2
$\begingroup$

I've been reading a book lately and this has come up several times. If $f$ is continuous, then $$ \lim_{h \to 0} \frac{1}{h} \int_{b}^{b+h} f = f(b). $$ Now, I know I can use the fundamental theorem of calculus to show this, and I just wanted to check that I'm not missing a much simpler argument. I ask because this is always said to 'follow from continuity' and the fundamental theorem of calculus is never explicitly mentioned.

$\endgroup$
  • 1
    $\begingroup$ I think the Fundamental theorem of Calculus would be a good resort indeed $\endgroup$ – imranfat Apr 14 '16 at 13:44
  • $\begingroup$ In fact the result you have mentioned above is called First Fundamental Theorem of Calculus namely if $f$ is Riemann integrable on $[a, b]$ and $$F(x) = \int_{a}^{x}f(t)\,dt$$ then $$F'(c) = \lim_{h \to 0}\frac{1}{h}\int_{c}^{c + h}f(t)\,dt = f(c)$$ for all points $c \in [a, b]$ at which $f$ is continuous. $\endgroup$ – Paramanand Singh Apr 17 '16 at 15:03
2
$\begingroup$

Let $\epsilon>0$. Choose $\delta>0$ such that etc. If $0<h<\delta$ then

$$\left|\frac1h\int_b^{b+h}f(t)\,dt-f(b)\right| =\left|\frac1h\int_b^{b+h}(f(t)-f(b))\,dt\right| \le \frac1h\int_b^{b+h}|f(t)-f(b)|\,dt\le\frac1h\int_b^{b+h}\epsilon\,dt=\epsilon.$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.