A) \[\frac{32}{3}\]
B) \[\frac{16}{3}\]
C) \[\frac{8}{3}\]
D) 0
Correct Answer: B
Solution :
Area\[=\int\limits_{0}^{4}{\left( 2\sqrt{x}-\frac{{{x}^{2}}}{4} \right)dx}\] \[=\left( 2\left( \frac{{{x}^{3/2}}}{3/2} \right)-\frac{{{x}^{3}}}{12} \right)_{0}^{4}\] \[=\frac{4}{3}\times 8-\frac{64}{12}\] \[=\frac{32}{3}-\frac{16}{3}=\frac{16}{3}\]You need to login to perform this action.
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