2
$\begingroup$

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

  • 5
    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.2012-02-12
  • 0
    Already did, final answer below, is that correct? if so, why doesn't it sum to 1? I know it is suppose to.2012-02-13

1 Answers 1

2

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 $(2)$?)

  • 0
    Hey, Brian. Thanks, it's pretty clear. I got a question on the $du$ part, where I got without the negative sign $$\[\frac{1}{2}{{(1-x)}^{\frac{1}{2}}}dx\]$$, did you let your u be the absolute value without negative signs?2012-02-12
  • 0
    @user1061210: I used $u$ as in $(1)$, so that there’s a factor of $-1$ from $(1-x)'$ in the derivative of the $x<1$ part.2012-02-12
  • 0
    $$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}} & 02012-02-13
  • 0
    Working rather hastily, I get $4-\frac2{e}$. Don’t forget that $dx=2u du$, so that you’re integrating $2e^{-u}du$.2012-02-13
  • 1
    I figured it out, thanks everyone for trying to help.2012-02-13
  • 0
    Is it necessary that the two primitives has the same limit when approaching 1?2014-03-04