### Video Transcript

Antiderivatives

At this point, you know how the
calculate derivatives of many functions. And have been introduced to a
variety of their applications. In this video, we want to ask a
question that turns that process around. Given a function π, how do we find
a function with the derivative π? And why would we be interested in
such a function?

The answer to the first part of
that question is the antiderivative. That question being, given a
function π, how do we find a function with the derivative π? The antiderivative of a function π
is a function with the derivative π. In other words, itβs a function
that reverses what the derivative does. Letβs consider the function π of
π₯ equals two π₯. To take the antiderivative of this
function, we need a function whose derivative is two π₯. We know that when we take the
derivative of π₯ squared, we get two π₯. And that means the antiderivative
of two π₯ would be π₯ squared. However, π₯ squared is not the only
antiderivative of two π₯. π₯ squared plus one is also an
antiderivative of two π₯. And thatβs because the derivative
of a constant is zero. To account for this, we say that
the antiderivative of two π₯ is π₯ squared plus π, where π is any constant
value.

Now letβs look at why we might be
interested in doing something like this. Letβs consider what we know about
motion. If we start with the position of an
object and the function π of π‘, we can find the velocity of that object, the π£ of
π‘, by taking the derivative of our position function. And if we take the derivative of
our velocity function, we can calculate the acceleration of our object, the π of
π‘. But what if we wanted to move in
the other direction? If we knew our acceleration and we
wanted to calculate the position, then we would need the antiderivative. This is just one of the many cases
when we have a need for an antiderivative. Letβs move on and look at some
examples of finding the antiderivative.

Find the most general
antiderivative capital πΉ of π₯ of the function lower case π of π₯ equals two π₯ to
the seventh power minus three π₯ to the fifth power minus π₯ squared.

To do this, weβll take the
antiderivative of each of these terms separately. We need the antiderivative of two
times π₯ to the seventh power. We need to know what function, when
we take the derivative, equals two times π₯ to the seventh power. And for π₯ to the π power, we can
find the antiderivative by taking π₯ to the π plus one power over π plus one. And then, in the general form, we
add π to represent any constant. Letβs apply this to two times π₯ to
the seventh power. Weβll leave two to the side, and
weβll take π₯ to the seven plus one power and divide by seven plus one.

The antiderivative of two times π₯
to the seventh power is two times π₯ to the eighth power over eight. And we can reduce this to one over
four. Two times π₯ to the seventh power
has an antiderivative of π₯ to the eighth power over four. And weβll repeat this process with
negative three π₯ to the fifth. We can keep the negative three and
we have π₯ to the five plus one power all over five plus one. Negative three times π₯ to the
sixth power over six, which will reduce to π₯ to the sixth power over two. And weβll make sure that we keep
that negative.

Weβll repeat the process one final
time. Weβre dealing with negative π₯
squared, so Iβll pull out a negative one. Then weβll have negative one times
π₯ to the two plus one power over two plus one, negative π₯ cubed over three. Because weβre looking for the most
general form, we canβt forget this constant at the end. Which makes our antiderivative πΉ
of π₯ equals π₯ to the eighth power over four minus π₯ to the sixth power over two
minus π₯ cubed over three plus π.

Now letβs look at a case where we
donβt want the most general form.

Determine the antiderivative
capital πΉ of the function lower case π of π₯ equals five π₯ to the fourth plus
four π₯ cubed where capital πΉ of one equals negative two.

Before we do anything else, weβll
calculate the general antiderivative. And that means weβll follow the
same process from the previous example. Weβll pull out the constant, add
one to our exponent, and then divide by the value of the new exponent. In this case, weβll have five times
π₯ to the fifth power divided by five. And weβll reduce that to π₯ to the
fifth. Now, for the second term, take out
that four, weβll raise π₯ cubed to π₯ to the fourth power, and then divide by
four. Which reduces to π₯ to the fourth
power. The four in the numerator and the
denominator cancel out.

If we were finding the general
form, we would add a constant π. And we say that capital πΉ of π₯
equals π₯ to the fifth power plus π₯ to the fourth power plus π. And we wanna plug in πΉ of one to
help us find the value of π. πΉ of one equals negative two. One to the fifth power plus one to
the fourth power. One plus one equals two. So two plus π has to equal
negative two. Subtract two from both sides. And we see that the constant value
is negative four. Weβll take that information and
plug it in to what we found for the general antiderivative. An antiderivative under these
conditions is π₯ to the fifth plus π₯ to the fourth minus four.

If the second derivative of π of
π₯ equals three π₯ to the fifth plus three π₯ cubed plus five π₯ squared plus two,
determine π of π₯.

