Science:Math Exam Resources/Courses/MATH105/April 2017/Question 04 (a)

MATH105 April 2017

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Question 04 (a)
If we use Simpson’s Rule to approximate the integral $\int _{1}^{2}x(1-\ln x)dx$, how large should $n$ be so that the error is not larger than ${\frac {1}{900000}}$?

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
For any function with continuous fourth derivative, the error in the Simpson's rule approximation is bounded above by ${\frac {K(b-a)}{180}}(\Delta x)^{4}$ where K is a constant such that $|f^{(4)}(x)|\leq K$ on the interval $[a,b].$ Make sure you know how to apply this formula. This formula is given on the formula sheet provided with this exam.

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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. For any function with continuous fourth derivative, the error in the Simpson's rule approximation is bounded above by ${\frac {K(b-a)}{180}}(\Delta x)^{4}$ where K is a constant such that $|f^{(4)}(x)|\leq K$ on the interval $[a,b].$ We need to determine an appropriate constant K. To do this we need to compute the fourth derivative of the integrand $x(1-\ln x).$ Letting $f(x)=x(1-\ln x)$, we have ${\begin{aligned}f'(x)&=1(1-\ln x)-x\cdot {\frac {1}{x}}=-\ln x,\\f''(x)&=-{\frac {1}{x}}\\f'''(x)&={\frac {1}{x^{2}}}\\f^{(4)}(x)&=-{\frac {2}{x^{3}}}\end{aligned}}$. On the interval [1,2], this fourth derivative has absolute value bounded above by 2. So we can select K = 2. Since b = 2 and a = 1, we get the error estimate $E_{s}\leq {\frac {2(2-1)}{180}}(\Delta x)^{4}\leq {\frac {1}{90}}(\Delta x)^{4}.$ We would like this error to be bounded above by ${\frac {1}{900000}}.$ This will happen if $(\Delta x)^{4}\leq {\frac {90}{900000}}={\frac {1}{10000}}$. We take the fourth root of both sides: $\Delta x\leq {\frac {1}{10}}$. So the size of our partitions must be smaller than 1/10. Because the interval over which we are integrating has length 1, a value of $\Delta x={\frac {1}{10}}$ corresponds to N = 10, so the error will be bounded above by ${\frac {1}{900000}}$ provided that $N\geq 10$. Answer: $\color {blue}10$