Science:Math Exam Resources/Courses/MATH101/April 2017/Question 03 (a)

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MATH101 April 2017

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Question 03 (a)
Find the area of the region between the curves $y=x-8$ and $y=-x^{2}+7x-16$ . A calculator-ready answer is sufficient.

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?

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Hint
Find the points of intersection of the two graphs to determine the integration limits for the area.

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Solution
Consider the two curves $y=x-8$ and $y=-x^{2}+7x-16$ . Note that $y=x-8$ is a line, and $y=-x^{2}+7x-16$ is a parabola whose leading coefficient is negative (so it is shaped like an up-side-down U). First, we find the intersection points of the two graphs. By solving $x-8=-x^{2}+7x-16\iff x^{2}-6x+8=0\iff (x-4)(x-2)=0,$ the intersection points are obtained at $x=2$ and $x=4$ . Then, the area we are looking for is $\int _{2}^{4}\|x-8-(-x^{2}+7x-16)\|\,dx$ . Either drawing the graphs or plugging a number in the interval $(2,4)$ , we can see that $-x^{2}+7x-16\geq x-8$ when $2\leq x\leq 4$ . Therefore, we have $\|x-8-(-x^{2}+7x-16)\|=-x^{2}+7x-16-(x-8)=-x^{2}+6x-8$ on $(2,4)$ and plug this into integral to have ${\begin{aligned}\int _{2}^{4}\|x-8-(-x^{2}+7x-16)\|\,dx&=\int _{2}^{4}-x^{2}+6x-8dx=-{\frac {1}{3}}x^{3}+3x^{2}-8x{\bigg \|}_{2}^{4}\\&=\left(-{\frac {1}{3}}\cdot 4^{3}+3\cdot 4^{2}-8\cdot 4\right)-\left(-{\frac {1}{3}}\cdot 2^{3}+3\cdot 2^{2}-8\cdot 2\right)=-{\frac {16}{3}}+{\frac {20}{3}}={\frac {4}{3}}\end{aligned}}$ Here, the Power Rule of differentiation is used. Therefore, the area between the two graphs is ${\frac {4}{3}}$ Answer: $\color {blue}{\frac {4}{3}}$