Functions defined by integrals: switched interval | AP Calculus AB | Khan Academy

The graph of f is shown below. Let G of X be equal to the definite integral from 0 to X of f of T DT.

Now, at first when you see this, you're like, "Wow, this is strange! I have a function that is being defined by an integral, a definite integral, but one of its bounds is X." You should just say, "Well, this is okay. A function can be defined any which way," and as we'll see, it's actually quite straightforward to evaluate this.

So, G of -2. G of -2, and I'll do the -2 in a different color. G of -2, well, what we do is we take this expression right over here, this definite integral, and everywhere we see an X, we replace it with a -2. So, this is going to be equal to the integral from 0 to X of, and I'll write X in a second, F of T DT. Well, X is now -2, this is now -2.

Now, how do we figure out what this is? Now, before we even look at this graph, you might say, "Okay, this is the region under the area, the region under the graph of F of T between -2 and 0." But you have to be careful. Notice our upper bound here is actually a lower number than our lower bound right over here.

So, it will be nice to swap those bounds so we can truly view it as the area of the region under F of T above the T-axis between those two bounds. When you swap the bounds, this is going to be equal to negative definite integral from -2 to zero of F of T DT.

And now what we have right over here, what I'm squaring off in magenta, this is the area under the curve F between -2 and 0. So, between -2 and zero, that is this area right over here that we care about.

Now, what is that going to be? Well, you could—there's a bunch of different ways that you could do this. You could split it off into a square and a triangle. The area of this square right over here is four; it's 2 by 2.

Just make sure to look at the unit; sometimes each square doesn't represent one square unit, but in this case it does, so that's four. Then up here, this is half of four, right? If it was all of this, that would be four. This triangle is half of four, so this is two right over there.

Or you could view this as base times height times 1/2, which is going to be 2 times 2 times 1/2. So, this area right over here is six. So, this part is six, but we can't forget that negative sign, so this is going to be equal to negative six. Thus, G of -2 is -6.