Question

Area enclosed between the curves $y=8-x^2$ and $y=x^2$, is:

• A. $32/3$
• B. $64/3$
• C. $30/4$
• D. $9$

Answer 1

The correct option is B $64/3$

$y=8-x^{2}y=x^2$

Intersection points are

$x^2=8-x^2$

$2x^2=8$

$x^2=4x=-2,2$

${\int }_{-2}^2(8-x^2-x^2)dx$

${\int }_{-2}^2(8-2x^2)dx$

$2{(8x-\frac{2x^3}{3})}_0^2$

$=2(8\left(2\right)-\frac{2\left(8\right)}{3})$

$=2(16-\frac{16}{3})$

$=2\left(\frac{32}{13}\right)$

$=\frac{64}{3}$ sq. units