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Area between a curve and and the _-axis | AP Calculus AB | Khan Academy


3m read
·Nov 11, 2024

So right over here I have the graph of the function y is equal to 15 / x, or at least I see the part of it for positive values of X. What I'm curious about in this video is I want to find the area not between this curve and the positive x-axis. I want to find the area up between the curve and the Y-axis, bounded not by two x values but bounded by two y values.

So, with the bottom bound of the horizontal line y is equal to e and an upper bound with y is equal to e to the 3rd power, pause this video and see if you can work through it.

One way to think about it, this is just like definite integrals we've done where we're looking between the curve and the x-axis. But now it looks like things are swapped around; we now care about the Y-axis. So, let's just rewrite our function here and let's rewrite it in terms of x.

So, if y is equal to 15/x, that means if we multiply both sides by x, xy is equal to 15, and if we divide both sides by y, we get x is equal to 15/y. These right over here are all going to be equivalent.

Now, how does this right over here help you? Well, think about the area; think about estimating the area as a bunch of little rectangles here. So, that's one rectangle, and then another rectangle right over there, and then another rectangle right over there. So, what's the area of each of those rectangles?

So, the width here that is going to be x, but we can express x as a function of y. So, that's the width right over there, and we know that that's going to be 15/y. And then, what's the height going to be? Well, that's going to be a very small change in y; the height is going to be dy.

So, the area of one of those little rectangles right over there, say the area of that one right over there, you could view as 15/y dy. And then we want to sum all of these little rectangles from y is equal to e all the way to y is equal to e to the 3rd power.

So, that's what our definite integral does. We go from y is equal to e to y is equal to e to the 3rd power. So, all we did—we're used to seeing things like this, where this would be 15/x dx; all we're doing here is this is 15/y dy.

So, let's evaluate this. We take the anti-derivative of 15/y and then evaluate at these two points. So, this is going to be equal to the anti-derivative of 1/y, which is the natural log of the absolute value of y.

So, it's 15 * the natural log of the absolute value of y, and then we're going to evaluate that at our endpoints. So, we're going to evaluate it at e to the 3 and at e.

So, let's first evaluate it at e to the 3. So that's 15 times the natural log of the absolute value of e to the 3rd power minus 15 times the natural log of the absolute value of e.

So, what does this simplify to? The natural log of e to the 3rd power. What power do I have to raise e to get to e to the 3? Well, that's just going to be three. And then the natural log of e—what power do I have to raise e to get e? Well, that's just one.

So, this is 15 * 3 minus 15. So, that is all going to get us to 30, and we are done: 45 minus 15.

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