0
$\begingroup$

This is first time I'm solving integral with absolute value and I have issue with example of this integral. Here is my solution, but textbook says it should be 1 so any input is greatly appreciated. I used $u$-substitution: $$u=-|x|$$ $$\int_0^\infty0.5e^{-|x|}\,dx=-0 .5\int_0^{-\infty}e^u\,du=0.5$$

$\endgroup$
4
  • $\begingroup$ Are you sure the question is not supposed to be: $$\int_{-\infty}^{\infty} 0.5e^{-|x|}~dx$$ If so, the answer is indeed $1$. The definite integral you wrote on the title indeed evaluates to $0.5$. $\endgroup$ May 14, 2017 at 11:50
  • $\begingroup$ |x| is positive on your interval so $ |x| =x$ $\endgroup$ May 14, 2017 at 11:50
  • $\begingroup$ Are you sure the original integral wasn't from $-\infty$ to $\infty$? That gives the desired answer and, as it stands, there is no point to the absolute value. $\endgroup$
    – lulu
    May 14, 2017 at 11:50
  • $\begingroup$ This is correct integral...so it is probably mistake and integral should range from $\infty$ to ${-\infty}$ ? $\endgroup$
    – econ
    May 14, 2017 at 11:53

1 Answer 1

2
$\begingroup$

Since $x\ge0$, $ |x|=x$ $$\int_{0}^{\infty} 0,5{e^{-|x|}} dx = 0,5\int_{0}^{\infty} {e^{-x}} dx = -0,5\lim_{M \to \infty}(e^{-M}-e^0)=-0,5(0-1)=0,5$$
If $x\in (-\infty$, $\infty)$, value converges to $1$, since $e^{-|x|}$ is even: $$ \int_{-\infty}^{\infty} 0,5{e^{-|x|}} dx = \int_{-\infty}^{0} 0,5{e^{x}} dx+ \int_{0}^{\infty} 0,5{e^{-x}} dx= 2\int_{0}^{\infty} 0,5{e^{-x}} dx $$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .