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By integrating the difference of two functions, you can find the area between them. Created by Sal Khan.
Video transcript
What I want to do in this video is find the area of the region in the first quadrant right over here that's below the graph of y is equals to square root of x, but above the graph of y is equal to x squared. So we're talking about this region right over here. And we know what the two endpoints are. The two endpoints are when these two functions equal each other. So x squared is equal to square root of x at x is equal to 0 and at x is equal to 1. So how do we think about this? Well, one way to do it is just to think about, well, what's the area between y is equal to square root of x and the x-axis? So that's going to be, and we're going from 0 to 1, so this is the area under square root of x from 0 to 1 dx. So this is literally denoting this entire region, right over here, all the way up to this boundary point. And then from that, subtract out this region. The area underneath y is equal to x squared, but above the x-axis. So from that, we could subtract from 0 to 1. That's the interval we care about. The area under x squared dx. And this would be completely legitimate. And you might say, hey, look, I have the same boundaries of integration, I have dx right over here. Couldn't I have also written this as the definite integral from 0 to 1 dx of square root of x minus x squared? And if you asked that, I would say absolutely. And this is actually a different way of conceptualizing it. Here, you're taking the area. And instead of, if you conceptualize dx as the width of each of your rectangles, now the height of the rectangle isn't just a function between the x-axis and the function itself. The height is the difference between the two functions. So in this case, this would be the width of a rectangle, and then the height would be the difference between the two functions. And then you would have another rectangle right over there. All of them have width dx, and then the height is the difference between the functions. And then as you take the limit as you have super thin, as these rectangles get thinner and thinner and thinner, and you have more and more and more them, you are essentially approximating the area. So this is one way of conceptualizing it. Or just using, I guess the tools we already had, the area under the curve of square root of x. Subtracting from that the area under the curve of x squared. These are completely legitimate ways of conceptualizing it. But now let's evaluate it. So what's the antiderivative? We know from the second fundamental theorem of calculus, we can say that this is right over here is just going to be the antiderivative of the square root of x. Well, square root of x is the same thing as x to the 1/2 power, so we just increment it. So we say x to the 3/2 power, and then divide by 3/2. So that's the same thing as multiplying by 2/3. And we're going to evaluate that from 0 to 1. And from that, we subtract the antiderivative of x squared, is x to the third over 3. And we're going to evaluate that from 0 to 1 as well. Now this first expression, if we evaluated it at 1. 2/3 times 1 and 3/2. That's just going to be 2/3. And then from that, we're going to subtract this thing evaluated at 0, which is just 0. So we're just left with 2/3. And then we're going to subtract this thing evaluated at 1, minus 1/3 third. And then, also, we're going to subtract this thing. Well, just to be clear, we're going to subtract this thing evaluated at 1, and then from that, it evaluated at 0, which is just 0. But this whole thing simplifies to just subtracting 1/3. And there you have it. We get our answer. 2/3 minus 1/3 is equal to 1/3. So this area right over here. Whatever units we're using is 1/3 square units of area.