Science:Math Exam Resources/Courses/MATH101/April 2013/Question 10 (b)/Solution 1

By part (a),

$\begin{array}{rl}\underset{0}{\overset{1/4}{\int }}\frac{1}{1+x^3}dx& ={[\left(\sum _{n=0}^{\mathrm{\infty }}\frac{(-1{)}^n}{3n+1}{x}^{3n+1}\right)+C]|}_0^{1/4}\\ & =\sum _{n=0}^{\mathrm{\infty }}\frac{(-1{)}^n}{3n+1}{\left(\frac{1}{4}\right)}^{3n+1}\end{array}$

This is an alternating sum with

$b_n=\frac{1}{3n+1}{\left(\frac{1}{4}\right)}^{3n+1}.$

Clearly $b_{n+1}\le b_n$ and $\underset{n\to \mathrm{\infty }}{\lim }{b}_n=0$. Then by Alternating Series Remainder Theorem, the error in the approximation is bounded as follows:

$|\text{error}|\le b_{n+1}$

Then we are guaranteed the error is less than $10^{-5}$ if $b_{n+1}\le 10^{-5}$, that is, we must solve

$\frac{1}{3(n+1)+1}{\left(\frac{1}{4}\right)}^{3(n+1)+1}\le 10^{-5}=\frac{1}{10^5}.$

For $n+1=1$,

$\frac{1}{3+1}{\left(\frac{1}{4}\right)}^{3+1}=\frac{1}{4^5}>\frac{1}{10^5}$

so we do not have enough terms yet. For $n+1=2$,

$\frac{1}{3(2)+1}{\left(\frac{1}{4}\right)}^{3(2)+1}\le \frac{1}{7}\frac{1}{16000}=\frac{1}{112000}<\frac{1}{10^5}$

where we have used $4^7=2^{14}=2^{10}\cdot 16=1024\cdot 16\ge 1000\cdot 16=16000$ to approximate. Thus, $n+1=2$ is sufficient. Since the series starts at 0, we need 2 terms (the $n=0$ and $n=1$ term, specifically).