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Suppose that $a(x)$ is continuous at $x=x_0$ and consider the function $$ F(x)=\int_{a}^{a(x)}f(t)\,dt $$ where $f(t)$ is continuous in the interval $[a,a(x_0)]$.

1) Is it true that $F(x)$ is continuous at $x=x_0$?

2) Is it true that if $\lim_{x\to x_0}a(x)=\infty$ and $f(t)$ is continues at $[a,\infty)$, then $$ \lim_{x\to x_0}F(x)=\int_{a}^{\infty}f(t)\,dt\ ? $$

My attempt for (1): Note that $$ |F(x)-F(x_0)|=\left|\int_{a(x_0)}^{a(x)}f(t)\,dt\right| $$ since $f$ is continuous in $[a,a(x_0)]$, there exist $c(x)\in(a(x_0),a(x))$ such that $$ |F(x)-F(x_0)|=\left|\int_{a(x_0)}^{a(x)}f(t)\,dt\right|=f(c(x))|a(x)-a(x_0)| $$ Since $f(t)$ and $a(x)$ are continuous, it follows that $\lim_{x\to x_0}c(x)=a(x_0)$, so $\lim_{x\to x_0}f(c(x))=f(a(x_0))$. Therefore $$ \lim_{x\to x_0}|F(x)-F(x_0)|=\lim_{x\to x_0}f(c(x))|a(x)-a(x_0)|=f(a(x_0))\cdot 0=0 $$ as required.

Chappers
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boaz
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    Define $g(x)=\int_a^xf(t),dt$. Then $F=g\circ a$. Now you can apply the first fundamental theorem of calculus (and a property of composition of continuous functions) to answer (1).

    For (2), assuming that $f$ is absolutely integrable on $[a,\infty)$, one has $$ \lim_{x\to x_0}F(x)=\lim_{x\to x_0}g\circ a(x)=\lim_{x\to\infty}g(x) $$ where the second equality is due to the continuity of $g$.

    The case when $f$ is not absolutely integrable on $[a,\infty)$ could be complicated and I don't know what one would like to do about that.

     –  Jun 13 '17 at 14:57

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