Prove integral of sequence of function uniformly converges

Question

Show that if $f_n \to f$ uniformly on $[a,b]$ and $f_n$ is integrable for each n then $\int_{a}^{x}f_n(t)dt\to \int_{a}^{x}f(t)dt$ uniformly in $x$ on [a,b].

I know how to prove the question $\int_{a}^{b}f_n(x)dx\to \int_{a}^{b}f(x)dx$ uniformly in $x$ on [a,b].

I want to make sure if it is same thing?

Answer 1

Nearly so. Did you understand the proof you have pasted above? $$ \left|\int_a^x (f_n(t) - f(t))\,dx\right| \leq \int_a^x |f_n(t) - f(t)| \,dt \leq \int_a^b \epsilon/(b-a) \,dt < \epsilon,$$ for all $n$ sufficiently large. The last inequality follows from the fact that $|f_n(t) - f(t)|$ is nonnegative.

(I've left out obvious details that are already present in what you've pasted above. )

Answer 2

"I know how to prove the question $\int_{a}^{b}f_n(x)dx\to \int_{a}^{b}f(x)dx$ uniformly in $x$ on $[a,b]$."
This is a sequence of numbers so we can not talk about uniform continuity. I think you can use the same method, just use that $x-a\le b-a$.