This topic is linked to Problem 3.
Y by Adventure10, Mango247
Here I think is a solution to the question: what is the expected area when we pick the second point from inside the largest interval?
Call the small first interval , call the leftmost piece from the second break. It's more useful to not define the the larger interval.
We can then say that the area of the triangle is obviously and so the average area is
From here we can make multiple simplifications:
We now look for a substitution to solve which was initially where I got stuck but works and makes the integral
It's pretty obvious when considering the substitution now that is a semicircle centered at (, ). The semicircle has area and so our equation changes to
The integral which is then formed, , would be pretty easy to solve by hand but I just plugged it into symbolab (https://www.symbolab.com/solver/definite-integral-calculator/2%5Cint_%7B0%7D%5E%7B%5Cfrac%7B1%7D%7B2%7D%7D%20%5Cfrac%7B%5Cpi%7D%7B8%7Dy%5Csqrt%7B%5Cfrac%7B1%7D%7B2%7D%5Cleft(%5Cfrac%7B1%7D%7B2%7D-y%5Cright)%7Ddy) which yields
Which is unsurprisingly pretty close to the other value we found of .
I am currently working on the other distribution discussed in problem three, hope this doesn't take too long.
Anyway, this seems right to me. Let me know if you see any errors.
Call the small first interval , call the leftmost piece from the second break. It's more useful to not define the the larger interval.
We can then say that the area of the triangle is obviously and so the average area is
From here we can make multiple simplifications:
We now look for a substitution to solve which was initially where I got stuck but works and makes the integral
It's pretty obvious when considering the substitution now that is a semicircle centered at (, ). The semicircle has area and so our equation changes to
The integral which is then formed, , would be pretty easy to solve by hand but I just plugged it into symbolab (https://www.symbolab.com/solver/definite-integral-calculator/2%5Cint_%7B0%7D%5E%7B%5Cfrac%7B1%7D%7B2%7D%7D%20%5Cfrac%7B%5Cpi%7D%7B8%7Dy%5Csqrt%7B%5Cfrac%7B1%7D%7B2%7D%5Cleft(%5Cfrac%7B1%7D%7B2%7D-y%5Cright)%7Ddy) which yields
Which is unsurprisingly pretty close to the other value we found of .
I am currently working on the other distribution discussed in problem three, hope this doesn't take too long.
Anyway, this seems right to me. Let me know if you see any errors.
This post has been edited 6 times. Last edited by adamodoherty, Nov 15, 2017, 8:44 PM