Science:Math Exam Resources/Courses/MATH101/April 2008/Question 02 (a)/Solution 1

A picture is included below.

First, we try to find the points of intersection. To do this, make the two curves equal so

$\displaystyle 2x^{2}=4+x^{2}$

and solving for x gives

$\displaystyle x=\pm 2.$

Now, from the picture (or by plugging in $x=0$ into both curves), it is clear that $4+x^{2}$ is the top most curve. Hence, we need to compute

$A=\int _{-2}^{2}(4+x^{2}-2x^{2})\,dx=\int _{-2}^{2}(4-x^{2})\,dx=2\int _{0}^{2}(4-x^{2})\,dx$

where the last equality holds above since the function is an even function over a symmetric interval. Thus,

$A=2\int _{0}^{2}(4-x^{2})\,dx=\left.(8x-{\frac {2x^{3}}{3}})\right|_{0}^{2}=16-{\frac {16}{3}}={\frac {32}{3}}$

completing the question.