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### 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

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?
• No need, i say NO NEED my friend
use the normal function, plug in y, solve for x
(1 vote)
• why are there two ends in the title? Typo?
• Seems to be fixed.
• I notice that the area between the curve and the y axis @ y=e^3,y=e is exactly the same as the area between the curve and the x axis where y=e and y=e^3. (ie @ x=15e^-1 and x=15e^-3). Any underlying principle here or just coincidence?
• It's not just a coincidence. The areas happen to be the same only if the inverse of the function is identical to the function itself. I don't have a rigorous proof but if you think about it, when you take the inverse you switch the y and x just like how you switch the bounds from y = e and e^3 to x = e and e^3. And y = 15/x just happens to be one of them.

Apparently, these functions are called involutions. You can find more about them here: https://en.wikipedia.org/wiki/Involution_%28mathematics%29
• 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.