Science:Math Exam Resources/Courses/MATH101/April 2006/Question 02 (a)/Solution 1

A picture is included below.

First, find the points of intersection.

$2-x^{2}=|x|$

Take the positive values of $x$ first and solving gives

$0=2-x^{2}-|x|=-x^{2}-x+2=-(x-1)(x+2)$

and so $x=1$ is a solution here (notice that we discard $x=-2$ since we started with only the positive x values first). Now, taking the negative x values gives

$0=2-x^{2}-|x|=-x^{2}+x+2=-(x+1)(x-2)$

and so $x=-1$ is the other solution. Now, we can draw the picture (see below). The picture clearly shows that the upper function is 2-x², and that the function $2-x^{2}-|x|$ is an even function. So we see that it suffices to compute

$A=2\int _{0}^{1}(2-x^{2}-|x|)\,dx=2\int _{0}^{1}(2-x^{2}-x)\,dx$

Evaluating gives

${\begin{aligned}A&=2\int _{0}^{1}(2-x^{2}-x)\,dx\\&=\left.2\left(2x-{\frac {x^{3}}{3}}-{\frac {x^{2}}{2}}\right)\right|_{0}^{1}\\&=2\left(2-{\frac {1}{3}}-{\frac {1}{2}}\right)\\&=3-{\frac {2}{3}}\\&={\frac {7}{3}}\end{aligned}}$

as required.