A) \[7/9\]
B) \[14/3\]
C) \[7/3\]
D) \[14/9\]
Correct Answer: D
Solution :
Required area \[=\int\limits_{y\,=\,1}^{4}{\sqrt{\frac{y}{9}}dy}\] \[=\frac{1}{3}\left. \int\limits_{y\,=\,1}^{4}{{{y}^{1/2}}dy=\frac{1}{3}\times \frac{2}{3}({{y}^{3/2}})} \right|_{1}^{4}\] \[=\frac{2}{9}[{{({{4}^{1/2}})}^{3}}-{{({{1}^{1/2}})}^{3}}]=\frac{2}{9}[8-1]\] \[=\frac{2}{9}\times 7=\frac{14}{9}\]sq. units
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