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

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?

• Consider the limits of integration: for all $x<0$, ${-|x|} = x$ – Brian Mar 4 at 22:34
• Many years ago, I tutored $2$nd year university engineering calculus for $2$ years. I was amazed that these students could often handle quite complicated $3$D transformations, integrals, etc., but for some reason, the one particular issue they had the most trouble with (at least in terms of asking me) was understanding that if $x \lt 0$, then $-|x| = x$. Note this only involved several students, & I'm not trying to disparage or put them down, as they were very intelligent, but I found it interesting, & perplexing (even now, I don't know why), that they had this particular math "blind" spot. – John Omielan Mar 5 at 6:05
• @JohnOmielan for me it is just the disconnect in seeing "-x" turn to "x" which usually implies dropping the negative sign altogether, and not realizing that in this case the negative sign will still be there because of the interval – blizz Mar 5 at 14:18
• @blizz What I referred to happened about $30$ years ago, so my recollection is rather limited, but your explanation sounds like it was what those engineering students had an issue with. Thanks for telling me this. – John Omielan Mar 5 at 16:59

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.
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 $$x$$.
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.