### Video Transcript

Applications of Indefinite
Integration

In this video, weβre going to learn
how we can use indefinite integration to express the function given an expression
for its rate of change. And weβll see several different
ways we can apply this in different situations. To do this, weβre going to first
need to recall what we mean by the indefinite integral of a function.

For example, letβs consider the
equation π¦ is equal to π₯ squared. Weβre going to integrate π₯ squared
with respect to π₯. And we know how to do this. We add one to our exponent of two
giving us a new exponent of three, and then we divide by this new exponent. This gives us π₯ cubed over three,
and then we add our constant of integration πΆ. But we need to remember exactly why
weβre calculating this expression. Remember, when we first defined
integration, we wanted to do this to find antiderivatives of different
expressions. So we defined integration to be the
reverse process of differentiation. In other words, what weβre
calculating when we integrate π₯ squared is an antiderivative of π₯ squared. And in fact, we call this the most
general antiderivative because itβs an antiderivative for any value of our constant
πΆ. And to see exactly what this means,
letβs differentiate this expression with respect to π₯.

If we wanted to differentiate π₯
cubed over three plus our constant of integration πΆ with respect to π₯, we would do
this term by term. To differentiate our first term, we
want to multiply it by the exponent of π₯ and then reduce this exponent by one. Of course, we can see this is going
to give us π₯ squared. Then, πΆ is a constant, so its rate
of change with respect to π₯ is going to be zero. It doesnβt vary as our value of π₯
varies. So the derivative of this
expression is exactly what we started with, π₯ squared. And of course, this is true for any
value of our constant πΆ. And this leads us to a very useful
property. What if instead of having π¦ is
equal to π₯ squared, we have instead started with dπ¦ by dπ₯ is equal to π₯
squared?

So now we start with dπ¦ by dπ₯ is
equal to π₯ squared. The derivative of π¦ with respect
to π₯ is equal to π₯ squared. We do exactly the same lines of
working out and we end up with the derivative of this expression is equal to π₯
squared. So we started with the derivative
of π¦ with respect to π₯ is equal to π₯ squared and we ended up with the derivative
of this expression with respect to π₯ is equal to π₯ squared. In other words, we found an
expression for π¦ up to our constant of integration πΆ. And because we found this
expression by integrating, youβll often see this as π¦ is equal to the integral of
our slope with respect to π₯.

So now if weβre given the slope of
a function, we know we can integrate this with respect to π₯ to try and find our
original equation. And as long as we know how to
integrate this, weβll be able to find an expression for π¦ up to our constant of
integration πΆ. The only question that remains is
how weβre going to find the value of πΆ. To find the value of πΆ, weβre
going to need to know a little bit more information. Usually, when weβre given the slope
of a curve, weβll be given a point on the curve. For example, we might be told that
the curve π¦ is equal to π of π₯ passes through the point one, one. This is often called the initial
value. Then, if weβve shown that π¦ is
equal to π₯ cubed over three plus our constant of integration πΆ, we can substitute
this point into our curve.

Since our curve passes through the
point one, one, we know when π₯ is equal to one, π¦ must be equal to one. Substituting these values into our
curve, we get one is equal to one cubed over three plus our constant of integration
πΆ. Simplifying and subtracting
one-third from both sides of the equation, we get that πΆ must be equal to two over
three. Then we can just write this value
of πΆ into the equation for our curve. And this gives us our equation for
our curve, π¦ is equal to π₯ cubed over three plus two over three. Therefore, what weβve actually
shown is if the slope of a curve is equal to π₯ squared and the curve π¦ is equal to
π of π₯ passes through the point one, one, then our curve must have the equation π¦
is equal to π₯ cubed over three plus two-thirds. And weβll be able to do this in
general, provided we can integrate the function in the slope. Letβs now see an example of how we
might do this in practice.

Find the equation of the curve that
passes through the point negative two, one given that the gradient of the tangent to
the curve is negative 11π₯ squared.

In this question, we want to find
the equation of a curve. And to do this, weβre given some
information about our curve. Weβre told that our curve passes
through the point negative two, one and the gradient of the tangent to this curve is
equal to the function negative 11π₯ squared. So letβs break down the two pieces
of information weβre given about our curve. First, if we wanted to write the
equation of this curve in the form π¦ is equal to π of π₯, then saying that our
curve passes through the point negative two, one means when π₯ is equal to negative
two, π¦ should be equal to one.

