Question

The area bounded by the parabola $y=4-x^2$ and $X$-axis in sq units is:

• A. $\frac{32}{3}$
• B. $\frac{16}{3}$
• C. $\frac{8}{3}$
• D. $\frac{1}{3}$

Answer 1

The correct option is A $\frac{32}{3}$

$y=4-x^2$

The above curve will intersect x-axis at two different points $-2$ and $2$

Then, the area bounded by the curve $y=4-x^2$ and x-axis is given by

$A=\stackrel{2}{\underset{-2}{\int }}(4-x^2)dx$

$=\stackrel{2}{\underset{-2}{\int }}4dx-\stackrel{2}{\underset{-2}{\int }}{x}^{2}dx$

$=4x{|}_{-2}^2-\left(\frac{x^3}{3}\right){|}_{-2}^2$

$=4(2+2)-[8/3-(-\frac{8}{3})]$

$=16-\left(\frac{16}{3}\right)=\frac{32}{3}$