### Video Transcript

Indefinite Integrals: The Power
Rule

In this video, we will learn how to
find the indefinite integrals of polynomials and general power functions using the
power rule for integration. Letβs start by recalling what the
antiderivative of a function is. We can say that capital πΉ is the
antiderivative of lowercase π if capital πΉ prime of π₯ is equal to lowercase π of
π₯. We can in fact say that this is
true for any function π of π₯ where π of π₯ is equal to capital πΉ of π₯ plus πΆ
for any constant πΆ. Now, this is very useful as weβll
be using it to define an indefinite integral.

We can say that the indefinite
integral of lowercase π of π₯ with respect to π₯ is equal to capital πΉ of π₯ plus
πΆ where capital πΉ is the antiderivative of lowercase π. Itβs very important to remember our
constant of integration when performing an indefinite integral. Letβs quickly consider why this
constant is here. If we perform the reverse operation
on this equation, so thatβs differentiating with respect to π₯, then since we are
simply performing the inverse operation on the left, the differential with respect
to π₯ of the integral of lowercase π of π₯ with respect to π₯ will just be
lowercase π of π₯.

On the right-hand side, when we
differentiate capital πΉ of π₯ with respect to π₯, we get πΉ prime of π₯ and our
constant πΆ will simply vanish because differentiating a constant gives zero. Now, when we go back the other way
to our integral, the constant will appear again. However, we do not know the value
of this constant, hence why we label it πΆ. Itβs just an unknown constant. Letβs now consider what happens
when we integrate some function with respect to π₯, for example, three π₯.

We know the equation for finding an
indefinite integral. We have that the integral of
lowercase π of π₯ with respect to π₯ is equal to uppercase πΉ of π₯ plus πΆ where
uppercase πΉ is the antiderivative of lowercase π. Now in our case, lowercase π of π₯
is equal to three π₯. So we simply need to find the
antiderivative of three π₯. Letβs try and do this by trial and
error. Weβre trying to work out what we
need to differentiate in order to get three π₯. Letβs start by differentiating π₯
squared with respect to π₯. This gives us two π₯ which is very
close to three π₯. However, weβre not quite there
yet.

Letβs try multiplying our
differential by one-half. Differentiating π₯ squared over two
with respect to π₯ give us π₯ which is one-third of what weβre trying to find. So letβs now try multiplying our
differential by three. And differentiating three π₯
squared over two with respect to π₯ gives us three π₯. Therefore, we found our
antiderivative. And itβs three π₯ squared over
two. So this is all capital πΉ of
π₯. Therefore, we can say that our
integral is equal to three π₯ squared over two plus πΆ. Now this is quite a lengthy method
for finding antiderivatives of power functions.

Letβs consider the power rule for
derivatives. We know that the derivative of π₯
to the power of π plus one with respect to π₯ is π plus one multiplied by π₯ to
the power of π since for the power rule of derivatives we multiply by the power and
then decrease the power by one. What this tells us is that π₯ to
the power of π plus one is the antiderivative of π plus one multiplied by π₯ to
the power of π. We can therefore recall π₯ to the
power of π plus one capital πΉ of π₯ and π plus one multiplied by π₯ to the power
of π lowercase π of π₯. And we can substitute these
functions into our equation for the indefinite integral of a function. And this gives us that the integral
of π plus one multiplied by π₯ to the power of π with respect to π₯ is equal to π₯
to the power of π plus one plus πΆ.

Now, we have a constant inside our
integral. And we can, in fact, factor this
out. Letβs quickly consider why we can
do this. We know that if we differentiate a
constant multiplied by some function π of π₯, then this is equal to the constant
multiplied by the differential of π of π₯ with respect to π₯. This, therefore, tells us that the
integral of π multiplied by π of π₯ with respect to π₯ is equal to π multiplied
by the integral of π of π₯ with respect to π₯. Therefore weβre able to factor out
our constant of π plus one. And then we can simply divide by
this constant. And this gives us that the integral
of π₯ to the power of π with respect to π₯ is equal to π₯ to the power of π plus
one over π plus one plus πΆ over π plus one.

However, since both πΆ and π plus
one are constants, this means that πΆ over π plus one will also be a constant. And we can call that constant
π·. Therefore, we arrive at our power
rule for integration. It tells us that the integral of π₯
to the power of π with respect to π₯ is equal to π₯ to the power of π plus one
over π plus one plus πΆ. Here I have relabeled our constant
of integration back to πΆ just for consistency with our definition of an indefinite
integral. However, it does not matter what
you label this constant. An easy way to remember this rule
is that when we integrate a power function, we increase the power by one and then
divide by the new power. And of course, we mustnβt forget to
add our constant of integration.

Now, this power rule for
integration works for any real value of π except for one specific value. And that is when π is equal to
negative one. If we tried using π is equal to
negative one, we get that the indefinite integral of π₯ to the power of negative one
with respect to π₯ is equal to π₯ to the power of negative one plus one over
negative one plus one plus πΆ. Since negative one plus one is
zero, this means that we have a zero in the denominator of our fraction. Therefore, this fraction is
undefined and so is the integral. And so we can say that our power
rule for integration works for all real values of π except negative one. Now it is, in fact, possible to
integrate π₯ to the power of negative one with respect to π₯. However, we will not be covering it
in this video. We will now be moving on to an
example of how we can use the power rule.

