Question

The area (in sq. units) of the region $A=\left\{\left(x,y\right):\frac{y^2}{2}\le x\le y+4\right\}$ is :

• A. $\frac{53}{3}$
• B. 18
• C. 30
• D. 16

Answer 1

Answer: (B)

$y^2=2x$
$x-y-4=0$
${\left(x-4\right)}^2=2x$
$x^2+16-8x-2x=0$
$x^2-10x+16=0$
$x=8,2$
$y=4,-2$
$A={\int }_{-2}^4\left(y+4-\frac{y^2}{2}\right)dy$
$=\frac{y^2}{2}{|}_{-2}^4+4y{|}_{-2}^4-\frac{y^3}{6}{|}_{-2}^4$
$=(8-2)+4(6)-\frac{1}{6}(64+8)$
$6+24-12=18$