This is a limit-esque question:
\begin{align}
&\mbox{If there is a}\ \epsilon > 0\
\mbox{such that}
\left\vert\operatorname{f}\left(x\right) -\operatorname{g}\left(x\right)\right\vert \leq \epsilon\
\mbox{for every}\ x \in \left(a,b\right),
\\[2mm] &\
\mbox{then does}\
\frac{1}{b-a}\,
\left\vert\,\int_{a}^{b}
\operatorname{f}\left(x\right)\,{\rm d}x -\int_{a}^{b}\operatorname{g}\left(x\right)
\,{\rm d}x\,\right\vert \leq \epsilon
\end{align}
Note that $\operatorname{f}\left(x\right), \operatorname{g}\left(x\right)$ are functions integrable and differentiable on $\left[a,b\right]$.
I'm not sure if I'm interpreting the question correctly, but what I see is a question asking if there's a limit for $\operatorname{f}\left(x\right) -\operatorname{g}\left(x\right)$, then does the average value of the absolute value of the function also smaller than the limit $?$.
This is an especially difficult question for me since it connects limits and integrals closely together, which I've previously not seen before.
Best Answer
Hint: $$\left|\int_a^b f(x)\mathrm{d}x - \int_a^b g(x)\mathrm{d}x\right| = \left|\int_a^b f(x)-g(x)\mathrm{d}x\right| \leq \int_a^b \left|f(x)-g(x)\right| \mathrm{d}x$$ This question is unrelated to limits.