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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.

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    Look $a$t my answer. If I am missing something feel free to comment.2012-02-12

2 Answers 2

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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 $F^{-1}(x)=\begin{cases}1-4x^2&x<1\\1+4x^2&x>1\end{cases}$

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    You could write the inverse function in terms of y, but then you would get $F^{-1}(y)=\frac{-1}{2}\sqrt{1-y}$ instead. I just wrote it in terms of x to get $F^{-1}(x)=1-4x^2$. Therefore the bounds are still valid.2012-02-12
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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. $f(x)=\frac{1}{4\sqrt{x}}$

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    I think the OP was trying to say that s/he wanted to do the inverse after integrating.2012-02-12