Worked examples: Definite integral properties 2 | AP Calculus AB | Khan Academy

So what we're going to do in this video is several examples where we evaluate expressions with definite integrals. Right over here we have the definite integral from -2 to 3 of 2 F of x DX plus the definite integral from 3 to 7 of 3 F of x DX. All we know about f of x is the graph of y = f of x from x = -6 to x = 7. They also give us the areas between f of x and the x-axis. The negative areas show that our function is below the x-axis.

So given that, can we evaluate that? Like always, pause the video and see if you can do it on your own. Well, the first thing that my brain wants to do is I want to take these constants out of the integral because then once they're out, I'm just taking the straight up definite integrals of F of x. I can relate that to the areas over here, and I know I can do that. This is a very common integration property, which applies to definite and indefinite integrals.

If I’m taking the integral of K F of x DX, this is the same thing as K times the integral of f of x DX. So let's just apply that property there, which is really you're taking the scalar outside of the integral to say this is going to be the same thing as 2 * the definite integral from -2 to 3 of f of x DX plus 3 * the integral from 3 to 7 of f of x DX.

All right, now can we evaluate these things? So what is this going to be, the definite integral from -2 to 3 of f of x DX? Well, we can view that as the area between the curves y = f of x and the x-axis between x = -2 and x = 3. Between x = -2 and x = 3, they give us the area, which is 7. So this is equal to 7.

Then we have the integral from 3 to 7 of f of x. So we're going to go from 3 to 7, and once again, this is going to evaluate to a negative value because f of x is below the x-axis there, and it's going to evaluate to -3.

So this is all going to be 2 * 7, which is 14, plus 3 * -3, so plus -9. Therefore, 14 - 9 is equal to 5. This is fun! Let's do more of these!

All right, okay, so here this first integral, the integral from 0 to 5 of f of x DX, is pretty straightforward. We are talking about the area there, which they tell us is 4. So that was pretty easy to evaluate.

Now we're going to subtract, we're going to subtract going from -8 to -4 of 2 F of x. Well, let's just take this 2 outside. If we just take this 2 outside, then this just becomes the integral from -8 to -4 of F of x. This area they're talking about evaluates to 5.

So this is all going to simplify to 4 minus that 2 that we brought out minus 2 * 5, which is equal to, let's see, 4 - 10, which is equal to -6.

All right, let's do another one of these! So here I have the integral from -7 to 5. I'm going from -7 to -5, which is going to be right around there. I want to find this area right over here.

Then I'm going to go from -5 to zero. This is going to be going from -5 to zero, which will be all of that. There are a couple of ways you could think about doing it. You could assume I have some symmetry here, and they don't tell it for sure, but it looks very symmetric around x = -5.

You can assume that this 8 is split between these two regions. But an easier way to do it is just to realize, look, I'm going from -7 to 5 and then from -5 to zero, and I'm integrating the same thing, F of x DX.

So this integral I can rewrite as the integral from -7 all the way to zero of F of x DX. This is really just going to be the net area between -7 and 0. So we have the positive 8 there, so this is going to be equal to the positive 8, and then we have the -1 there, so minus 1, which is equal to 7.