Science:Math Exam Resources/Courses/MATH101/April 2011/Question 06 (b)/Solution 1

We saw from part (a) that we have the series expansion

${\frac {1}{1+x^{3}}}=1-x^{3}+x^{6}-x^{9}+\ldots$

and so we can write that

${\begin{aligned}\int _{0}^{0.1}{\frac {1}{1+x^{3}}}\,dx&=\int _{0}^{0.1}dx-\int _{0}^{0.1}x^{3}\,dx+\int _{0}^{0.1}x^{6}\,dx-\int _{0}^{0.1}x^{9}\,dx+\ldots \\&=0.1-{\frac {(0.1)^{4}}{4}}+{\frac {(0.1)^{7}}{7}}-{\frac {(0.1)^{10}}{10}}+\ldots \end{aligned}}$

where the $(n+1)$^st term is

${\frac {(-1)^{n}(0.1)^{3n+1}}{3n+1}}$

Since this is an alternating sequence which converges to zero, we have that the error when neglecting everything past the $(n+1)$^st term. Thus

${\begin{aligned}|{\textrm {error}}|&\leq \left|{\frac {(-1)^{n}(0.1)^{3n+1}}{3n+1}}\right|\\&={\frac {0.1^{3n+1}}{3n+1}}\\&={\frac {10^{-(3n+1)}}{3n+1}}\end{aligned}}$

if we want this to be less than $10^{-9}$, then we can choose $n=3$, where the error will be less than

${\frac {10^{-(9+1)}}{9+1}}=10^{-11}$

which works fine since it is clearly even less than $10^{-9}$.