I need to approximate the following integral with six exact decimals using power series:
$$\int_{0}^{0.2} dx/(1+x^5) $$
What I have so far :
$$\int_{0}^{0.2} dx/(1+x^5) = (0.2)-(0.2)^6/6+(0.2)^{11}/11- ... $$
But the answer in the manual is 0,044522 so I must have done something very wrong here...
On
Consider the series $$\frac 1 {1+x}=\sum_{i=0}^\infty (-1)^n x^n$$ Let $x=t^5$ to make $$\frac 1 {1+t^5}=\sum_{i=0}^\infty (-1)^n t^{5n}$$ Integrate termwise $$\int \frac {dt} {1+t^5}=\sum_{i=0}^\infty (-1)^n \frac{t^{5n+1}}{5n+1}$$ This is an alternating series, so you need $p$ such that $$\frac{(0.2)^{5p+6}}{5p+6} \leq \epsilon$$ I do not think that you need many terms.
The power series for $f(x)=1/(x+1)$ is $$ f(x)=\frac1{x+1} = 1 - x + x^2 - x^3 + \cdots $$ and thus $$ f(x^5) = \frac1{x^5+1}=1-x^5+x^{10}-x^{15}+\cdots. $$ Therefore, the integral becomes $\int_0^{1/5}f(x^5)dx$, which is $$\begin{split} \int_0^{1/5}f(x^5)dx & = \int_0^{1/5}(1-x^5+x^{10}-x^{15}+\cdots)dx\\ & = x -\frac16x^6+\frac1{11}x^{11}-\frac1{16}x^{16}+\cdots\bigg]^{1/5}_0\\ & = \frac15-\frac16\left(\frac15\right)^6+\frac1{11}\left(\frac15\right)^{11}-\frac1{16}\left(\frac15\right)^{16}+\cdots. \end{split}$$ This is indeed the same as what you got. You are right; the answer book is wrong.