Proving of continuous function about an integral I have $f:\Bbb R \rightarrow \Bbb R$ is continuous and $a,b \in \Bbb R$, then $$\lim_{h\to0} \int_a^bf(x+h)\,dx = \int_a^bf(x)\,dx$$
I don't know why, but I can't seem to figure this one out. I'not sure by looking at it if I need to go a proof route or more of a counter example route. It's frustrating because it seems relatively elementary, but i've been staring at it for 2 hours. Please, any explanation/help would be amazing! Just now starting to go about abstract/calc algebra and it's really getting the best of me!
 A: By the definition of the limit we want to show that for any $\epsilon > 0$ there exists $\delta > 0$ such that $\lvert \int_a^b f(x)\, dx - \int_a^b f(x+h) \, dx \rvert < \epsilon$ whenever $\lvert h \rvert < \delta$. We have
$$\left|\int_a^b f(x)\, dx - \int_a^b f(x+h) \, dx \right| = \left|\int_a^b f(x) -  f(x+h) \, dx \right| \leq \int_a^b \lvert f(x) - f(x+h) \rvert \, dx.$$
Since $f$ is uniformly continuous on any closed interval, we may choose a $\delta > 0$ independently of $x$ such that $\lvert f(x) - f(x+h) \rvert < \frac{\epsilon}{b-a}$ whenever $\lvert h \rvert < \delta$. Hence
$$\int_a^b \lvert f(x) - f(x+h) \rvert \, dx \leq \int_a^b \frac{\epsilon}{b-a} \, dx = \epsilon$$
as desired.
A: Defining $f_h: \mathbb{R} \rightarrow \mathbb{R}$ by $f_h(x)=f(x+h)$, we want to prove that
$$\lim_{h \rightarrow 0}\int_a^b f_h=\int_a^b f.$$
By change of variables, we have that
$$\int_a^b f_h=\int_{a+h}^{b+h} f.$$
If $F: \mathbb{R} \rightarrow \mathbb{R}$ is $F(x)=\int_a^x f$, then
$$\int_a^b f_h=F(b+h)-F(a+h).$$
It follows from the FTC that $F$ is continuous. Therefore, 
$$\lim_{h \rightarrow 0}\int_a^b f_h=\lim_{h \rightarrow 0} \big(F(b+h)-F(a+h)\big)=F(b)-F(a)=\int_a^bf.$$

Sidenote: Note that nowhere was continuity of $f$ used. COV holds without continuity of $f$, and continuity of $F$ holds also without continuity of $f$. We only need $f$ to be Riemann-integrable. (c.f. Rudin)
