# Simple question II

Discussion in 'Math' started by boks, Dec 13, 2008.

1. ### boks Thread Starter Active Member

Oct 10, 2008
218
0
Is my book right to write

$\int ^{0}_{- \infty}e^{-a|x|}dx = \int ^{0}_{- \infty}e^{ax}dx$

?

In case, why?

2. ### studiot AAC Fanatic!

Nov 9, 2007
5,005
515
Remembering that exp(x) and exp (-x) are mirror image functions and that both are positive definite you should be able to see that the dotted area is your first integral and the shaded one the second.

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3. ### Ratch New Member

Mar 20, 2007
1,068
4
boks,

For the same reason that -a|x| = -a(-x) = ax when x <= 0, which is the range of the integral. In other words |x| = -x when x is negative or zero.

Ratch