Question

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. $\frac{5}{2}$

Answer 1

Answer: (C)

$\frac{9}{5}$

Answer 2

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
$=\frac{2}{3}+\frac{1}{2}+\frac{4\sqrt{2}}{3}+\frac{16}{3}-\frac{4\sqrt{2}}{3}-2$
$=\frac{2}{3}+\frac{1}{2}+\frac{16}{3}-2$
$=\frac{27}{6}=\frac{9}{2}s\,unit$