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. $\frac{5}{2}$
• E. $\frac{13}{2}$

Answer 1

Answer: (C)

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 $BCO={\int }_{-1}^0-xdx={\left(-\frac{x^2}{2}\right)}_{-1}^0$
$=\frac{1}{2}squnit$ 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, $ODE=areaofODEF-areofOFE$ ${\int }_0^2\sqrt{x+2}dx-{\int }_0^2(-x)dx$
$={\left\{\frac{2}{3}{(x+2)}^{3/2}\right\}}_0^2-{\left(-\frac{x^2}{2}\right)}_0^2$
$=\left(\frac{16}{3}-\frac{4\sqrt{2}}{3}\right)-(2)$ (neglecting the negative sign)
$=\left(\frac{16}{3}-\frac{4\sqrt{2}}{3}\right)-(2)$
$\therefore$ Required area
$=\frac{2}{3}+\frac{1}{2}+\frac{4\sqrt{2}}{3}+\frac{16}{3}-\frac{4\sqrt{2}}{3}-2$
$=\frac{2}{3}+\frac{1}{2}+\frac{16}{3}-2$
$=\frac{27}{6}=\frac{9}{2}sq.unit$