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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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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
    
If this is homework, please use the homework tag. –  Ben Crowell Feb 12 '12 at 4:51

2 Answers 2

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

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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? wolframalpha.com/input/?i=1%2F%284*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. $f(x)=\frac{1}{4\sqrt{x}}$

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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

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