A) \[\frac{128}{3}\] sq. unit
B) \[\frac{32}{3}\] sq. unit
C) \[\frac{64}{3}\] sq. unit
D) 32 sq. unit
Correct Answer: B
Solution :
\[{{y}^{2}}=8x\] and \[y=x\Rightarrow {{x}^{2}}=8x\Rightarrow x=0\],8 \ Required area =\[\int_{0}^{8}{(2\sqrt{2}\sqrt{x}-x)dx}\] \[=\left[ \frac{4\sqrt{2}}{3}{{x}^{3/2}}-\frac{{{x}^{2}}}{2} \right]_{0}^{8}=\frac{128}{3}-\frac{64}{2}=\frac{32}{3}sq.\]unit.
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