Determine the indefinite
integral of negative π₯ to the power of nine with respect to π₯.

In order to find this integral,
we need to use the power rule for integration. We have that the integral of π₯
to the power of π with respect to π₯ is equal to π₯ to the power of π plus one
over π plus one plus πΆ. And weβre trying to find the
integral of negative π₯ to the power of nine with respect to π₯. Letβs start by factoring out
the negative one. We have that our integral is
equal to the negative integral of π₯ to the power of nine with respect to
π₯. Now, when we look at our
integral and we compare it to the integral in the formula, we can see that in
our case π is equal to nine. Therefore, we simply substitute
π is equal to nine into the formula. Therefore, we obtain negative
π₯ to the power of nine plus one all over nine plus one plus πΆ. This gives negative π₯ to the
power of 10 over 10 minus πΆ.

Now, negative πΆ is just
another constant. So we can call it π·. And now weβve reached our
solution. And that is that the indefinite
integral of negative π₯ to the power of nine with respect to π₯ is equal to
negative π₯ to the power of 10 over 10 plus π·.

Letβs now work out how we will
integrate a polynomial. We know that the differential of a
sum of functions, so π of π₯ plus π of π₯ with respect to π₯, is equal to the sum
of the differential of the functions. So π by dπ₯ of π of π₯ plus π by
dπ₯ of π of π₯. From this we obtain that the
integral of a sum of functions is equal to the sum of the integrals of the
functions.

Using this rule, weβre able to
split integrals of polynomials down into integrals of power functions. For example, the integral of π₯
squared plus π₯ with respect to π₯ is equal to the integral of π₯ squared with
respect to π₯ plus the integral of π₯ with respect to π₯. And we already know how to
integrate these power functions on the right. Using this, weβre able to integrate
any polynomial. Letβs now look at an example.

Determine the indefinite
integral of 25π₯ squared minus 65π₯ plus 36 with respect to π₯.

Letβs start by splitting this
integral up using the fact that the integral of a sum of functions is equal to
the sum of the integrals of the functions. We obtain that our integral is
equal to the integral of 25π₯ squared with respect to π₯ plus the integral of
negative 65π₯ with respect to π₯ plus the integral of 36 with respect to
π₯. We can factor out the constant
in each integral. Now, we can use the power rule
for integration which tells us that the integral of π₯ to the power of π with
respect to π₯ is equal to π₯ to the power of π plus one over π plus one plus
πΆ.

In the case of our first
integral, π is two. Therefore, itβs equal to 25
multiplied by π₯ cubed over three plus some constant which weβll call πΆ
one. For our second integral, weβre
integrating π₯. π₯ is equal to π₯ to the power
of one. Therefore, π is equal to
one. Therefore, itβs equal to
negative 65 multiplied by π₯ squared over two plus some constant which weβll
call πΆ two. For our third integral, weβre
integrating one. One is also equal to π₯ to the
power of zero. Therefore, our value of π is
zero. And so itβs equal to 36
multiplied by π₯ to the power of one over one plus some constant which weβll
call πΆ three.

Expanding and simplifying, we
obtain that our integral is equal to 25π₯ cubed over three minus 65π₯ squared
over two plus 36π₯ plus 25πΆ one minus 65πΆ two plus 36πΆ three. Now, since πΆ one, πΆ two, and
πΆ three are all constant, 25πΆ one minus 65πΆ two plus 36πΆ three is also a
constant. And we can relabel this as
πΆ. And so we reach our
solution. And that is that the indefinite
integral of 25π₯ squared minus 65π₯ plus 36 with respect to π₯ is equal to 25π₯ cubed over three minus 65π₯ squared over two plus 36π₯ plus πΆ.

In the next example, weβll see how
we can simply integrate a polynomial without splitting it up into separate
integrals.

Determine the indefinite
integral of π₯ minus six multiplied by π₯ minus five multiplied by π₯ minus
three.

Letβs start by expanding the
brackets. Expanding the first two sets of
brackets, we get π₯ squared minus 11π₯ plus 30. And we then multiply this by π₯
minus three. We obtain the indefinite
integral of π₯ cubed minus 14π₯ squared plus 63π₯ minus 90 which is in fact a
polynomial. And we can use the power rule
for integration to integrate this term by term. The power rule tells us that
the integral of π₯ to the power of π with respect to π₯ is equal to π₯ to the
power of π plus one over π plus one plus πΆ. If instead we were integrating
π₯ to the power of π multiplied by some constant π, then since we can factor a
constant out of our integral, then this would simply be equal to π multiplied
by π₯ to the power of π plus one over π plus one plus πΆ.

Now, you may be wondering why
we havenβt multiplied the πΆ by π. And thatβs because π is also a
constant. Therefore, π multiplied by πΆ
is a constant. And we can just rename this new
constant to be πΆ. So now, letβs apply this rule
to our integral term by term. The first term is π₯ cubed. Therefore, π is equal to
three. We increase the power by one
and divide by the new power to get π₯ to the power of four over four. The next term is negative 14π₯
squared. Negative 14 is just a
constant. So that will remain. Our power is two. So π is two. We increase the power by one to
get π₯ cubed and divide by the new power.

