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ISI2015-DCG-51

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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.

 

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