So we can substitute these values
into the equation for our curve. We should have that one is equal to
π evaluated at negative two. Next, weβre told the gradient of
the tangent to our curve is negative 11π₯ squared. And remember, we know the gradient
of the tangent to our curve at a value of π₯ is as rate of change of π¦ with respect
to π₯. In other words, this is telling us
dπ¦ by dπ₯ is equal to negative 11π₯ squared. So we need to find the equation of
a curve given its slope and a point on the curve. And to do this, weβre going to need
to find an antiderivative of negative 11π₯ squared. And we know how to find
antiderivatives by using integration. The integral of negative 11π₯
squared with respect to π₯ will give us the most general antiderivative of negative
11π₯ squared.

And remember, the reason weβre
finding this is we know the derivative of our expression for π¦ with respect to π₯
has to be equal to negative 11π₯ squared. So we need to integrate negative
11π₯ squared with respect to π₯. We can do this by using the power
rule for integration. We recall this tells us for any
real constants π and π, where π is not equal to negative one, the integral of π
times π₯ to the πth power with respect to π₯ is equal to π times π₯ to the power
of π plus one divided by π plus one plus a constant of integration πΆ. We add one to our exponent and then
divide by this new exponent.

In our case, the value of π is
negative 11 and the value of π is two. So we add one onto our exponent of
two to get an exponent of three and then divide by this new exponent. We get negative 11π₯ cubed over
three plus a constant of integration πΆ. And of course, the derivative of
this expression with respect to π₯ is equal to negative 11π₯ squared. So far, for our curve π¦ is equal
to negative 11π₯ cubed over three plus a constant of integration πΆ, we have dπ¦ by
dπ₯ is equal to negative 11π₯ squared. However, remember, we still need
this curve to pass through the point negative two, one. And to do this, we need to
substitute π₯ is equal to negative two and π¦ is equal to one in to our curve.

Substituting π₯ is equal to
negative two and π¦ is equal to one into the equation for our curve, we get one is
equal to negative 11 times negative two cubed all over three plus our constant of
integration πΆ. Now all we need to do is solve this
equation for πΆ. Rearranging this equation, we see
that πΆ should be equal to negative 85 divided by three. Now all we need to do is use this
value of πΆ in the equation for our curve, and this gives us our final answer. Therefore, we were able to show the
equation of the curve that passes through the point negative two, one and which has
the gradient of its tangent line at π₯ given by negative 11π₯ squared has the
equation π¦ is equal to negative 11 over three times π₯ cubed minus 85 over
three.

Letβs now see an example where the
gradient is given by a more complicated function.

Find the equation of the curve
given the gradient of the tangent is five times the sin squared of π₯ over two and
the curve passes through the origin.

In this question, weβre asked to
find the equation of a curve. And to do this, weβre given some
information about our curve. First, weβre told that the gradient
of the tangent line to our curve at π₯ is given by five times the sin squared of π₯
over two. Next, weβre also told that our
curve passes through the origin. To answer this question, letβs
start by saying we can write the equation of our curve in the form π¦ is equal to
some function π of π₯. Then we need two things to be true
about this equation. First, weβre told that our curve
must pass through the origin. And remember, the origin is the
point with both π₯- and π¦-coordinate equal to zero.

And if our curve π¦ is equal to π
of π₯ passes through the origin, then this equation must be satisfied when π₯ is
equal to zero and π¦ is equal to zero. In other words, we must have zero
is equal to π evaluated at zero. This is not the only piece of
information weβre given. Weβre also told the gradient of the
tangent to our curve. Remember, though, the gradient of
the tangent to our curve, our value of π₯, is the rate of change of π¦ with respect
to π₯. In other words, weβre told that dπ¦
by dπ₯ must be equal to five times the sin squared of π₯ over two. And itβs also worth pointing out,
we couldβve wrote this down as π prime of π₯. Itβs personal preference which one
you would use.

So to find the equation of this
curve, we need both of these pieces of information to be true. In particular, weβre first looking
for a function which differentiates to give us five times the sin squared of π₯ over
two. In other words, weβre looking for
an antiderivative of this function. And we know how to find
antiderivatives by using integration. If we integrate this with respect
to π₯, weβll find an antiderivative of this. In other words, π¦ will be equal to
the integral of five times the sin squared of π₯ over two with respect to π₯ up to a
constant of integration. However, we can see a problem. Itβs very difficult to integrate
the sin squared function. So weβre going to want to simplify
this expression first. And to do this, weβre going to use
the double-angle formula for cos.

