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### Course: AP®︎/College Calculus AB > Unit 9

Lesson 2: AP Calculus AB 2015 free response- 2015 AP Calculus AB/BC 1ab
- 2015 AP Calculus AB/BC 1c
- 2015 AP Calculus AB/BC 1d
- 2015 AP Calculus AB 2a
- 2015 AP Calculus AB 2b
- 2015 AP Calculus 2c
- 2015 AP Calculus AB/BC 3a
- 2015 AP Calculus AB/BC 3b
- 2015 AP Calculus AB/BC 3cd
- 2015 AP Calculus AB/BC 4ab
- 2015 AP Calculus AB/BC 4cd
- 2015 AP Calculus AB 5a
- 2015 AP Calculus AB 5b
- 2015 AP Calculus AB 5c
- 2015 AP Calculus AB 5d
- 2015 AP Calculus AB 6a
- 2015 AP Calculus AB 6b
- 2015 AP Calculus AB 6c

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# 2015 AP Calculus AB 5d

Constructing expression for f with integral.

## Want to join the conversation?

- At0:40, when you take the anti-derivative of f'(x), wouldn't it be f(x) + C?(1 vote)
- Here Sal is taking a definite integral. when we take definite integral between "a" and 'b" then the anti-derivative of f'(x) becomes f(b)-f(a) ( the constant cancels out). For indefinite integrals, anti derivative of f'(x) has the the form f(x) + c.(7 votes)

- Would it have been valid to leave the equation for f(x) in terms of f(a), without substituting in the 3 for f(a) and 1 for a?(1 vote)

## Video transcript

- [Voiceover] Part d, given
that f of one is equal to three, write an expression for f of
x that involves an integral. It says it involves an integral. We can assume it's going
to involve f prime. Thus you can see given us so
much information about f prime including its graph and the area under or above the curve
of the different integrals. And it wanted us to find f of
four and f of negative two. So let's think about how we can connect f prime, an integral and f of x. Well, if we have the integral from a to b of f prime of x, dx. What is this going to be equal to? Well, this is going to be
equal to the antiderivative of f prime of x which is f of x. And we're going to
evaluate that at b and a and subtract the difference. So this is going to be f of b minus f of a. This is straight out of the
fundamental theorem of calculus. All right, well this is interesting. Well, what if, instead of having a b, what if we had an x there? And if we're going to put
an x as one of our bounds of integration, well, we'll
use a different variable for our integration here. So let me write it this way. So find the integral from a to x of f prime of u, du. Well, what is this going to be equal to? Well, this is going to
be, by the same logic, the antiderivative of
this evaluated with x, so f of x minus f of a, minus f of a. Or, if we wanted to solve for f of x, we could add f of a to bought sides and we would get f of x
is equal to the integral from a to x of f prime of u, du. Once again, I wanted the u one. This is some letter other than x since I already used x as one
of my bounds of integration. And I'm adding f of a to both sides. I swapped the sides too. So f of x is going to be
equal to this plus f of a. Plus f of a. So this is a general form
if a is a lower bound but they gave us some information. They said that f of one, f of one is equal to three. So if we choose a to be equal to one, if we say this is one right over here. This is one. Then we know that this is
going to be equal to three. So we can write f of x, f of x is equal to the integral. I'm using one as my a since
I know what f of one is. From one to x of f prime of u, du, plus f of one, they told us what that is. That is going to be equal to three. So this is the first part right here. This is it. That is the first part of the problem. Now, let's try to find f of
four and f of negative two. All right, f of four. Well, whenever see an x,
we substitute to four. There's gonna be integral from one to four of f prime of u, du plus three. So, what is gonna be the
integral from one to four of f prime of u, du? So let's look up here. So the integral from one to four of f prime, that's the curve right over here. Well, that's going to give this area but it's going to be
the negative of the area because it's the, if you
just take the integrals, it's the area that you
could view above the x-axis and below where for the intervals of u, above the u axis and below the function. Well, this is the other way around. The function is below
the horizontal axis here. So this area, which they
told us in the problem, the area bound by the x-axis
in the graph of f prime over the interval one, four is 12. So this area over here is 12 but the integral is going to be a negative because, once again, our
function is below the x-axis. So this integral, this right over here is going to be a negative 12. This is negative 12. So negative 12 plus
three is negative nine. All right, now let's
evaluate f of negative two. F of negative two is equal to the integral from one to negative two of f prime of u, du plus three. Well, it's kind of feels a little strange to have the upper bound to being
lower than the lower bound. So we could swap the bounds and then add a negative about here. So this is going to be
equal to the negative of, if we swap the bounds here. So from negative two to one, f prime of u, du plus three. So that's the integral
from negative to the one. Well, they gave us that. From negative two to one. They told us that this
area over here is nine. But once again, because it's
below the horizontal axis and above the curve, we
would say that the integral would evaluate to be negative nine. So the area is nine. But once again, the curve
is below the x-axis. So the integral would
give us negative nine. This would evaluate to negative nine. So take the negative of negative
nine which is positive nine plus three which is equal to, so nine plus three is equal to 12, and we're done.