Suppose that the domain of convergence of the power series $\sum_{k=0}^{\infty} c_{k}x^{k}$ contains the interval $(-r, r)$. Define $$f(x) = \sum_{k=0}^{\infty} c_{k}x^{k} \hspace{1cm} \text{ > if } |x| < r. $$
Let $[a, b] \subseteq (-r, r).$ Prove that
$$\int_{a}^{b} f(x) \mathop{dx} = \sum_{k = 0}^{\infty} \frac{c_{k}}{k + 1}\left(b^{k + 1} - a^{k + 1}\right).$$
Here's the solution I have. It might be wrong because it's not official.
Recall Theorem $5$, which states that if a sequence of integrable functions $\{f_{n} : [a, b] \rightarrow \mathbb{R}\}$ converges uniformly to the function $f : [a, b] \rightarrow \mathbb{R}$, then the limit function is also integrable.
So,
$$\int_{a}^{b} f(x) \mathop{dx} = \lim_{n\to\infty} \int_{a}^{b} \sum_{k = 0}^{n} c_{k}x^{k} = \lim_{n\to\infty} \sum_{k=0}^{n}\int_{a}^{b} c_{k}x^{k} \mathop{dx} = \lim_{n\to\infty} \sum_{k=0}^{n} \left(\frac{c_{k}}{k + 1}\right)\left(b^{k + 1} - a^{k + 1}\right) $$
$$= \sum_{k=0}^{\infty} \left(\frac{c_{k}}{k + 1}\right)\left(b^{k + 1} - a^{k + 1}\right), $$
which is what we wanted to show. (Switching integral/summation is justified by Fubini's Theorem).
My misunderstanding comes from them citing Theorem $5$. Why is that Theorem necessary here?
Notice that $\forall x \in (-r,r),$
$$f(x) = \sum_{k = 0}^{\infty} c_kx^k = \lim_{n \to \infty} \sum_{k = 0}^{n} c_kx^k.$$
Hence
$$\int_{a}^{b} f(x) dx = \int_{a}^{b} \lim_{n \to \infty} \sum_{k = 0}^{n}c_kx^k dx.$$
Now apply Theorem $5$ to pull out the limit.