Science:Math Exam Resources/Courses/MATH101/April 2007/Question 02 (a)/Solution 1

A picture is included below.

First, we determine the points of intersection. Setting the two curves equal to each other, we have

$4-x^{2}=(x-2)^{2}=x^{2}-4x+4$

Bringing the terms to the right hand side, we have

$0=2x^{2}-4x=2x(x-2)$

and so the points of intersection are $x=0$ and $x=2$. Notice either form the picture or testing a point between 0 and 2 (say $x=1$) and plugging it into both equations, we see that the curve $4-x^{2}$ is bigger between 0 and 2. Hence, our area is

${\begin{aligned}\int _{0}^{2}(4-x^{2}-(x-2)^{2})\,dx&=\int _{0}^{2}(4-x^{2}-x^{2}+4x-4)\,dx\\&=\int _{0}^{2}(-2x^{2}+4x)\,dx\\&=\left.{\frac {-2x^{3}}{3}}+2x^{2}\right|_{0}^{2}\\&={\frac {-2(2)^{3}}{3}}+2(2)^{2}\\&={\frac {8}{3}}\end{aligned}}$

completing the question.