Science:Math Exam Resources/Courses/MATH101/April 2005/Question 02 (a)

MATH101 April 2005

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Question 02 (a)
Full-Solution Problem. Justify your answer and show all your work. Simplification of your answer is not required. Let $\displaystyle R$ be the finite region bounded above by the curve $\displaystyle y=4-x^{2}$ and bounded below by $\displaystyle y=2-x$. Carefully sketch $\displaystyle R$ and find its area explicitly.

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
There are two key things to figure out. One is the points of intersection and the other is which curve is larger on the region of integration.

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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. A picture can be found below. First, we find the points of intersection. Set the curves equal to each other so that $\displaystyle 4-x^{2}=2-x$ Then isolating to one side gives $\displaystyle 0=x^{2}-x-2=(x+1)(x-2)$ and so the x coordinates of the points of intersection are $x=-1,2$. Now, plugging in the point $x=0$ into both curves shows that $4-x^{2}$ is the upper most curve. Hence, the area we seek is ${\begin{aligned}A&=\int _{-1}^{2}(4-x^{2}-(2-x))\,dx\\&=\int _{-1}^{2}(-x^{2}+x+2)\,dx\\&=\left.\left({\frac {-x^{3}}{3}}+{\frac {x^{2}}{2}}+2x\right)\right|_{-1}^{2}\\&=\left({\frac {-8}{3}}+{\frac {4}{2}}+2(2)\right)-\left({\frac {1}{3}}+{\frac {1}{2}}+2(-1)\right)\\&={\frac {-8}{3}}+6-{\frac {1}{3}}-{\frac {1}{2}}+2\\&=-3+{\frac {15}{2}}\\&={\frac {9}{2}}\end{aligned}}$ completing the question.

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