Science:Math Exam Resources/Courses/MATH101/April 2014/Question 01 (e)

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MATH101 April 2014

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Question 01 (e)

Short Problem. Show your work. No credit will be given for the answer without the correct accompanying work. Consider the Trapezoid Rule for making numerical approximations to $\int _{a}^{b}f(x)dx$. The error for the Trapezoid Rule satisfies $|E_{T}|\leq {\frac {K(b-a)^{3}}{12n^{2}}}$, where $|f''(x)|\leq K$ for $a\leq x\leq b$. If $-2<f''(x)<0$ for $1\leq x\leq 4$, find a value of $n$ to guarantee the Trapezoid Rule will give an approximation for $\int _{1}^{4}f(x)dx$ with absolute error, $|E_{T}|$, less than $0.001$.

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Hint
What should $a$, $b$, and $K$ be in the Trapezoid Rule?

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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. First, we have $-2<f''(x)<0$, so $|f''(x)|<2$ and thus $K=2$ in the Trapezoid rule formula. Also, from the integral bounds, $1\leq x\leq 4$, and so $a=1$, and $b=4$ in the formula. Thus, using the Trapezoid Rule with these values, ${\begin{aligned}|E_{T}|\leq {\frac {2(4-1)^{3}}{12n^{2}}}={\frac {2\cdot 27}{12n^{2}}}={\frac {9}{2n^{2}}}.\end{aligned}}$ Now, find $n$ such that ${\begin{aligned}|E_{T}|={\frac {9}{2n^{2}}}<10^{-3}.\end{aligned}}$ This implies ${\begin{aligned}{\frac {2n^{2}}{9}}>10^{3}\implies n^{2}>{\frac {9\cdot 10^{3}}{2}}=4500.\end{aligned}}$ Thus, $n>67$ works.

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