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## AP®︎ Calculus BC (2017 edition)

### Unit 11: Lesson 4

Area: horizontal areas between curves# Area between a curve and the 𝘺-axis

AP.CALC:

CHA‑5 (EU)

, CHA‑5.A (LO)

, CHA‑5.A.2 (EK)

We can use a definite integral in terms of 𝘺 to find the area between a curve and the 𝘺-axis.

## Video transcript

- [Instructor] So right over here, I have the graph of the function
y is equal to 15 over x, or at least I see the part of
it for positive values of x. And 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 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 third power. So pause this video, and see
if you can work through it. So 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 over 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 over y. These right over here are
all going to be equivalent. Now how does this right over 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 over y. And then what's the height gonna 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, let me do it over here, as 15 over 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 third 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 third power. So all we did, we're used
to seeing things like this, where this would be 15 over x, dx. All we're doing here is,
this is 15 over y, dy. So let's evaluate this. So we take the antiderivative of 15 over y and then evaluate at these two points. So this is going to be equal to antiderivative of one over y is going to be the natural log
of the absolute value of y. So it's 15 times 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 third and at e. So let's first evaluate at e to the third. So that's 15 times the natural log, the absolute time, the natural,
(laughs) the natural log of the absolute value of
e to the third 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 third power, what power do I have to raise e to, to get to e to the third? Well, that's just going to be three. And then the natural log of e, what power do I have to
raise e to, to get e? Well, that's just one. So this is 15 times three minus 15. So that is all going to get us to 30, and we are done, 45 minus 15.