Question

1. -/1 pointsfind a power series representation for the function. (give your power series representation centered at ( x = 0 ).)( g(x)=\frac{x^{4}}{1+x^{5}} )( g(x)=sum_{n = 0}^{infty}(quad) )use a power series to approximate the definite integral, ( i ), to six decimal places.( int_{0}^{0.3} \frac{x^{4}}{1+x^{5}} d x )( i= )

Explanation:

Step1: Recall the geometric series formula

The geometric series formula is \(\sum_{n = 0}^{\infty}r^{n}=\frac{1}{1 - r}\), for \(|r|\lt1\).
For the function \(g(x)=\frac{x^{4}}{1 + x^{5}}\), we can rewrite it as \(x^{4}\cdot\frac{1}{1-(-x^{5})}\).
Let \(r=-x^{5}\), then \(\frac{1}{1-(-x^{5})}=\sum_{n = 0}^{\infty}(-x^{5})^{n}=\sum_{n = 0}^{\infty}(- 1)^{n}x^{5n}\), for \(|x^{5}|\lt1\) (i.e., \(|x|\lt1\)).
So \(g(x)=x^{4}\sum_{n = 0}^{\infty}(-1)^{n}x^{5n}=\sum_{n = 0}^{\infty}(-1)^{n}x^{5n + 4}\).

Step2: Integrate the power - series term - by - term

We know that \(\int_{0}^{0.3}\frac{x^{4}}{1 + x^{5}}dx=\int_{0}^{0.3}\sum_{n = 0}^{\infty}(-1)^{n}x^{5n+4}dx\).
By the power - rule for integration \(\int x^{m}dx=\frac{x^{m + 1}}{m+1}+C(m
eq - 1)\), we have \(\int_{0}^{0.3}\sum_{n = 0}^{\infty}(-1)^{n}x^{5n+4}dx=\sum_{n = 0}^{\infty}\frac{(-1)^{n}}{5n + 5}x^{5n+5}\big|_{0}^{0.3}=\sum_{n = 0}^{\infty}\frac{(-1)^{n}(0.3)^{5n + 5}}{5n+5}\).

Step3: Calculate the sum of the series to six - decimal places

When \(n = 0\): \(\frac{(-1)^{0}(0.3)^{5}}{5}=\frac{(0.3)^{5}}{5}=\frac{0.00243}{5}=0.000486\).
When \(n = 1\): \(\frac{(-1)^{1}(0.3)^{10}}{10}\approx\frac{-5.9049\times10^{-6}}{10}=- 5.9049\times10^{-7}\).
When \(n\geq2\), the terms are even smaller.
\(\int_{0}^{0.3}\frac{x^{4}}{1 + x^{5}}dx\approx0.000486-0.00000059049\approx0.000485\).

Answer:

The power - series representation of \(g(x)\) is \(\sum_{n = 0}^{\infty}(-1)^{n}x^{5n + 4}\) and \(I\approx0.000485\).