A) \[\frac{32}{3}\,{{a}^{2}}\] sq. unit
B) \[\frac{16}{3}\] sq. unit
C) \[\frac{32}{3}\] sq. unit
D) \[\frac{16}{3}\,{{a}^{2}}\] sq. unit
Correct Answer: D
Solution :
Solving the two equations, we have\[{{x}^{4}}=64{{a}^{3}}x\] \[\Rightarrow x=0,\,\,4a\] Required area =\[\int_{0}^{4a}{2{{a}^{1/2}}{{x}^{1/2}}dx-\int_{0}^{4a}{\frac{{{x}^{2}}}{4a}dx}}\] \[=\frac{32}{3}{{a}^{2}}-\frac{16}{3}{{a}^{2}}=\frac{16}{3}{{a}^{2}}\]sq. unit.You need to login to perform this action.
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