2015 AP Calculus AB 5d | AP Calculus AB solved exams | AP Calculus AB | Khan Academy

Part D given that F of one is equal to three, write an expression for f f of x that involves an integral. Since it involves an integral, we can assume it's going to involve F prime somehow, especially since they've given us so much information about F prime, including its graph and the area under or above the curve at different intervals.

Then they want us to find F of four and F of -2. So let's think about how we can connect F prime, an integral, and F of x. Well, if we took the integral from A to B of F prime of F prime of x dx, what is this going to be equal to? Well, this is going to be equal to the anti-derivative of F prime of x, which is F of x, and you'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, so this is interesting. Well, instead of having A and B, what if we had an X there? And if we're going to put an X as one of our bounds of integration, we'll use a different variable for our integration here. So let me write it this way. If I have 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 anti-derivative of this evaluated at X, so F of x minus F of A.

Or if we wanted to solve for F of x, we could add F of A to both sides, and we would get F of x is equal to the integral from A to X of F prime of U du. And once again, why did I pick U? Well, I just needed some letter other than x since I already used X as one of my bounds of integration.

I'm adding F of A to both sides, I swap 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 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 is equal to the integral, I'm using one as my A since I know what F of one is, from 1 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. We just did it. That is the first part of the problem.

Now let's try to find F of four and F of -2. All right, F of four. Well, wherever we see next, we substitute a four there; it's going to be the integral from one to four of F prime of U du + 3. So what is going to be the integral from 1 to four of F prime of U du?

So let's look up here. So the integral from 1 to four of F prime, that's that curve right over here. Well, that's going to give this area, but it's going to be the negative of the area because if you just take the integral, it's the area that you could view above the x-axis and below or, if we're thinking in terms of U, above the U axis and below the function. But 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 bounded by the x-axis and the graph of F prime over the interval 1 to 4 is 12. So this area over here is 12, but the integral is going to be negative because once again our function is below the x-axis. So this integral, this right over here, is going to be -12. This is -12, so -12 + 3 is 9.

All right, now let's evaluate F of -2. F of -2 is equal to the integral from 1 to -2 of F prime of U du + 3. Well, it kind of feels a little strange to have the upper bound being lower than the lower bound, so we can swap the bounds and then add a negative out here. So this is going to be equal to the negative of, if we swap the bounds here, from -2 to 1 of F prime of U du + 3.

So what's the integral from -2 to 1? Well, they give us that from -2 to 1, 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 will evaluate to be -9. So the area is 9, but once again the curve is below the x-axis, so the integral would give us -9. So this would evaluate to 9.

So you take the negative of 9, which is positive 9 + 3, which is equal to 12, and we're done.