$\int_a^b\sum_{n=1}^\infty f_n(x) dx=\sum_{n=1}^\infty \int_a^b f_n(x) dx$ $\{f_n\}$ is a sequence of continuous (so also integrable) real functions on $[a,b]$ that is uniformly convergent and uniformly equicontinuous. I want to show $$\int_a^b\sum_{n=1}^\infty \frac{f_n(x)}{{\color{red}{n^2}}} dx=\sum_{n=1}^\infty \frac{1}{{\color{red}{n^2}}}\int_a^b f_n(x) dx,$$
and I cannot figure out how to do this. I think to take the limit as $n\to\infty$ and use uniform convergence to the limit $f$ which will somehow allow the sum and integrals to switch, but I cannot figure out how to do this. Help would be greatly appreciated.
Edit: In red.

Let $f(x) = \sum\limits_{n=1}^\infty f_n(x)$. We need $$\int_a^b f(x)dx = \lim\limits_{N \to \infty} \int_a^b \sum\limits_{n=1}^N f_n(x)dx$$ Fix $\epsilon>0$, for $N$ big enough, $|f(x)-\sum\limits_{n=1}^N f_n(x)|\leq\epsilon$. Therefore $$|\int_a^b f(x)dx -\int_a^b \sum\limits_{n=1}^N f_n(x)dx| \leq \int_a^b |f(x) - \sum\limits_{n=1}^N f_n(x)|dx \leq \epsilon (b-a)$$
