Science:Math Exam Resources/Courses/MATH101/April 2017/Question 04 (b)

MATH101 April 2017

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Question 04 (b)

Let $M_{6}$ denote the Midpoint Rule approximation with $n=6$ points. Find an upper bound for the difference between the integral $\int _{0}^{\pi }x\sin xdx$ and its approximation $M_{6}$. You may use the fact that, when approximating $\int _{a}^{b}f(x)dx$ with the Midpoint Rule using $n$ points, the absolute value of the error is at most ${\frac {K(b-a)^{3}}{24n^{2}}}$, where $|f''(x)|\leq K$ for all $a\leq x\leq b$.

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
The difference between the integral and the midpoint rule of order 6 will at most be the error, given by the formula above. To bound the 2nd derivative of $f(x)=x\sin x$ on $[0,\pi ]$ use the inequalities $|x+y|\leq |x|+|y|,\,|\cos x|\leq 1,\,|\sin x|\leq 1$. Note you do not need to calculate the midpoint rule approximation.

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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. Using the information given above all we need to find is $K$ and replace the values of $a=0,\,b=\pi ,\,n=6$ in the formula for the error ${\frac {K(b-a)^{3}}{24n^{2}}}$, as the difference between the 2 numbers will, at most, be the error. Thus we must first find the 2nd derivative of $f(x)=x\sin x$: $f^{\prime }(x)=\sin x+x\cos x$ $f^{\prime \prime }(x)=\cos x+\cos x-x\sin x$. We must now bound the derivative. Using the inequatility $|x+y|\leq |x|+|y|$, we have: $|2\cos x-x\sin x|\leq 2|\cos x|+|x||\sin x|$. Furthermore we know that $|\cos x|\leq 1$, $|\sin x|\leq 1$ and, in our domain, $|x|\leq \pi$, which makes us conclude that $|f^{\prime \prime }(x)|\leq \pi +2$ on $[0,\pi ]$. (Note that this does not necessarily mean the maximum of $|f''(x)|$ is $\pi +2$. The value $\pi +2$ is only an upper bound; other bounds are possible.) Thus the difference is given by ${\frac {(\pi +2)(\pi -0)^{3}}{24(6)^{2}}}$ Answer: $\color {blue}{\frac {(\pi +2)\pi ^{3}}{24(6)^{2}}}$ (note that this is only one possible answer; see the note above)