Question

Area bounded by curve $y=x^2$ and $y=2-x^2$ is ?

• A. $\frac{8}{3}$ sq units
• B. $\frac{3}{8}$ sq units
• C. $\frac{3}{2}$ sq units
• D. None of these

Answer 1

The correct option is A $\frac{8}{3}$ sq units

$y=x^2$ and $y=2-x^2$

now both he curve intersect each other ,

from both the equation

$x^2=2-x^2$

$x=+1,-1$

now area bounded by both curve is

$A={\int }_{-1}^1(2-x^2)-{\int }_{-1}1\left(x^2\right)$

because both are even function hence

$A=2[{\int }_0^1(2-x^2)-{\int }_{0}1\left(x^2\right)$]

$A=2\left[\right(2x-\frac{x^3}{3})-(\frac{x^3}{3}\left)\right]$

now on putting upper and lower value of limit ,we will get

$A=\frac{8}{3}$