Question

The area (in sq units) of the plane region bounded by the curve $x=y^2-2$ and the line $y=-x$ is

• A. $\frac{13}{3}$
• B. $\frac{2}{5}$
• C. $\frac{9}{2}$
• D. $\frac{5}{2}$
• E. $\frac{13}{2}$

Answer 1

The correct option is C

$\frac{9}{2}$

Explanation for the correct option:

Find the area (in sq units) of the plane region bounded by the curve .

Consider the given Equation as,

$x=y^2-2\to \left(1\right)$

$y=-x\to \left(2\right)$

Put $y=-x$ in the Equation (1) we get

$$\begin{gathered} \Rightarrow x={\left(-x\right)}^2-2 \\ \Rightarrow x=x^2-2 \\ \Rightarrow x^2-x-2=0 \\ \Rightarrow x^2-2x+x-2=0 \\ \Rightarrow x\left(x-2\right)+1\left(x-2\right)=0 \\ \Rightarrow \left(x+1\right)\left(x-2\right)=0 \\ \Rightarrow x=-1,x=2 \end{gathered}$$

These are the points of intersection

The graph of the given curves is shown below,

The area of ABC is Equal to

$$\begin{gathered} A_1={\int }_{-2}^{-1}\sqrt{x+2}dx \\ \Rightarrow A_1={\left[\frac{2}{3}{\left(x+2\right)}^{\frac{3}{2}}\right]}_{-2}^{-1} \\ \Rightarrow A_1=\left[\frac{2}{3}{\left(-1+2\right)}^{\frac{3}{2}}-\frac{2}{3}{\left(-2+2\right)}^{\frac{3}{2}}\right] \\ \Rightarrow A_1=\frac{2}{3}\text{sq units} \end{gathered}$$

Area of BOC is Equal to

$$\begin{gathered} A_2={\int }_{-1}^0-xdx \\ \Rightarrow A_2=-{\left[\frac{x^2}{2}\right]}_{-1}^0 \\ \Rightarrow A_2=\frac{1}{2}\left(0-1\right) \\ \Rightarrow A_2=\frac{1}{2}\text{sq units} \end{gathered}$$

Area of AOD is Equal to

$$\begin{gathered} A_3={\int }_{-2}^0\sqrt{x+2}dx \\ \Rightarrow A_3=\frac{2}{3}{\left[{\left(x+2\right)}^{\frac{3}{2}}\right]}_{-1}^0 \\ \Rightarrow A_3=\frac{2}{3}\left[{\left(0+2\right)}^{\frac{3}{2}}-{\left(-2+2\right)}^{\frac{3}{2}}\right] \\ \Rightarrow A_3=\frac{2}{3}\left(2^{\frac{3}{2}}\right) \\ \Rightarrow A_3=\frac{4}{3}\sqrt{2}\text{sq units} \end{gathered}$$

Area of ODE is Equal to area of ODEF-area of OEF

$$\begin{gathered} A_4={\int }_0^2\sqrt{x+2}dx-\frac{1}{2}\times 2\times 2\left[\text{Area of}∆OEF=\frac{1}{2}\times b\times h\text{where}b=2,h=2\right] \\ \Rightarrow A_4=\frac{2}{3}{\left[{\left(x+2\right)}^{\frac{3}{2}}\right]}_0^2-2 \\ \Rightarrow A_4=\frac{2}{3}\left[{\left(4\right)}^{\frac{3}{2}}-{\left(2\right)}^{\frac{3}{2}}\right]-2 \\ \Rightarrow A_4=\frac{16}{3}-\frac{4\sqrt{2}}{3}-2\text{sq units} \end{gathered}$$

Now the required area of a shaded region is

$$\begin{gathered} A=A_1+A_2+A_3+A_4 \\ \Rightarrow A=\frac{2}{3}+\frac{1}{2}+\frac{4\sqrt{2}}{3}+\frac{16}{3}-\frac{4\sqrt{2}}{3}-2 \\ \Rightarrow A=\frac{2}{3}+\frac{1}{2}+\frac{16}{3}-2 \\ \Rightarrow A=\frac{18}{3}-\frac{1-4}{2} \\ \Rightarrow A=\frac{18}{3}-\frac{3}{2} \\ \Rightarrow A=\frac{36-9}{6} \\ \Rightarrow A=\frac{27}{6} \\ \Rightarrow A=\frac{9}{2}\text{Sq units} \end{gathered}$$

Hence, option (C) is the correct answer.