ziphumulegn

2022-07-02

Is it true that if f is continuous, $\underset{h\to 0}{lim}\frac{1}{h}{\int }_{x}^{x+h}f\left(t\right)dt=f\left(x\right)$

Alexia Hart

Expert

There is no need to invke the FTC for this. Note that, by the monotonicity of the integral,
$h\underset{x\le s\le x+h}{min}f\left(s\right)={\int }_{x}^{x+h}\underset{x\le s\le x+h}{min}f\left(s\right)\phantom{\rule{thinmathspace}{0ex}}dt\le {\int }_{x}^{x+h}f\left(t\right)\phantom{\rule{thinmathspace}{0ex}}dt$
Similarly,
$h\underset{x\le s\le x+h}{max}f\left(s\right)={\int }_{x}^{x+h}\underset{x\le s\le x+h}{max}f\left(s\right)\phantom{\rule{thinmathspace}{0ex}}dt\ge {\int }_{x}^{x+h}f\left(t\right)\phantom{\rule{thinmathspace}{0ex}}dt$
Now divide these inequalities by h and let $h\to 0$. By continuity,
$\underset{h\to 0}{lim}\underset{x\le s\le x+h}{min}f\left(s\right)=f\left(x\right)$
with a similar result for the max. Combinining the resulting inequalities yields the result.

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