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?

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    $\begingroup$ Consider the limits of integration: for all $x<0$, ${-|x|} = x$ $\endgroup$ – Brian Mar 4 at 22:34
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    $\begingroup$ 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. $\endgroup$ – John Omielan Mar 5 at 6:05
  • $\begingroup$ @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 $\endgroup$ – blizz Mar 5 at 14:18
  • $\begingroup$ @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. $\endgroup$ – 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.

  • $\begingroup$ I was overthinking this one... :\ $\endgroup$ – blizz Mar 4 at 22:46
  • $\begingroup$ Easy thing to do! :) $\endgroup$ – R_B Mar 4 at 22:47

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.


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