Question

Let $A=\left\{\left(x,y\right):y^2\le 4x,y-2x\ge -4\right\}.$ The area (in square units) of the region A is :

• A. 8
• B. 9
• C. 10
• D. 11

Answer 1

Answer: (B)

solve for $y;y^2=4x$ as $y-2x=-4$
gives $y=-2,4$
$\Rightarrow$ Area $=\underset{-2}{\overset{4}{\int }}$$\left(\frac{y+4}{2}-\frac{y^2}{4}\right)dy={\left[\frac{y^2}{4}+2y-\frac{y^3}{12}\right]}_{{}_{{}_{-2}}}^{{}^4}$
$=9$