
and we want to find the profit or the loss from three units so
we need to plug in p of three we need to find p of three so let’s just
plug in three everywhere times nine minus forty times what is that four

01:27
so four to the three three halves and then subtract 300 from there and so
for all that you get minus 260 so that’s certainly a loss let’s say dollars
and p of the other one they ask us for here part b
would be p of eight so now we’re going to plug in eight
so forty times the eight squared and then minus forty times the
um nine to the three halves now and then minus the three hundred
and so for all that we’re going to get eleven eighty dollars
okay so there we we we worked it out twice we found marginal cost they gave us
marginal costs we found the cost it gave us marginal revenue so we found
the revenue we put together the profit function and

01:28
then we plugged in some values to find out how much money we made
or how much money we lost all right so very good so let’s look at some exercises
okay so these exercises are for you to try and we have some um
problems here to help you get started down the path of being able to find
anti-derivatives on your own and so here i give us three problems at a
time so number one has an f a g and an h and we want to go find
the anti-derivative of it we want to write out an indefinite integral
equation for each one of these and then we have some more here four
more problems here so i recommend trying those problems out

01:29
on your own thinking hard about them and finding your solution to them
and in the comments below tell me how you did tell me if you did a great job
tell me if you need some more help let me know
and we can make a video over these exercises if you want um so
i want to say thank you for watching and see you on the next video which
is about in definite indefinite integrals but we’re going to do
substitution with them so we’re going to gain a lot of strength
in terms of the number of types of integrals that we can use
and find and so we gotta we got up really close to using the
integration by substitution in this video but
we didn’t do it and we’ll do it next time and i look forward to seeing you then

01:30
bye-bye 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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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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