If weβre given a second derivative,
we can take the antiderivative which will give us the first derivative of π of
π₯. And then weβll take the
antiderivative of that value which will give us our original π of π₯. The process wonβt be any different
than our previous examples. Weβll just have to do that
twice. Letβs take the antiderivative of
three π₯ to the fifth which would become three times π₯ to the sixth power over
six. And the second term, three times π₯
cubed, has an antiderivative of three times π₯ to the fourth over four. Five π₯ squared becomes five π₯
cubed over three. And the antiderivative of two is
two π₯. Make sure you add your constant
term.

Now, at this point, we have the
first derivative of this function. And we could simplify some of the
coefficients here. But because weβre going to take the
antiderivative again, Iβll wait and simplify in the last step. Now, we need the antiderivative of
three times π₯ to the sixth power over six. Three-sixths times π₯ to the
seventh power over seven. Three-fourths π₯ to the fourth is
three-fourths times π₯ to the fifth over five. Five-thirds times π₯ cubed. Five-thirds times π₯ to the fourth
over four. Two π₯ becomes two times π₯ squared
over two. The antiderivative of a constant is
that constant times π₯. And weβll need an additional
constant on the end, which we can call π·.

Weβve found our π of π₯ value, but
we want to simplify. We can reduce this three over six
to one-half. And then, weβll have π₯ to the
seventh power over 14. We multiplied the denominators in
our second term. And we get three times π₯ to the
fifth over 20. Our third term is five times π₯ to
the fourth over 12. In our fourth term, the twos cancel
out and we have π₯ squared. Our final two terms, the ππ₯ plus
π· canβt be simplified any further. Which means the general
antiderivative of the function we were given is π₯ to the seventh power over 14 plus
three times π₯ to the fifth power over 20 plus five times π₯ to the fourth power
over 12 plus π₯ squared plus ππ₯ plus π·.

For our last two examples, weβre
going to consider what at first might seem like an irregular form.

Determine the most general
antiderivative capital πΉ of π₯ of the function lower case π, given that lower case
π of π₯ equals five over two plus four over π₯.

This first term, five over two,
acts just like a constant. And we take its antiderivative by
saying five over two times π₯. At first, it might not seem very
clear what we can do with four over π₯. But what if we wrote it like this,
four times of one over π₯? Now weβre saying, what function has
a derivative of one over π₯? The natural log of π₯ has a
derivative of one over π₯. This means the antiderivative of
four times one over π₯ is going to be four times the natural log of π₯. Because four times the natural log
of π₯ has a derivative of four times one over π₯. So weβll bring down four times the
natural log of π₯. And since weβre considering a
general form, weβll still need to add a constant value π. Capital πΉ of π₯ equals five over
two π₯ plus four times the natural log of π₯ plus π.

By considering the product rule,
find the function π so that π prime of π₯ equals π to the π₯ power over the
square root of π₯ plus two times π to the π₯ power times the square root of π₯.

Weβll first need to remember the
product rule for derivatives. That product rule tells us the
derivative of the function π of π₯ times the function π of π₯ equals π of π₯
times the derivative of π of π₯ plus π of π₯ times the derivative of π of π₯. Before we try to find an π of π₯
and a π of π₯, letβs rewrite this function. We have π prime of π₯ equals π to
the π₯ power. And we know that itβs being
multiplied by one over the square root of π₯. We can write that as π₯ to the
negative one-half power. Weβre multiplying π to the π₯
power times π₯ to the negative one-half power plus two times π to the π₯ power
times π₯ to the one-half power.

Something that we know is that the
derivative of π to the π₯ power equals π to the π₯ power. If we say that π of π₯ equals π
to the π₯ power, then π prime of π₯ also equals π to the π₯ power. This means that π₯ to the negative
one-half power equals π prime of π₯. And it means that π of π₯ equals
two times π₯ to the one-half power. π of π₯ equals two times π₯ to the
one-half power. If we check that derivative, we get
two times one-half times π₯ to the one-half minus one power, which is in fact π₯ to
the negative one-half power. But what does this mean for us? Well, in the product rule, this
value is the derivative of π of π₯ times π of π₯. And that means the antiderivative
is going to be π of π₯ times π of π₯. We know π of π₯ and we know π of
π₯, which means the antiderivative equals two times π₯ to the one-half power times
π to the π₯ power. And we can put that back in the
form it was given to us in. Two times the square root of π₯
times π to the π₯ power.

Letβs briefly recap this key point
about antiderivatives. If youβre given some function π
prime of π₯, its antiderivative will be the function π of π₯ which derivative would
then be π prime of π₯. Our very first example was when π
prime of π₯ equals two π₯, we can take the antiderivative which gives us π₯ squared
plus any constant π. And π₯ squared plus any constant
π, if you take that derivative, would then give you two π₯.