Use a power series to approximate the definite integral to six decimal places.
$\int_{0}^{0.3} \frac{x^2}{1+x^4}$
I'm not sure how to find the sum of this for solving when it has x's in the numerator, this is what I assumed however.
$\frac{1}{1-x} $ $\sum_{n=0}^\infty $ x^n
$\sum_{n=0}^\infty \frac{1}{1-x^4}$
$\sum_{n=0}^\infty {(-x^4)}^n$ ==> $\frac{x}{1+x^4}$
I'm not sure if this is right nor what the next step is to get the approximate definite integral.
Best Answer
$$ I = \int_{0}^{\frac{3}{10}}\frac{x^2}{1+x^4}\,dx = \int_{0}^{\frac{3}{10}}\frac{x^2-x^6}{1-x^8}\,dx = \sum_{n\geq 0}\left(\frac{\left(\frac{3}{10}\right)^{8n+3}}{8n+3}-\frac{\left(\frac{3}{10}\right)^{8n+7}}{8n+7}\right) \tag{1}$$ and the last series is a series with positive terms. Since $$ \sum_{n\geq 2}\left(\frac{\left(\frac{3}{10}\right)^{8n+3}}{8n+3}-\frac{\left(\frac{3}{10}\right)^{8n+7}}{8n+7}\right)<\frac{1}{19}\sum_{n\geq 2}\left(\frac{3}{10}\right)^{8n+3}<10^{-11} \tag{2}$$ the first eight figures of $I$ are given by the sum appearing in the RHS of $(1)$ restricted to $n=0$ and $n=1$: $$ \sum_{n=0}^{1}\left(\frac{\left(\frac{3}{10}\right)^{8n+3}}{8n+3}-\frac{\left(\frac{3}{10}\right)^{8n+7}}{8n+7}\right)=\frac{3453033133161387}{385000000000000000}=\color{green}{0.00896891}72289906\ldots\tag{3}$$