How would you evaluate $\int_{0}^{0.4} \sqrt{1+x^4} dx$ using the functions power series.
I have figured out that the power series for $\sqrt{1+x^4}= 1 +\sum_{n=1}^{\infty} \frac{(2n-2)!x^{4n}}{2^{2n-1}(n-1)!n!}$ but how would i differentiate this considering I have the trailing $1$ as part of the series
On
We have
$$\begin{eqnarray*} \int_{0}^{2/5}\sqrt{1+x^4}\,dx &=&\frac{2}{5}\sqrt{1+\left(\frac{2}{5}\right)^4}-\int_{0}^{2/5}\frac{2x^4}{\sqrt{1+x^4}}\,dx\\&=&\frac{2}{5}\sqrt{1+\left(\frac{2}{5}\right)^4}-\sum_{n\geq 0}\frac{(-1)^n\binom{2n}{n}}{4^n(4n+5)}\left(\frac{2}{5}\right)^{4n+5}\end{eqnarray*}$$
where the last series has terms with alternating signs, behaving like $\frac{C}{n^{3/2}}\left(\frac{16}{625}\right)^{n}$ in absolute value.
This ensures a very fast convergence. The initial step of integration by parts is performed in order to boost such convergence speed.
After proving the uniform convergence of the series in $[0,0.4]$ (less required, actually): $$\int_0^{0.4}\sqrt{1 + x^4}\,dx = \int_0^{0.4}1\,dx + \sum_{n=1}^{\infty}\int_0^{0.4}\frac{(2n-2)!x^{4n}}{2^{2n-1}(n-1)!n!}\,dx. $$