Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I need to integrate, then find the inverse.

The function I am working on, $$f(x)=\frac{1}{4\sqrt{|1-x|}}, x\in[0,2]$$

I tried to solve it on wolfram. It looks pretty complicated, am I doing this right? Could I use the [0,2] bounds to make the problem easier? Thanks for any kind of help.

share|cite|improve this question
Do you want to integrate the inverse with bounds $[0,2]$ or integrate $f(x)$ and then find the inverse of $f(x)$? – E.O. Feb 12 '12 at 3:45
Yes, I would like to integrate with bounds [0,2], and it should integration should sum to 1. Then find its inverse. Two separate things. – user1061210 Feb 12 '12 at 4:13
Look at my answer. If I am missing something feel free to comment. – E.O. Feb 12 '12 at 4:14
up vote 0 down vote accepted

You could break it up into cases. Then you get $$f(x)=\begin{cases}\frac{1}{4\sqrt{1-x}}&x<1\\\frac{1}{4\sqrt{x-1}}&x>1\end{cases}$$ Now you can just integrate as normal and get $$F(x)=\int f(x)dx=\begin{cases}\frac{-1}{2}\sqrt{1-x}+c&x<1\\\frac{1}{2}\sqrt{x-1}+c&x>1\end{cases}$$ So to integrate from 0 to 2 you get $$\int_0^2f(x)dx=\int_0^1\frac{1}{4\sqrt{1-x}}dx+\int_1^2\frac{1}{4\sqrt{x-1}}dx$$ $$=\left[\frac{-1}{2}\sqrt{1-x}\right]_0^1+\left[\frac{1}{2}\sqrt{x-1}\right]_1^2=0+\frac{1}{2}+\frac{1}{2}+0=1$$ As for the inverse $$F^{-1}(x)=\begin{cases}1-4x^2&x<1\\1+4x^2&x>1\end{cases}$$

share|cite|improve this answer
wow, that looks pretty awesome, thanks! – user1061210 Feb 12 '12 at 4:14
@user1061210 No problem! You can accept the answer by clicking the green tick in the top left corner of the answer. – E.O. Feb 12 '12 at 4:17
will do! Why does wolfram integrator give such a complicated answer?*sqrt%28abs%281-x%29%29 – user1061210 Feb 12 '12 at 4:21
My guess is because it does not like piecewise functions and instead puts into one function. This makes it slightly more complicated. – E.O. Feb 12 '12 at 4:27
Should the last line be $$F^{-1}(y)=\begin{cases}1-4y^2&x<1\\1+4y^2&x>1\end{cases}$$ Aren't we finding the bounds for y then? Just wondering. – user1061210 Feb 12 '12 at 19:11

1.This function is NOT one to one on the interval $[0,2]$, because $f(1-x)=f(1+x)$, therefore there is no inverse on $[0,2]$ , consider $f(.5),f(1.5) and f(.9),f(1.1)$ but there are inverses for $[0,1]$ and $[1,2]$ separately.

2.Draw the funtcion on $[0,2]$, then rotate it 90 degrees, you can see the inverse of the function, and what the integral of the function and it's inverse add up to ( hint adding them up should give you area of a rectangle).

3.Try a simpler function instead, for example $f(x)=\frac{1}{4x}$, then try to move up to a function that looks more similar to this one e.g. $f(x)=\frac{1}{4\sqrt{x}}$

share|cite|improve this answer
I think the OP was trying to say that s/he wanted to do the inverse after integrating. – Ben Crowell Feb 12 '12 at 4:52

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.