Why is absolute value of negative exponent equal to positive value?

Question

I am asked to integrate the following: $\int_{-\infty}^{0}e^{-\left\lvert 3x\right\rvert}dx$

And I am told that $e^{-\left\lvert x\right\rvert}=e^{x}$

How is it that an absolute value (the exponent) multiplied by -1 is still equal to a positive number?

Answer 1

Your integral is over only negative numbers (and zero). If $x$ is negative, then $|x|$ is positive and $-|x|$ is negative again, so $x=-|x|$. This is of course not true in general, but absolutely fine if you only deal with negative numbers.

Answer 2

Consider $x=-2$. Then $|x| = 2$ and $-|x|= -2 = x$. We know from the bounds of the integral that $x<0$, so the example with $x = -2$ works for all $x$.

Answer 3

Another possibility is to change $t=-x$ such that the integral is now on positive numbers.

$\displaystyle \int_{-\infty}^0 e^{-|3x|}\mathop{dx}=\int_{+\infty}^0 e^{-|-3t|}(-\mathop{dt})=\int_0^{+\infty}e^{-3t}\mathop{dt}$

With now $-|-3t|=-(3t)=-3t$ and the sign before $\mathop{dt}$ is cancelled with reordering the bounds of integration.