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

MATH101 April 2016

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Question 09 (b)

Both parts of this question concern the integral $I=\int _{0}^{2}(x-3)^{5}\,dx$. (b) Which method of approximating $I$ results in a smaller error bound: the Midpoint Rule with $n=100$ intervals, or Simpson’s Rule with $n=10$ intervals? Justify your answer. You may use the formulas $|E_{M}|\leq {\frac {K(b-a)^{3}}{24n^{2}}}\qquad {\text{and}}\qquad |E_{S}|\leq {\frac {L(b-a)^{5}}{180n^{4}}},$ where $K$ is an upper bound for $|f''(x)|$ and $L$ is an upper bound for $|f^{(4)}(x)|$.

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
Find the upper bounds $K$ and $L$. Then, plug these with the given $n$ into the formulas, and compare the two error bounds.

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. 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.