Absolute value sign moving from around an integral to the limits of integration?

Question

Saw something today that bugged me a little:

Is it possible, through some theorem/law/etc, for a definite integral to make this manipulation?

$$\left|\int^{x+h}_{x}f(t)\,\mathrm{d}t\right| \leq \int^{|x|+|h|}_{|x|}f(t)\,\mathrm{d}t$$

I see the (possible) end result of the triangle inequality there, so I guess this may be the same as asking if

$$\left|\int^{x+h}_{x}f(t)\,\mathrm{d}t\right| \leq \int^{|x+h|}_{|x|}f(t)\,\mathrm{d}t$$

For context, I was looking at this question. Perhaps it was a typo, as a comment there points out?

Answer 1

This is not true. It's not too difficult to find counterexamples. Notice that $$\left\vert \int_{x}^{x + h}f(t)dt \right\vert \geq 0,$$ but we know nothing about the sign of $$\int_{\vert x \vert}^{\vert x \vert + \vert h \vert}f(t)dt.$$ In particular any function satisfying $f(t) < 0$ for all $t$ is an automatic counterexample.

On the other hand, if we assume that $f$ is increasing and nonnegative, then the answer is yes, for the interval $[x,x+h]$ (resp. $[x+h,x]$ if $h < 0$) lies to the left of the interval $[\vert x \vert, \vert x \vert + \vert h \vert]$. Since both intervals have length $\vert h \vert$, the inequality follows since $f$ is increasing.