Is $\int_a^b \frac{f(x)}{g(x)}g(x)\,dx$ improper integral when $g(a)=0$ ?
Or is it same with $\int_a^b f(x)\,dx$ ?
I think it is improper integral, so
$\int_a^b \frac{f(x)}{g(x)}g(x)\,dx = \lim\limits_{t \to a+}\int_t^b f(x)\,dx$
But if $f(x)$ is continuous at $[a,b]$
$g(t) =\int_t^bf(x)\,dx$ is continous because
$g(t) =\int_t^bf(x)\,dx=-\int_b^tf(x)\,dx$
$g'(t) = -f(t)$
$f(t)$ is continuous -> $g'(t)$ continuous -> $g(t)$ is continous
so $\lim\limits_{t \to a+}\int_t^b f(x)\,dx=\lim\limits_{t \to a+}g(t)=g(a)=\int_a^b f(x)\,dx$
Is this correct? how about when $f(x)$ is not continuous at $[a,b]$ ?