Recall one version of this tells us
the cos of two π is equivalent to one minus two times the sin squared of π. And this will be true for any value
of π. We want to use this to find an
expression for the sin squared of π₯ over two. And to do this, weβre going to need
to substitute π is equal to π₯ over two. Doing this, we get the cos of two
times π₯ over two is equivalent to one minus two sin squared of π₯ over two. And of course, we can simplify
this, since two divided by two is equal to one. Next, weβll rearrange this
equation; weβll add two sin squared of π₯ over two to both sides and subtract the
cos of __ from both sides. This gives us two sin squared of π₯
over two is equivalent to one minus the cos of π₯.

Thereβs a few different options of
what we could do next. Since weβre integrating five sin
squared of π₯ over two, we should try and make our coefficient five. And to do this, weβll multiply both
sides of the equivalence through by five over two. And if we do this and simplify, we
get five sin squared of π₯ over two is equivalent to five over two minus five over
two cos of π₯. And we can see that this expression
is much easier to integrate. So weβll substitute this into our
integral. This gives us that π¦ is equal to
the integral of five over two minus five over two times the cos of π₯ with respect
to π₯ up to our constant of integration.

Now we could just evaluate this
integral term by term. First, five over two is a
constant. When we integrate this with respect
to π₯, we get five over two times π₯. Next, to integrate our second term,
we recall the following rule. For any real constant π, the
integral of negative π cos of π₯ with respect to π₯ is negative π sin of π₯ plus
the constant of integration πΆ. We can use this with π equal to
negative five over two to get the integral of this term. So we have π¦ is five over two π₯
minus five over two times the sin of π₯ plus our constant of integration πΆ. Now all thatβs left to do is find
the value of πΆ. And to do this, we need to use the
fact that zero is equal to π evaluated at zero.

In other words, because our curve
passes through the origin, we can just substitute π₯ is equal to zero and π¦ is
equal to zero into our expression for the curve. And this equation must be
satisfied. This gives us that zero is equal to
five over two times zero minus five over two times the sin of zero plus πΆ. Of course, five over two times zero
is equal to zero and the sin of zero is also equal to zero. So this entire expression just
simplifies to give us that πΆ is equal to zero. So weβve shown that πΆ must be
equal to zero. All we need to do now is substitute
this value of πΆ into our expression for our curve. And by substituting πΆ is equal to
zero, we get our final answer. The equation of this curve was π¦
is equal to five over two times π₯ minus five over two times the sin of π₯ over
two.

Letβs now see an example involving
exponential functions.

Given that the slope at the point
π₯, π¦ is three π to the power of three π₯ and π evaluated at zero is negative
three, determine π evaluated at negative three.

In this question, we need to
determine what π evaluated at negative three is. And weβre told some information
about our function π. Weβre told that the slope at the
point π₯, π¦ is given by three π to the power of three π₯. And weβre also told that π
evaluated at zero is negative three. So weβre given two pieces of
information about π of π₯. First, weβre told the slope is
three π to the power of three π₯. And another way of saying this is
π prime of π₯ is equal to three π to the power of three π₯. So to find π of π₯, we need to
find an antiderivative of π prime of π₯. And we know how to do this by using
integration.

We can find an antiderivative of π
prime of π₯ by integrating it with respect to π₯. We have π of π₯ will be equal to
the integral of three π to the power of three π₯ with respect to π₯ up to a
constant of integration. And to evaluate this integral, we
need to recall the following. For any real constant π, the
integral of π to the power of ππ₯ with respect to π₯ is equal to one over π
times π to the power of ππ₯ plus a constant of integration πΆ. We need to divide by the
coefficient of π₯ in our exponent. And in our function, thatβs equal
to three. So we get three multiplied by
one-third times π to the power of three π₯. And remember, we need to add a
constant of integration πΆ.

Of course, we can simplify
this. Three multiplied by one-third is
just equal to one. So this simplifies to give us π of
π₯ is equal to π to the power of three π₯ plus πΆ. Now we want to find the value of
πΆ. And to do this, we need to use the
fact that π evaluated at zero is equal to negative three. So we substitue π₯ is equal to
zero. We know π of zero is negative
three, and this is equal to π to the power three times zero plus πΆ. And now we just solved this
equation for πΆ. π to the zeroth power is just
equal to one. And then we rearrange to get πΆ is
equal to negative four. So if πΆ is equal to negative four,
we can substitute this into our expression for π of π₯.

We have that π of π₯ is equal to
π to the power of three π₯ minus four. The question wants us to find π
evaluated at negative three. So we substitute π₯ is equal to
negative three into this expression. We get π to the power of three
times negative three minus four. And if we evaluate this expression
and rearrange, we get our final answer of negative four plus one over π to the
ninth power.

Letβs now see an example of a
question involving the maxima and minima of a curve.

