I gave a two liner proof on a test for this as follows:
Proof: By the Fundamental Theorem of Calculus (First Form) $F'(x)=f(x)$, thus $F$ is differentiable. By an earlier Theorem, $F$ is continuous.
I looked back after submitting, and the earlier theorem states:
Theorem: Let $f:(a,b)\to\mathbb{R}$ and let $c\in(a,b)$. If $f$ is differentiable at $c$ then $f$ is continuous at $c$.
Now I am pretty sure I messed this up because the bounds of the theorem are open, not closed. What should I have done to prove this correctly?
UPDATE: I received my exam back, and missed five points on this question (only 5 points that I missed) here is the feedback my professor gave:
Excellent!
The only problem is:
Problem 6: 5. The First Fundamental Theorem of Calculus only works is f is continuous, not if f is merely integrable. It is not, in general, true that F is differentiable for an integrable f.
Best Answer
I suspect that whoever asked you this question expected this answer: since $f$ is Riemann integrable, then, in particular, it is bounded. Take $M>0$ such that $(\forall t\in[a,b]):|f(t)|<M$. If $\varepsilon>0$, take $\delta=\frac\varepsilon M$. Then, if $x_0\in[a,b]$ and $x\in[a,b]$ is such that $|x-x_0|<\delta$, the$$|F(x)-F(x_0)|=\left|\int_{x_0}^xf(t)\,\mathrm dt\right|\leqslant M.|x-x_0|<\delta|x-x_0|<\varepsilon.$$
However, what you did is not silly. For any interval (with more than one point) $I$, every differentiable function from $I$ to $\Bbb R$ is continuous.