The area bounded by the curves $y^2 = 4 + x$ and $x + 2y = 4$, is
• A
$9$
• B
$18$
• C
$72$
• D
$36$

#### Solution

Area when $x > 0 = \displaystyle 2\int_{-4}^{0}\sqrt{4+x}$
$=\left.\begin{matrix}\dfrac{4}{3}(x+4)^{\frac{3}{2}}\end{matrix}\right|_{-4}^{0}$

$=\dfrac{4}{3}\times 8-\dfrac{4}{3}\times 0$

$=\dfrac{32}{3}$
Area when $x > 0$
$\displaystyle \int_{0}^{12}(\dfrac{4-x}{2})-\int_{0}^{12}(-\sqrt{4}+x)$

$\displaystyle \int_{0}^{12}\frac{4-x}{2}+\int_{0}^{12}\sqrt{4+x}$

$=-12+\dfrac{112}{3}$

$\therefore$ Total area $=-12+\dfrac{144}{3}=36$