Use Maclaurin Series to evaluate the definite integral correct to within an error $\lt 0.0001$ Definite integral:

$$\int_0^{0.2} \dfrac{1}{1+x^5}\text{d}x$$ 

So I did series expansion of 
$$\sum_{n=0}^{\infty}\dfrac{(-1)^n\cdot x^{5n}}{5n+1}+C$$ and when I plug in $0.2$ that makes it $$\dfrac{(-1)^n\cdot 0.2^{5n}}{5n+1}$$
But from this point, I'm not sure what I'm supposed to do. I know it has something to do with alternate series error thing, but not sure where to go from here.
 A: You were on the right track, but you made a small error.  The indefinite integral is
$$\int{1\over1+x^5}dx=\sum_{n=0}^\infty{(-1)^nx^{5n+1}\over5n+1}+C$$
(You had the sum with an $x^{5n}$ instead of $x^{5n+1}$.)  Thus
$$\int_0^{0.2}{1\over1+x^5}dx=\sum_{n=0}^\infty{(-1)^n(0.2)^{5n+1}\over5n+1}={1\over5}-{1\over6\cdot5^6}+{1\over11\cdot5^{11}}-\cdots$$
Now the nice things about alternating series in which each term is smaller than the one before it (which is certainly the case here) is that no matter where you stop, the error is no greater than the next term.  So if you want the error to be less than $0.0001$, you just have to look for the first term in the series that is smaller than that.  But this happens almost immediately:
$${1\over6\cdot5^6}={1\over93750}\lt{1\over10000}=0.0001$$
So
$$\int_0^{0.2}{1\over1+x^5}dx\approx{1\over5}$$
is good to within $0.0001$.
A: I hope I am not too late to the party. Before I answer, let me give you a bit of background on what you are doing. 
The goal of your problem is to approximate a function (this time an integral) using an infinite series expansion. In most cases, you can write: 
$f(x) = \sum_0^\infty c_n (x-a)^n$. If you we are talking about a McLaurin series, $a=0$. Remember that before you use the expansion of your function, you will need to know the interval of convergence. 
If instead of adding terms from $0$ to $\infty$, you decide to do some partial addition (from the $0^\text{th}$ to the $N^\text{th}$ term), there will be an error term creeping up. Usually this is written as:
$ f(x) =\sum_0^N c_n(x-a)^n + R_N(x)$. Knowing the magnitude of $R_N(x)$ will allow you to know how close you are from the real answer. In your particular problem, you want $R_N(x) \leq 10^{-4}$. This is equivalent to saying "how many terms of the infinite series can I use to get an answer that is within $10^{-4}$ of the real answer?". 
The series expansion you have gotten for your problem is correct; we can derive an expression for the error a similar way. 
If I let $f(x)=\frac{1}{1+x^5}=\frac{1}{1-(-x^5)}=\sum_0^\infty (-1)^n x^{5n}$ or if I am doing a partial sum I can write $f(x)= \sum_0^N (-1)^n x^{5n} + R_N(x)$ then we can write
\begin{align*}
\int_0^{0.2} f(x) \ dx &=\int_0^{0.2}\sum_0^\infty (-1)^n x^{5n} \ dx = \int_0^{0.2}\sum_0^N (-1)^n x^{5n} +R_N(x)\ dx \\
&=\int_0^{0.2}\sum_0^N (-1)^n x^{5n} \ dx+\int_0^{0.2} R_N(x)\ dx\\
 \int_0^{0.2} f(x) \ dx  &=\underbrace{\sum_0^N(-1)^n\frac{x^{5n+1}}{5n+1}\Big|_0^{0.2}}_{\text{this is what you got so far}} +\underbrace{\int_0^{0.2} R_N(x)\ dx}_{\text{you want this to be $\leq 10^{-4}$}}.
\end{align*}
The way I have it written, $R_N(x)=\frac{f^{(N+1)}(\xi)x^{N+1}}{(N+1)!}$, with $0\leq \xi \leq x$. With the above, you will have an expression for the error in your integral. That error will be
\begin{align*}
\text{Error} = \Big|\int_0^{0.2} \frac{f^{(N+1)}(\xi)x^{N+1}}{(N+1)!} \ dx \Big|,
\end{align*}
where $0\leq \xi \leq x \leq 0.2$. The goal now is to find the $N$ so that the error is less than $10^{-4}$. It can be a little difficult to find a general formula for $R_N{x}$, so what you can do is to simply go through a few values of $N$ such as $0, 1, 2, 3,$ etc.. and see which will give you an Error less than $10^{-4}$. 
For example, if $N=0$, you get:
\begin{align*}
\text{Error} &= \Big|\int_0^{0.2} f^{'}(\xi) x\ dx\Big| \\
&\leq \Big|\int_0^{0.2} f^{'}(0.2) x\ dx\Big|, \text{ since $f^{'}{(x)}$ is increasing on $[0, 0.2]$}\\
&\leq\int_0^{0.2}\frac{5\times0.2^4}{(1+0.2^5)^2} x\ dx = \frac{5\times0.2^4}{(1+0.2^5)^2}\times (0.2^2)/2 \approx 0.0002.
\end{align*}
You can already see that by simply taking 1 term in the infinite series the error is already $2\times 10^{-4}$. Try going through the same derivation but for $N=1,2$ etc... Once you find the $N$ that gives you an error less or equal to $10^{-4}$, you would have answered your question.
Good luck. 
