Worked example: area between curves | AP Calculus AB | Khan Academy

What we're going to do using our powers of calculus is find the area of this yellow region. If at any point you get inspired, I always encourage you to pause the video and try to work through it on your own.

So, the key here is you might recognize, "Hey, this is an area between curves; a definite integral might be useful!" I'll just set up the definite integral sign.

First, we need to think about what are our left and right boundaries of our region. Well, it looks like the left boundary is where the two graphs intersect right over here, and the right boundary is where they intersect right over there.

Well, what is this point of intersection? It looks like it is negative one comma negative two. We can verify that in this red curve. If x is negative one, let's see, you square that, you'll get 1 minus 3; you do indeed get y is equal to negative 2. In this blue function, when x is equal to negative 1, you get 1 minus 4 plus 1. Once again, you do indeed get y is equal to negative 2.

The same thing is true when x is equal to one: one minus three is negative two, and one minus four plus one is negative two. So our bounds are indeed from x equals negative one to x equals positive one.

Now, let's think about our upper and lower bounds over that interval. This blue graph is our upper bound, and so we would subtract the lower bound from the upper bound. We would have x to the fourth minus four x squared plus 1, and from that, we will subtract x squared minus 3 dx.

In many other videos, we have talked about why you do this—why this makes sense to just subtract the lower graph from the upper graph when you're finding the area between them. But now we just have to evaluate this definite integral. So let's just get down to business.

All right, so we have the integral from negative one to one. We have x to the fourth, and now we have minus four x squared, and then when you distribute this negative sign, you're gonna have to subtract another x squared, so you're gonna have minus 5 x squared, and then you have plus 1 and then you're going to subtract a negative 3. So it's going to be 1 plus 3; so it's going to be plus 4 dx.

Just to be clear, I should put parentheses right over there, because it's really the dx that is being multiplied by this entire expression. Let's see, let's find the anti-derivative of this. This should be pretty straightforward. We're just going to use the reverse power rule multiple times.

The anti-derivative of x to the fourth is x to the fifth over five. We just incremented the exponent and divided by that incremented exponent. Minus, same idea here, 5 x to the third over 3 plus 4x, and then we're going to evaluate it at 1 and then subtract from that it evaluated at negative 1.

So let's first evaluate it at 1. We're going to get one fifth minus five thirds plus four. Now, let us evaluate it at negative one. So minus let's see if this is negative one: we're going to get negative one-fifth, negative one-fifth, and this is going to be plus five-thirds, plus five-thirds, and then this is going to be minus four, minus four.

When you distribute the negative sign, we're going to distribute this over all of these terms. If we make this positive, this will be positive, this one will be negative, and then this one will be positive.

So you have one-fifth plus one-fifth, which is going to be two-fifths. That and that, and then minus five-thirds minus five-thirds, so minus ten over three, and then four plus four, so plus 8. Also, we just need to simplify this; this is going to be, let's see, it's going to be 8.

If I write these two with a denominator of 15, because that's the common denominator of 3 and 5, let's see, two-fifths is 6 15. Yeah, that's right: 5 times 3 is 15, 2 times 3 is 6. And then 10 thirds, let's see, if we multiply the denominator times 5, we have to multiply the numerator times 5.

So it's going to be 50 15. So what's 6 15 minus 50 15? This is going to be equal to 8 minus 6 minus 50, which is minus 44. So, minus 44 over 15. What is 44 over 15? 44 over 15 is equal to 2 and 14 15.

So that's really what we're subtracting; we're going to subtract 2 and 14 15. If you subtract 2 from this, you would get 6 minus 14 over 15, because we still have to subtract the 14 15, and then 6 minus 14 15 is going to be equal to 5 and 1 15.

So just like that, we were indeed able to figure out this area.