Proof check that $\int\limits_0^1\left(\sum\limits_{k=0}^\infty a_kx^k\right)dx=\sum\limits_{k=0}^\infty\frac{a_k}{k+1}$ Prove the following:  If $\sum |a_{k}| &lt; \infty$, prove that $\int\limits_{0}^{1} \left( \sum\limits_{k=0}^{\infty} a_{k}x^{k} \right)dx=\sum\limits_{k=0}^{\infty}\dfrac{a_{k}}{k+1}$.  

$\textbf{My attempt}$: Fix $x \in [0,1]$. By the Weierstrass M-Test consider $\sum\limits M_{k}=|a_{k}| &lt; \infty$; then it can be seen that $|a_{k}x^{k}| \leq M_{k}$,  $\forall x \in [0,1]$ and $k \in \mathbb{N}$. Therefore it follows that $\sum\limits_{k=0}^{\infty} a_{k}x^{k}$ converges uniformly on $[0,1]$. 

Let $f(x)=\sum\limits_{k=0}^{\infty} a_{k}x^{k}=\sum\limits_{k=0}^{\infty} f_{k}(x)$. Since the sequence of functions converges uniformly, then $f_{k} \in \mathscr{R}[0,1], \forall k \in \mathbb{N}$.  

Therefore, $f(x) \in \mathscr{R}[0,1]$ as well with the following:
\begin{align*}
\int_{0}^{1}f(x)dx &amp;= \sum_{k=0}^{\infty}\int_{0}^{1}f_{k}(x)dx \\ \\
&amp;= \sum_{k=0}^{\infty}\int_{0}^{1} \left[a_{k}x^{k} \right]dx \\ \\
&amp;= \sum_{k=0}^{\infty} a_{k} \int_{0}^{1} \left[x^{k} \right]dx \\ \\
&amp;= \sum_{k=0}^{\infty} \dfrac{a_{k}}{k+1}
\end{align*}
$\mathbf{QED}$  

I was wondering if my proof is correct; I was wanting to get some feedback! Prepping for a final this Thursday, and our professor kind of skimmed through this section quickly.
You are complicating too much.  We know that $ \sum a_{k}x^{k} $ is a power series and so we can integrate a power series term by term inside its interval of convergence. Since $ \sum \mid a_{k}\mid $ converges, 1 is in the interval of convergence. Then, continue...

I don't thing they want you to prove the cited theorem.
You are complicating too much.  We know that $ \sum a_{k}x^{k} $ is a power series and so we can integrate a power series term by term inside its interval of convergence. Since $ \sum \mid a_{k}\mid $ converges, 1 is in the interval of convergence. Then, continue...

I don't thing they want you to prove the cited theorem.