So I was trying to prove the first part of the Fundamental Theorem of Calculus in a different way using the Riemann sum definition of the definite integral, rather than the way it was presented in my high school calculus class (squeeze theorem, mean value theorem). There's one step in my proof that I'm not confident is legit, and I'm unsure if the proof breaks down.

The theorem says that $\displaystyle F(x) = \int \limits_a ^x f(t) dt \rightarrow \frac{dF(x)}{dx} = f(x)$.

Using the definition of the derivative, $ dF(x)/dx $ equals

$ \displaystyle \lim \limits_{h \rightarrow 0} \frac{F(x+h)-F(x)}{h} = \lim\limits_{h \rightarrow 0}\frac{\int\limits_a^{x+h} f(t) dt - \int\limits_{a}^xf(t) dt}{h} = \lim \limits_{h \rightarrow 0} \frac{1}{h} \int \limits_x ^{x+h} f(t)dt $.

In my attempt I replaced the definite integral with its Riemann Sum definition with regular partitions, so the latter equals

$ \displaystyle \lim \limits_{h\rightarrow 0} \hspace{2mm} \frac{1}{h} \lim \limits_{n \rightarrow \infty} \sum \limits_{i = 1}^n(\frac{x+h - x}{n}) \hspace{2mm}f(x+(\frac{x+h-x}{n})i)$ =

$\displaystyle \lim \limits_{h\rightarrow 0} \hspace{2mm} \lim \limits_{n \rightarrow \infty} \sum \limits_{i = 1}^n\frac{1}{n} f(x+\frac{h}{n}i)] = $

(At this step, I'm not sure if switching the order of the limits is allowed; does my proof break down?) :

$ \displaystyle \lim \limits_{n \rightarrow \infty}\hspace{2mm}\lim \limits_{h \rightarrow 0}\sum \limits_{i = 1}^n\frac{1}{n} f(x+\frac{h}{n}i)]=$

$\displaystyle \lim \limits_{n \rightarrow \infty}\hspace{2mm}\sum \limits_{i = 1}^n\frac{1}{n} f(x)=$


If the limit switching step (or any other) is incorrect, could you explain to me why? Thanks.

  • $\begingroup$ Is switching the order of th elimits allowed? Not in general, maybe you should make use of a property of $f$ that belongs to the conditions for the theorem and that youforgot ot mention. $\endgroup$ – Hagen von Eitzen Feb 14 '14 at 6:00
  • 1
    $\begingroup$ I don't see how you justify the limit exchange. You need to assume $f$ is continuous or something. It is good that you are trying a different approach. $\endgroup$ – copper.hat Feb 14 '14 at 6:02

Your Answer

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

Browse other questions tagged or ask your own question.