Question
Area of the region $$\left\{ {\left( {x,\,y} \right) \in {{R}^2}\,:\,y \geqslant \sqrt {\left| {x + 3} \right|} ,\,5y \leqslant x + 9 \leqslant 15} \right\}$$ is equal to :
A.
$$\frac{1}{6}$$
B.
$$\frac{4}{3}$$
C.
$$\frac{3}{2}$$
D.
$$\frac{5}{3}$$
Answer :
$$\frac{3}{2}$$
Solution :
Area $$ABE$$ (under parabola) $$ = \int_{ - 4}^{ - 3} {\sqrt { - x - 3} dx} = \frac{2}{3}$$
Area $$BCD$$ (under parabola) $$ = \int_{ - 3}^1 {\sqrt {x + 3} dx} = \frac{{16}}{3}$$
Area of trapezium $$ACDE = \frac{1}{2}\left( {1 + 2} \right)5 = \frac{{15}}{2}$$
Required area $$ = \frac{{15}}{2} - \frac{{16}}{3} - \frac{2}{3} = \frac{3}{2}$$
Area $$ABE$$ (under parabola) $$ = \int_{ - 4}^{ - 3} {\sqrt { - x - 3} dx} = \frac{2}{3}$$
Area $$BCD$$ (under parabola) $$ = \int_{ - 3}^1 {\sqrt {x + 3} dx} = \frac{{16}}{3}$$
Area of trapezium $$ACDE = \frac{1}{2}\left( {1 + 2} \right)5 = \frac{{15}}{2}$$
Required area $$ = \frac{{15}}{2} - \frac{{16}}{3} - \frac{2}{3} = \frac{3}{2}$$