question_answer

If the area bounded by the parabola \[y=2-{{x}^{2}}\] and the line \[x+y=0\] is \[A\] sq unit, then \[A\] equals:

A) \[\frac{1}{2}\]                                   

B) \[\frac{1}{3}\]

C) \[\frac{2}{9}\]                                   

D) \[\frac{9}{2}\]

Correct Answer: D

Solution :

Given equation of parabola can be rewritten as\[{{x}^{2}}=-(y-2)\]and equation of line is\[y=-x\]. \[\therefore \]Point of intersection are \[A(-1,\,\,1)\] and\[D(2,\,\,-2)\]. \[\therefore \]Required area\[=\int_{-1}^{2}{({{y}_{1}}-{{y}_{2}})dx}\]                                 \[=\int_{-1}^{2}{(2-{{x}^{2}}+x)}\,dx\]                                 \[=\left[ 2x-\frac{{{x}^{3}}}{3}+\frac{{{x}^{2}}}{2} \right]_{-1}^{2}\]                                 \[=\left[ 4-\frac{8}{3}+2-\left( -2+\frac{1}{3}+\frac{1}{2} \right) \right]\]                                 \[=8-\frac{8}{3}-\frac{1}{3}-\frac{1}{2}\]                                 \[=\frac{9}{2}sq\,\,unit\]