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

$f(x)={{\left| 1-x \right|}^{-0.5}}\exp (-{{\left| 1-x \right|}^{0.5}})$ for $x>0$

I was thinking to do a u-sub, $u={{\left| 1-x \right|}^{0.5}}$

but what would your $du$ be?

should I consider the negative sign inside the absolute value?

share|cite|improve this question
One definition of the absolute value is that abs($x$) $=x,$ if $x>0$ or $-x$ if $x<0.$ Try to determine where the terms inside the absolute value change sign. Once you've done this, you can break up the integral at those places and remove the absolute values. – Mike B Feb 12 '12 at 22:23
Already did, final answer below, is that correct? if so, why doesn't it sum to 1? I know it is suppose to. – user1061210 Feb 13 '12 at 0:44
up vote 2 down vote accepted

If $u=\sqrt{|1-x|}$, then in fact $$u=\begin{cases} \sqrt{1-x},&\text{if }x\le 1\\\\ \sqrt{x-1},&\text{if }x>1\;, \end{cases}\tag{1}$$

since by definition $$|1-x|=\begin{cases} 1-x,&\text{if }1-x\ge 0\\\\ -(1-x)=x-1,&\text{if }1-x<0\;. \end{cases}$$

Thus, $$du=\begin{cases} -\frac12(1-x)^{-1/2}dx,&\text{if }x<1\\\\ \frac12(x-1)^{-1/2}dx,&\text{if }x>1\;. \end{cases}\tag{2}$$

Thus, you’ll need to split the integral in two, one for $x<1$ and one for $x>1$. (Why did I change $x\le 1$ in $(1)$ to $x<1$ in $(2)$?)

share|cite|improve this answer
Hey, Brian. Thanks, it's pretty clear. I got a question on the $du$ part, where I got without the negative sign $$\[\frac{1}{2}{{(1-x)}^{\frac{1}{2}}}dx\]$$, did you let your u be the absolute value without negative signs? – user1061210 Feb 12 '12 at 23:15
@user1061210: I used $u$ as in $(1)$, so that there’s a factor of $-1$ from $(1-x)'$ in the derivative of the $x<1$ part. – Brian M. Scott Feb 12 '12 at 23:27
$$F(a)=\int{f(x)dx }=\left\{ \begin{array}{*{35}{l}} \frac{1}{2}{{e}^{-{{\left( 1-a \right)}^{0.5}}}}-\frac{1}{2}{{e}^{-1}} & 0<a\le 1 \\ 1-\frac{1}{2}{{e}^{-{{(a-1)}^{0.5}}}}-\frac{1}{2}{{e}^{-1}} & 1<a \\ \end{array} \right.\ $$ I got the above answer after integration, it's a cumulative function, which should sum to 1, but it doesn't, do you know what's wrong with it? – user1061210 Feb 13 '12 at 0:38
Working rather hastily, I get $4-\frac2{e}$. Don’t forget that $dx=2u du$, so that you’re integrating $2e^{-u}du$. – Brian M. Scott Feb 13 '12 at 3:14
I figured it out, thanks everyone for trying to help. – user1061210 Feb 13 '12 at 4:05

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.