Question

The area (in sq.units) of the region enclosed by the curves $y=x^2-1$ and $y=1-x^2$ is equal to

• A. $\frac{4}{3}$
• B. $\frac{7}{2}$
• C. $\frac{16}{3}$
• D. $\frac{8}{3}$

Answer 1

The correct option is D

$\frac{8}{3}$

Explanation for the correct option:

Equation of the curves are :

$y=x^2-1$

$y=1-x^2$

on solving both the equations,

$$\begin{gathered} \Rightarrow 1-x^2=x^2-1 \\ \Rightarrow 2x^2-2=0 \\ \Rightarrow x^2-1=0 \\ \Rightarrow x=1,-1 \end{gathered}$$

points of the intersection are $(0,1)$ and $(-1,0)$

Now,the area of the shaded portion is,

$$\begin{gathered} \Rightarrow A={\int }_{-1}^1\left(\right(1-x^2)-(x^2-1\left)\right)dx \\ \Rightarrow A={\int }_{-1}^1(1-x^2-x^2+1)dx \\ \Rightarrow A={\int }_{-1}^1(2-2x^2)dx \\ \Rightarrow A={[2x-\frac{2}{3}{x}^3]}_{-1}^1 \\ \Rightarrow A=2-\frac{2}{3}-(-2+\frac{2}{3}) \\ \Rightarrow A=4-\frac{4}{3} \\ \Rightarrow A=\frac{8}{3}sq.units \end{gathered}$$

Hence,option (D) is the correct answer.