# Simple question II

#### boks

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

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

?

In case, why?

#### studiot

Joined Nov 9, 2007
5,003
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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#### Ratch

Joined Mar 20, 2007
1,068
boks,

Is my book right to write

?

In case, why?
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