Science:Math Exam Resources/Courses/MATH101/April 2016/Question 09 (b)/Solution 1

Here $f(x)=(x-3)^{5}$, the derivatives of which are

${\begin{aligned}f'(x)&=5(x-3)^{4}&f''(x)&=20(x-3)^{3}\\f^{(3)}(x)&=60(x-3)^{2}&f^{(4)}(x)&=120(x-3).\end{aligned}}$

For $0\leq x\leq 2$, the functions $|f^{''}(x)|$ and $|f^{(4)}(x)|$ attain their maximum values at $x=0$. Thus,

$K=|f^{''}(0)|=20\cdot 3^{3}=540$ ,

and

$L=|f^{(4)}(0)|=120\cdot 3=360$.

Consequently, for the Midpoint Rule with $n=100$:

$|E_{M}|\leq {\frac {K(b-a)^{3}}{24n^{2}}}={\frac {540\times 2^{3}}{24\times 10^{4}}}={\frac {180}{10^{4}}},$

whereas for Simpson’s Rule with $n=10$:

$|E_{S}|\leq {\frac {360\times 2^{5}}{180\times 10^{4}}}={\frac {64}{10^{4}}}.$

Therefore, the fact that $64<180$ implies that Simpson’s Rule results in a smaller error bound.