question_answer

The area bounded by the curve \[y=2x-{{x}^{2}}\] and the straight line \[y=-x\] is given by   AIEEE  Solved  Paper-2002

A) \[\frac{9}{2}\] sq units     

B)           \[\frac{43}{6}\] sq units      

C) \[\frac{35}{6}\] sq units       

D)           None of these

Correct Answer: A

Solution :

The equations of given curve and a line are                                 \[y=2x-{{x}^{2}}\]                            ... (i) and            \[y=-x\]                                ... (ii) On solving Eqs. (i) and (ii), we get the points of intersection of curves which are (0, 0) and (3, - 3).              \[\therefore \] Required area \[=\int_{0}^{3}{\{(2x-{{x}^{2}})-(-x)\}dx}\]                                    \[=\int_{0}^{3}{(3x-{{x}^{2}})dx}\]                                    \[=\left[ \frac{3{{x}^{2}}}{2}-\frac{{{x}^{3}}}{3} \right]_{0}^{3}\]                                    \[=\frac{27}{2}-\frac{27}{3}=\frac{27}{2}-9\]                                    \[=\frac{9}{2}\] sq units