Science:Math Exam Resources/Courses/MATH105/April 2012/Question 08 (f)

MATH105 April 2012

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Question 08 (f)
Short-answer questions. No credit will be given for the answer (even if it is correct) without the accompanying work. Find a bound for the error in approximating $\int _{0}^{1}\left[e^{-2x}+3x^{3}\right]\,dx$ using Simpson’s rule with n = 6 subintervals. There is no need to simplify your answer. Do not write down the Simpson’s rule approximation Sn.

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Hint
Recall that if $\left|f^{(4)}(x)\right|\leq K$ on the interval $[a,b]$, then the error in using $S_{n}$ to approximate $\int _{a}^{b}f(x)dx$ has absolute value less than or equal to $K(b-a)^{5}/180n^{4}$.

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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. The error bound for Simpson's rule requires us to find the fourth derivative of the integrand. For $f(x)=e^{-2x}+3x^{3}$, we find: $\displaystyle f'(x)=-2e^{-2x}+9x^{2}$ $\displaystyle f''(x)=4e^{-2x}+18x$ $\displaystyle f'''(x)=-8e^{-2x}+18$ and $\displaystyle f^{(4)}(x)=16e^{-2x}$ To find our K, we need to know the largest $|f^{(4)}|$ can be over the integration range $[0,1]$. $|f^{(4)}(x)|=16e^{-2x}$ is a decreasing function, with its largest value on $[0,1]$ being $16$ at $x=0$, so K = 16. The error with $n=6$ is bounded by ${\frac {K(b-a)^{5}}{180n^{4}}}={\frac {16(1-0)^{5}}{180\times 6^{4}}}={\frac {16}{180\times 6^{4}}}$. Note: the question did not ask us to evaluate the $n=6$ Simpson's rule approximation; we were only asked to bound its error.

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