Question
The area (in sq. units) of the region $$A = \left\{ {\left( {x,\,y} \right):{x^2} \leqslant y \leqslant x + 2} \right\}$$ is :
A.
$$\frac{{10}}{3}$$
B.
$$\frac{{9}}{2}$$
C.
$$\frac{{31}}{6}$$
D.
$$\frac{{13}}{6}$$
Answer :
$$\frac{{9}}{2}$$
Solution :
Required area is equal to the area under the curves $$y \geqslant {x^2}$$ and $$yd''x + 2$$
$$\eqalign{ & \therefore {\text{Required area :}} \cr & {\text{ = }}\int\limits_{ - 1}^2 {\left( {\left( {x + 2} \right) - {x^2}} \right)dx} \cr & {\text{ = }}\left( {\frac{{{x^2}}}{2} + 2x - \frac{{{x^3}}}{3}} \right)_{ - 1}^2 \cr & {\text{ = }}\left( {2 + 4 - \frac{8}{3}} \right) - \left( { + \frac{1}{2} - 2 + \frac{1}{3}} \right) \cr & {\text{ = }}\frac{9}{2} \cr} $$
Required area is equal to the area under the curves $$y \geqslant {x^2}$$ and $$yd''x + 2$$
$$\eqalign{ & \therefore {\text{Required area :}} \cr & {\text{ = }}\int\limits_{ - 1}^2 {\left( {\left( {x + 2} \right) - {x^2}} \right)dx} \cr & {\text{ = }}\left( {\frac{{{x^2}}}{2} + 2x - \frac{{{x^3}}}{3}} \right)_{ - 1}^2 \cr & {\text{ = }}\left( {2 + 4 - \frac{8}{3}} \right) - \left( { + \frac{1}{2} - 2 + \frac{1}{3}} \right) \cr & {\text{ = }}\frac{9}{2} \cr} $$