Help with the fundamental theorem of calculus

Question

In my single variable calculus book the derivation of the fundamental theorem of calculus only covered a special case and I would like help with generalisering the proof.

In my book the following was proven: If a function $f$ is continuous and differentiable on $(a, b)$ then it may be integrated over $[\alpha, \beta]$ satisfying $[\alpha, \beta] \subset (a,b)$ and its value is given by $F(\beta) - F(\alpha)$.

However, I believe the general case allows for $f$ to be integrated over $[a, b]$ given that $f$ is also continuous on $[a, b]$. This case was not proven. So I suppose that if i let $\alpha \to a^+$ and $\beta \to b^-$ and if I somehow show that the continuity of $f$ implies the continuity of $F$ on $[a, b]$ then perhaps that is enough to show that the integral of $f$ over $[a, b]$ is given by $F(b) - F(a)$ as well. How do I prove this?

Answer 1

Consider the counterexample $f(x) = \frac{1}{x}$, letting the interval be $(0,1)$. Certainly, $f$ is continuous and differentiable on that interval. And, indeed, if $\alpha = \epsilon > 0 = a$ and $\beta > \alpha$, we get its value as promised.

The problem is that $f$ is not defined on that closed interval $[0,1]$. In fact, for $f$ to be continuous, you need it to be defined on an open interval around $0$ (hence, the limit).

So, you could still sort-of apply that limit rule if you can prove $f$ is continuous and differentiable on $(a-\epsilon, b + \epsilon)$ for all reasonably small $\epsilon > 0$. Then you can then have your closed interval $[a, b]$ for your integral.