question_answer

The area bounded by \[\mathbf{y}=\mathbf{2}-{{\mathbf{x}}^{\mathbf{2}}}\]and \[\mathbf{x}+\mathbf{y}=\mathbf{0}\]is:

A)  \[\frac{9}{2}\] sq. unit              

B)  \[\frac{7}{2}\]sq. unit    

C)  7 sq. unit                     

D)  9 sq. unit

Correct Answer: A

Solution :

Required area \[=\left| \int\limits_{0}^{2}{(2x-{{x}^{2}}).dx} \right|+\left| \int\limits_{0}^{3}{(-x).dx} \right|\left| \int\limits_{2}^{3}{(2x-{{x}^{2}}).dx} \right|\] \[=\left| {{x}^{2}}-\frac{{{x}^{3}}}{3} \right|+\left| \left( \frac{-{{x}^{2}}}{2} \right) \right|_{0}^{3}-\left| \left( {{x}^{2}}-\frac{{{x}^{3}}}{3} \right) \right|_{2}^{3}\] \[=\left| 4-\frac{8}{3} \right|+\left| \left( \frac{-9}{2} \right) \right|_{0}^{3}-\left| \left( 9-\frac{27}{3} \right)-\left( 4-\frac{8}{3} \right) \right|_{2}^{3}\] \[=\frac{4}{3}+\frac{9}{2}-\frac{4}{3}=\frac{9}{2}\] Hence, option [a] is correct.