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# 2015 AP Calculus AB 5d

Constructing expression for f with integral.

## Want to join the conversation?

• At , when you take the anti-derivative of f'(x), wouldn't it be f(x) + C?
(1 vote) • 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) • 1. at , ∫ f'(x) dx (a, b are bounds)=f(x)|(a, b are bounds).
while in the previous video, it's defined that F(x)=∫ f(t) dt (bounds are a and x).

I'm confused with the t and x as well as the bounds in the above equations.

2. I suppose the equation of ∫ f'(x) dx=f(x) without bounds is not correct, but why? And how do you get the antiderivative of f'(x) to be f(x)? Thanks.
(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.