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

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

$\begin{array}{rlrl}{f}^{\prime }(x)& =5(x-3{)}^4& f^{″}(x)& =20(x-3{)}^3\\ f^{(3)}(x)& =60(x-3{)}^2& f^{(4)}(x)& =120(x-3).\end{array}$

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

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

and

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

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

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

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

$|E_S|\le \frac{360\times 2^5}{180\times 10^4}=\frac{64}{10^4}.$

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