A) \[\frac{4}{3}\]
B) 1
C) \[\frac{2}{3}\]
D) \[\frac{1}{3}\]
Correct Answer: A
Solution :
\[{{y}^{2}}=x\]and \[2y=x\Rightarrow {{y}^{2}}=2y\Rightarrow y=0,\,2\] \[\therefore \,\]Required area\[=\int_{0}^{2}{({{y}^{2}}-2y)dy=\left( \frac{{{y}^{3}}}{3}-{{y}^{2}} \right)_{0}^{2}=\frac{4}{3}}\]sq. unit.
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