Science:Math Exam Resources/Courses/MATH101/April 2005/Question 02 (a)/Solution 1

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.