Science:Math Exam Resources/Courses/MATH101/April 2017/Question 04 (b)/Solution 1

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)