# Use a power series to approximate the definite integral to six decimal places.

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.

• This is definitely in the right direction. Then multiply the last fraction by $x$ to make it equal to the integrand. – Simply Beautiful Art Oct 26 '16 at 20:04
• so it becomes x^2/1+x^4? isn't this back in the same direction? how do i find the intergrand – Megan Byers Oct 26 '16 at 20:14
• To be more clear, $$\frac1{1+x^4}=\sum_{n=0}^\infty(-x^4)^n$$ $$\implies\frac{x^2}{1+x^4}=x^2\sum_{n=0}^\infty(-x^4)^n$$ and expand the sum as many digits as you need. – Simply Beautiful Art Oct 26 '16 at 21:10

$$I = \int_{0}^{\frac{3}{10}}\frac{x^2}{1+x^4}\,dx = \int_{0}^{\frac{3}{10}}\frac{x^2-x^6}{1-x^8}\,dx = \sum_{n\geq 0}\left(\frac{\left(\frac{3}{10}\right)^{8n+3}}{8n+3}-\frac{\left(\frac{3}{10}\right)^{8n+7}}{8n+7}\right) \tag{1}$$ 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 $n=1$: $$\sum_{n=0}^{1}\left(\frac{\left(\frac{3}{10}\right)^{8n+3}}{8n+3}-\frac{\left(\frac{3}{10}\right)^{8n+7}}{8n+7}\right)=\frac{3453033133161387}{385000000000000000}=\color{green}{0.00896891}72289906\ldots\tag{3}$$
That's the best you can do when $\ds{\texttt{mp} \sim 10^{-16}}$.