Find the local maximum and minimum
values of the curve that passes through the point negative one, seven where the
gradient of the tangent is six times π₯ squared plus four π₯ plus three.

In this question, we need to find
the local maximum and minimum values of a curve, but weβre not given the equation
for our curve directly. Instead, weβre only given
information about our curve. Weβre told that our curve passes
through the point negative one, seven. And weβre also given the gradient
of the tangent line at the point π₯, π¦. Itβs given by six times π₯ squared
plus four π₯ plus three. Of course, to find the local
maximum and minimum values of a function, weβre going to need to know its output at
these values. So weβre going to want to find an
expression for our function. And to do this, weβre going to need
to use the two pieces of information weβre given about our curve.

First, weβre given an expression
which represents the gradient of the tangent lines to our curve at a value of
π₯. And we recall this is exactly the
same as saying the rate of change of π¦ with respect to π₯ is equal to this
expression. In other words, we have dπ¦ by dπ₯
is equal to six times π₯ squared plus four π₯ plus three. But this is not the only piece of
information weβre given. Weβre also told that our curve
passes through the point negative one, seven. So when π₯ is equal to negative
one, we must have that π¦ is equal to seven. Weβre given an expression for the
gradient of our curve.

So differentiating our curve with
respect to π₯, we get the following expression. This means we want to find an
antiderivative of this expression. And we know how to do this by using
integration. We must have that π¦ is equal to
the integral of six times π₯ squared plus four __ plus three with respect to π₯ up
to a constant of integration because this will give us the most general
antiderivative of this expression. Thereβs a few different ways of
solving this. Weβre going to distribute six over
our parentheses. Doing this, we get the integral of
six π₯ squared plus 24π₯ plus 18 with respect to π₯. And now this is just the integral
of a quadratic. We can do this term by term by
using the power rule for integration.

We recall for any real constants π
and π, where π is not equal to negative one, the integral of ππ₯ to the πth
power with respect to π₯ is equal to π times π₯ to the power of π plus one divided
by π plus one plus a constant of integration πΆ. We add one to our exponent of π₯
and then divide by this new exponent. Applying this to six π₯ squared, we
add one to the exponent of two giving us a new exponent of three and then divide by
the exponent of three. Doing this and simplifying, we get
two π₯ cubed. Weβll do the same for 24π₯, which
we can write as 24π₯ to the first power. We add one to our exponent of one
and then divide by this new exponent. This gives us 12π₯ squared. Finally, we could do the same for
18 by writing it as 18 times π₯ to the zeroth power. However, we can just write the
antiderivative of this as 18π₯. And donβt forget, we need to add
our constant of integration πΆ.

So now we found an expression for
the curve given to us in the question up to a constant of integration πΆ. We now want to find the value of
our constant πΆ. And to do this, weβre going to need
to use the fact that our curve passes through the point negative one, seven. If our curve passes through this
point, we can substitute these into the equation for our curve. And then the equation has to be
true for our curve to pass through this point. Substituting π₯ is equal to
negative one and π¦ is equal to seven into the equation for our curve, we get seven
is equal to two times negative one cubed plus 12 times negative one squared plus 18
times negative one plus πΆ.

Now by simplifying this expression
and solving for πΆ, we get that πΆ is equal to 15. And if πΆ is equal to 15, we can
substitute this into the expression for our curve. So by clearing some space and using
πΆ is equal to 15, we have the following expression for our curve. But remember, we need to find the
local maximum and minimum values of this curve. And to find this, we need to recall
the following piece of information. In polynomials, our local extrema
will occur at the turning points, in other words, when the gradient is equal to
zero. And weβre given an expression for
the gradient.

So to find our local extrema, we
need to solve six times π₯ squared plus four π₯ plus three is equal to zero. And to do this, all we need to do
is factor our quadratic. We see this factors to give us π₯
plus one times π₯ plus three. Solving this is equal to zero, we
get π₯ is negative one or π₯ is negative three. To find the values of these, we
need to substitute these into the equation for our curve. Substituting negative one into our
curve, we get π¦ is equal to seven. Similarly, when π₯ is equal to
negative three, we see that π¦ is equal to 15. We now need to determine which of
these is a maximum and which is a minimum. To do this, we notice our curve is
a cubic with a positive leading coefficient. It will look something like
this. In fact, we know it has two turning
points, and we know the coordinates of these two turning points. So weβve shown the local maximum
value is 15 and the local minimum value is seven.

The key point of this video is if
weβre given the slope of a function dπ¦ by dπ₯, we can find an equation for π¦ by
using integration.