If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

## Integral Calculus

### Course: Integral Calculus>Unit 3

Lesson 5: Area: horizontal area between curves

# Area between a curve and the 𝘺-axis

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?
• Because logarithmic functions cannot take negative inputs, so the absolute value sign ensures that the input is positive.
• 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?
• 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?
• Seems to be fixed.
• 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).
• 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)
• If you got anything other than exactly 30, its wrong.
(1 vote)
• I thought of it as a 270 deg rotation and got x = -15y because a 270 deg rotation would get you from (x, y) to (y, -x) and got -30.
(1 vote)
• It doesn't though. A 270 degree rotation goes from (x,y) to (x, -y). So, x stays positive and hence, so should the integral. Plus, a negative answer wouldn't make sense here as volume should be positive.
(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.