A) 3 sq. unit
B) \[\frac{7}{5}\]sq. unit
C) \[\frac{7}{3}\]sq. unit
D) None of these
Correct Answer: C
Solution :
Required area = \[\int_{1}^{4}{x\,dy=\int_{1}^{4}{\frac{\sqrt{y}}{2}dy}}\] = \[\frac{1}{2}.\frac{2}{3}|{{y}^{3/2}}|_{1}^{4}=\frac{7}{3}\] sq. unit.You need to login to perform this action.
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