A) \[\frac{5}{3}\]
B) \[\frac{5}{4}\]
C) \[\frac{5}{12}\]
D) \[\frac{12}{5}\]
Correct Answer: C
Solution :
Given curves are, \[y={{x}^{3}}\] and \[y=\sqrt{x}\] On solving, we get \[x=0,x=1\] Therefore, required area =\[\int_{0}^{1}{({{x}^{3}}-\sqrt{x})}\,dx\] \[=\left[ \frac{{{x}^{4}}}{4}-\frac{2x\sqrt{x}}{3} \right]_{0}^{1}=\left[ \frac{1}{4}-\frac{2}{3} \right]=\frac{5}{12}\],(Area can?t be negative).
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