On next term, we have 63π₯. So we can start by writing in
our constant of 63. We then note that π₯ is equal
to π₯ to the power of one. So π is equal to one. So when we integrate this, we
get π₯ squared over two. For our final term, we have
negative 90. And we know that this can also
be written as negative 90 multiplied by π₯ to the power of zero since π₯ to the
power of zero is just one. So now, we can integrate
it. We start by writing our
constant of negative 90. Since our power of π₯ is zero,
we increase the power by one giving us π₯ to the power of one and divide by the
new power. So thatβs dividing by one. And we miss then adding our
constant of integration πΆ.

We can write this out a little
neater for our solution which is π₯ to the power of four over four minus 14π₯
cubed over three plus 63π₯ squared over two minus 90π₯ plus πΆ.

Letβs quickly note that this
integration power rule works for any real π except negative one. So that includes both negative and
noninteger powers of π₯.

Letβs see how this works in the
following examples.

Determine the indefinite
integral of negative two over seven multiplied by π₯ to the power of negative
nine with respect to π₯.

In our integral, we simply have
a power function. So we can use the power rule
for integration in order to find this integral. It tells us that the indefinite
integral of π₯ to the power of π with respect to π₯ is equal to π₯ to the power
of π plus one over π plus one plus πΆ. In our case, weβre integrating
negative two-sevenths π₯ to the power of negative nine with respect to π₯. So our power of π₯ is negative
nine. We can start by writing down
our constant which is negative two-sevenths.

Then since our value of π is
negative nine, we next need to write π₯ to the power of π plus one over π plus
one. So thatβs π₯ to the power of
negative nine plus one which is π₯ to the power of negative eight over negative
nine plus one. So thatβs negative eight. Then we mustnβt forget to add
our constant of integration πΆ. For our final step here, we
just need to simplify. And we obtain our solution
which is that the indefinite integral of negative two-seventh multiplied by π₯
to the power of negative nine with respect to π₯ is equal to π₯ to the power of
negative eight over 28 plus πΆ.

In our final example, weβll see how
to integrate a function with noninteger powers of π₯.

Determine the indefinite
integral having negative four multiplied by the fifth root of π₯ to the power of
nine plus eight all multiplied by the fifth root of π₯ squared with respect to
π₯.

Letβs start by writing this
roots as powered. We know that the πth root of
π₯ is equal to π₯ to the power of one over π. Once weβve written our roots as
powers, we can then combine them with the existing powers, using the fact that
π₯ to the power of π to the power of π is equal to π₯ to the power of π
multiplied by π. Therefore, π₯ to the power of
nine to the power of one-fifth becomes π₯ to the power of nine-fifths. And π₯ squared to the power of
one-fifths becomes π₯ to the power of two-fifths. Now, we can expand the
brackets, using the fact that π₯ to the power of π times π₯ to the power of π
is equal to π₯ to the power of π plus π. So our integral becomes the
integral of negative four π₯ to the power of eleven-fifths plus eight multiplied
by π₯ to the power of two-fifths with respect to π₯.

Here, we can use the power rule
for integration which tells us that the indefinite integral of π₯ to the power
of π with respect to π₯ is equal to π₯ to the power of π plus one over π plus
one plus πΆ. We can apply this rule to our
integral term by term. For the first term, we have
negative four π₯ to the power of eleven-fifths. Therefore, π is equal to
eleven-fifths. When we integrate this term, we
get negative four multiplied by π₯ to the power of π plus one. And π plus one is simply
sixteen-fifths. So itβs π₯ to the power of
sixteen-fifths. And then we need to divide by
π plus one. So thatβs dividing by
sixteen-fifths.

For the second time, we have
eight multiplied by π₯ to the power of two-fifths. Therefore, π is
two-fifths. So we add eight multiplied by
π₯ to the power of π plus one which is π₯ to the power of seven-fifths. And we then divide by
seven-fifths. And we mustnβt forget to add
our constant of integration πΆ. Now, all that remains to do is
to simplify. And so we obtain a solution
that the indefinite integral of negative four multiplied by the fifth root of π₯
to the power of nine plus eight all multiplied by the fifth root of π₯ squared
with respect to π₯ is equal to negative five multiplied by π₯ to the power of
sixteen-fifths over four plus 40 multiplied by π₯ to the power of seven-fifths
over seven plus πΆ.

Weβve now covered a variety of
examples of indefinite integrals of power functions. Letβs recap some key points of the
video. Key points. The indefinite integral of
lowercase π of π₯ with respect to π₯ is equal to capital πΉ of π₯ plus πΆ where
capital πΉ of π₯ is the antiderivative of lowercase π of π₯ and πΆ is a
constant. The power rule for integration. The indefinite integral of π₯ to
the power of π with respect to π₯ is equal to π₯ to the power of π plus one over
π plus one plus πΆ for any real number π except negative one.