Horizontal area between curves | Applications of definite integrals | AP Calculus AB | Khan Academy

So I have two curves graphed here, and we're used to seeing things where Y is a function of X. But here we have X as a function of Y. In fact, we can write this top expression as being a function of Y, and this second one, just to make it different, we could view this as G of Y. Once again, it is a function of Y.

What we're concerning ourselves with in this video is how do we find this area in this light blue color between these two curves. I encourage you to pause the video and try to work through it.

All right, so a huge hint here is we're going to want to integrate with respect to Y—a definite integral where our bounds are in terms of Y. So, for example, this is this lower point of intersection right over here. This would be our lower bound in terms of Y. Let's call that y1. And then this up here, this would be our upper y bound.

If we think about where these two curves intersect and we look at the y-coordinates of those intersections, well, that gives us two nice bounds for our integral. So we're going to take our integral from y1 to Y2. From y1 to Y2, we're going to integrate with respect to Y, dy.

And so what are we going to sum up? Well, when we integrate, we can think about taking the sum of infinitely thin rectangles, and in this case, it would be infinitely flat rectangles since we're thinking about dy. So dy would be the height of each of these rectangles.

And what would be, in this case, the width or the length of this rectangle right over here? Well, over this interval from y1 to Y2, our blue function f of y takes on larger X values than G of Y. So this length right over here, this would be f of y, this x value, minus this x value, minus G of Y.

So this is going to be f of Y minus G of Y. Well, we know what F of Y and G of Y are. Really, the trickiest part is figuring out these points of intersection. So let's think about where these two curves intersect. They are both equal to X, so we can set the two y expressions equal to each other.

So we know that negative... let me do it in that other color... so we know that -y^2 + 3y + 11 is going to be equal to, this is going to be equal to y^2 + y - 1. So let's just subtract all of this from both sides so that on the right side we have zero, and on the left side we just have a quadratic.

So let's subtract y's; let's subtract y and then subtract -1, which is just adding one. And over here, we're going to do the same thing: minus y + 1. What we are left with is going to be, hopefully, a straightforward quadratic.

So let's see, this is going to be -2y² + 2y. Am I doing that right? Yeah, plus 2y + 12 is equal to 0. And then this over here, I can factor out a -2 and I get -2 * (y² + y + 6) is equal to zero.

This we can factor from inspection. What two numbers, when we add, equal 1, and when we take their product, we get -6? Well, that would be -3 and 2. So this is going to be -2 * (y - 3)(y + 2). That's just straightforward factoring of a polynomial, a quadratic. Did I do that right? Right? Yep, that looks right.

So what are the points of intersection? The points of intersection are going to be y = 3 and y = -2. So this right over here is y = -2, and then the upper bound is y = 3. So now we just have to evaluate this from -2 all the way until 3.

So let's do that. I'm going to clear this out so I get a little bit of real estate. So this is equal to the integral from -2 to 3 of (y² + 3y + 11) minus all of this stuff. So if we just distribute a negative sign here, it's minus (y² + y - 1) and then we have a dy.

This is equal to the definite integral from -2 to 3 of, let's see, y² - y² - 2y² and then 3y - y is going to be +2y, and then 11 + 1 gives us +12, we saw this just now when we were trying to solve for y dy.

And so what is that going to be equal to? Well, we just take the anti-derivative here. This is going to be... let's see... -2, let's increment the exponent, y³, divide by that exponent, reverse power rule + 2y²/2, which is just y², just the reverse power rule.

And then + 12y, and we're going to evaluate that at 3 and at -2. So if we evaluate that at 3, we are going to get... let's see... -2 * (27/3) + 9 + 36.

And then we are going to want to subtract... minus all of this evaluated at -2. So it's going to be -2 * (8/3) + 4 - 24. So we just have a little bit of mathematics to do.

So let's see, this is going to be 27. And at 3 is... 9, so this is -8 - 8 + 9 is going to be -9 + 36. All of that is going to be equal to... so the stuff in blue is equal to 27, right? Did I do that right? Yeah, -8 + 9... yep, 27.

And then all the stuff in red, over here we have... this is going to be negative times a negative, so it's 16/3 + 4 - 24. So that is going to be 16/3, and then minus -20. But then we have this negative out here.

So if we distribute that, we'll get +20. Instead of 16/3, we could rewrite that as 5 and 1/3 or 5 1/3. And so what is that going to get us? Let me scroll down a little bit, or let me go to the right so I have a little bit more real estate here.

So then that is going to be equal to... we get a minor drum roll here... 47 - 5 - 1/3, which is equal to, let's see, 47 - 5, that's 42 - 1/3 is... now we get a drum roll: 41 and 2/3. And we are done!