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## AP®︎/College Calculus BC

### Course: AP®︎/College Calculus BC > Unit 8

Lesson 5: Finding the area between curves expressed as functions of y# 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.

## Want to join the conversation?

- Why do you have to do the ln of the absolute value of y as the integral of a constant divided by y? In other words, why 15ln|y| and not 15lny?(6 votes)
- Because logarithmic functions cannot take negative inputs, so the absolute value sign ensures that the input is positive.(12 votes)

- In the video, Sal finds the inverse function to calculate the definite integral. What if the inverse function is too hard to be found? Is there an alternative way to calculate the integral?(8 votes)
- Bit late but if anyone else is wondering the same thing, you will always be able to find the inverse function as an implicit relation if not an explicit function of the form y = f(x). From there on, you have to find the area under the curve for that implicit relation, which is extremely difficult but here's something to look into if you're interested:

https://math.stackexchange.com/questions/1019452/finding-the-area-of-a-implicit-relation(1 vote)

- why are there two ends in the title? Typo?(6 votes)
- Seems to be fixed.(0 votes)

- Would finding the inverse function work for this? I know the inverse function for this is the same as its original function, and that's why I was able to get 30 by applying the fundamental theorem of calculus to the inverse, but I was just wondering if this applies to other functions (probably not but still curious).(3 votes)
- The way I did it initially was definite integral 15/e^3 to 15/e of (15/x - e)dx + 15/e^3(20-e) I got an answer that is very close to the actually result, I don't know if I did any calculation errors(1 vote)
- Can you just solve for the x coordinates by plugging in e and e^3 to the function? It seems like that is much easier than finding the inverse.(0 votes)
- how can I fi d the area bounded by curve y=4x-x and a line y=3(0 votes)
- Could you please specify what type of area you are looking for? What are the bounds?(2 votes)

## 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.