A) 7/9
B) 14/3
C) 7/3
D) 14/9
Correct Answer: D
Solution :
[d] Required area \[=\int\limits_{y=1}^{4}{\sqrt{\frac{y}{9}}dy=\frac{1}{3}\int\limits_{y=1}^{4}{{{y}^{1/2}}}dy}\] \[=\frac{1}{3}\times \frac{2}{3}\left| ({{y}^{3/2}}) \right|_{1}^{4}=\frac{2}{9}[{{({{4}^{1/2}})}^{3}}-{{({{1}^{1/2}})}^{3}}]\] \[=\frac{14}{9}\] sq. units.You need to login to perform this action.
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