ISI2015-DCG-51

Question

The area bounded by $y=x^2-4$, $y=0$ and $x=4$ is

• A. $\frac{64}{3}$
• B. $6$
• C. $\frac{16}{3}$
• D. $\frac{32}{3}$

Answer 1

The area bounded by $y=x^2-4,y=0$ and $x=4$ is as shown below

Area $= \int^4_2 ydx = \int^4_2 (x^2-4)dx = [\frac{x^3}{3} - 4x]^4_2= \frac{64}{3}-16-\frac{8}{3}+8= \frac{56}{3}-8=\frac{32}{3}$

$\therefore$ Option $D.$ is correct.