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Prove that if $f$ is a continuous function, then $\frac{d}{dx}\left( \int_x^b f(t)dt\right)= -f(x).$ Any help would be appreciated. Thanks!

2 Answers 2

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First, using the properties of the integral as in the other answer, we write

$ F(x)=\int_{b}^{x}f(t) dt \implies F(x+h)= \int_{b}^{x+h}f(t) dt= \int_{b}^{x}f(t) dt + \int_{x}^{x+h}f(t) dt $

Now, we have $ \left|\frac{F(x+h)-F(x)}{h}-f(x)\right| = \left|\frac{1}{h}\int_{x}^{x+h}f(t) dt - \frac{1}{h}\int_{x}^{x+h}f(x) \right| $

$\left|\frac{F(x+h)-F(x)}{h}-f(x)\right| = \left|\frac{1}{h}\int_{x}^{x+h}(f(t)-f(x)) dt \right| \leq \frac{1}{h}\int_{x}^{x+h}|f(t)-f(x)| dt $

$ x

Use the continuity of the function $f$ on the right hand side of the above inequality to finish the proof.

  • 0
    Did you reverse the upper and lower limits on purpose?2012-12-08
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Hint: $\int_{x}^{b} f(t)dt=-\int_{b}^{x}f(t)dt$