Find definite integral as power series I need to find the following indefinite integral as power series and then compute the definite integral
$$\int_0^{1/2}\sqrt{1+x^3}dx$$
with the accuracy of $10^{-4}$.
My attempt:
Using the binomial theorem, we can express $\sqrt{1+x^3}$ as
$$\sum_{k=0}^\infty{\frac{1}{2}\choose k}x^{3k}.$$
Thus, we can rewrite the integral:
\begin{align}
\int_0^{1/2}\sqrt{1+x^3}dx&=\int_0^{1/2}\sum_{k=0}^\infty{\frac{1}{2}\choose k}x^{3k}\\
&=\sum_{k=0}^\infty{\frac{1}{2}\choose k}\left[\frac{x^{3k+1}}{3k+1}\right]_0^{1/2}\\
&=\sum_{k=0}^\infty{\frac{1}{2}\choose k}\frac{\left(\frac{1}{2}\right)^{3k+1}}{3k+1}
\end{align}
How can I approximate the definite integral from here?
 A: Use the well-known Taylor series for the function $\sqrt{1+x}$ and substitute $x\to x^3$ to get
$$\sqrt{1+x^3}=\sum^{\infty}_{n=0} \frac{(-1)^{n-1}(2n)!}{4^n (n!)^2 (2n-1)} x^{3n}.$$
This series converges for $\lvert x\rvert <1$. Now integrating the series yields
$$\int_0^{1/2}\sum^{\infty}_{n=0} \frac{(-1)^{n-1}(2n)!}{4^n (n!)^2 (2n-1)} x^{3n}\,\mathrm{d}x=\sum^{\infty}_{n=0} \frac{(-1)^{n-1}(2n)!}{4^n (n!)^2 (2n-1)}\frac{2^{-3n - 1}}{3n + 1}.$$

Theorem (Alternating series approximation). Let $S=\sum\limits_{n=0}^\infty a_n$ be an alternating series where $\vert a_{n+1} \vert < \vert a_n \vert$ and $\displaystyle\lim_{n \rightarrow \infty}a_n=0$. Let $S_n$ be the $n$th partial sum of the series. Then $\vert S-S_n \vert \leq \vert a_{n+1} \vert$.

We want to find the biggest $a_{n+1}$ which is smaller than $10^{-4}$. This is the case for $n=2$, so
$$\lvert \sum^{\infty}_{n=0} \frac{(-1)^{n-1}(2n)!}{4^n (n!)^2 (2n-1)}\frac{2^{-3n - 1}}{3n + 1}-\sum^{2}_{n=0} \frac{(-1)^{n-1}(2n)!}{4^n (n!)^2 (2n-1)}\frac{2^{-3n - 1}}{3n + 1}\rvert \le \frac{1}{163840}<10^{-4}.$$
So the final answer is $$\int_0^{1/2}\sqrt{1+x^3}dx\approx \sum^{2}_{n=0} \frac{(-1)^{n-1}(2n)!}{4^n (n!)^2 (2n-1)}\frac{2^{-3n - 1}}{3n + 1}= \frac{3639}{7168}$$
where the approximation error is at most $10^{-4}$. See here for numerical verification.
