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