question_answer

The area of the plane region bounded by the curve\[x={{y}^{2}}-2\]and the line\[y=-x\]is (in square units)

A)  \[\frac{13}{3}\]                               

B)  \[\frac{2}{5}\]

C)  \[\frac{9}{5}\]                  

D)         \[\frac{5}{2}\]

E)  \[\frac{13}{2}\]

Correct Answer: C

Solution :

Given curves\[x={{y}^{2}}-2\]and\[y=x\] Thus, interection point are \[(-1,1)\]and\[(2,-2)\] We are to find the area of shaded part Area of \[ABC=\int_{-2}^{-1}{\sqrt{x+2}}dx\]                 \[=\left[ \frac{2}{3}{{(x+2)}^{3/2}} \right]_{-2}^{-1}=\frac{2}{3}\]sq unit Area of \[BCO=\int_{-1}^{0}{-x}\,dx=\left( -\frac{{{x}^{2}}}{2} \right)_{-1}^{0}\] \[=\frac{1}{2}sq\text{ }unit\] Area of ADO                 \[=\int_{-2}^{0}{\sqrt{x+2}}dx=\left[ \frac{2}{3}{{(x+2)}^{3/2}} \right]_{-2}^{0}\]                 \[=\frac{4}{3}\sqrt{2}\] Area of,\[ODE=area\text{ }of\text{ }ODEF-are\text{ }of\text{ }OFE\] \[\int_{0}^{2}{\sqrt{x+2}}dx-\int_{0}^{2}{(-x)}dx\] \[=\left\{ \frac{2}{3}{{(x+2)}^{3/2}} \right\}_{0}^{2}-\left( -\frac{{{x}^{2}}}{2} \right)_{0}^{2}\]                 \[=\left( \frac{16}{3}-\frac{4\sqrt{2}}{3} \right)-(2)\] (neglecting the negative sign) \[=\left( \frac{16}{3}-\frac{4\sqrt{2}}{3} \right)-(2)\] \[\therefore \]Required area                 \[=\frac{2}{3}+\frac{1}{2}+\frac{4\sqrt{2}}{3}+\frac{16}{3}-\frac{4\sqrt{2}}{3}-2\]                 \[=\frac{2}{3}+\frac{1}{2}+\frac{16}{3}-2\]                 \[=\frac{27}{6}=\frac{9}{2}sq.\,unit\]