A) 16 sq unit
B) 32 sq unit
C) \[\frac{1}{4}\]sq unit
D) \[\frac{1}{2}\]sq unit
E) \[\frac{1}{8}\]sq unit
Correct Answer: B
Solution :
Given curves\[y=4{{x}^{3}}\]and\[y=16x\] intersect at point satisfying \[16x=4{{x}^{3}}\] \[\Rightarrow \] \[4{{x}^{3}}-16x=0\] \[4x({{x}^{2}}-4)=0\] \[\Rightarrow \] \[x=0,\pm 2\] \[\Rightarrow \] \[y=0,\pm 32\] \[\therefore \]Required area \[=2\int_{0}^{2}{(16x-4{{x}^{3}})}dx\] \[=2\left[ \frac{16{{x}^{2}}}{2}-\frac{4{{x}^{4}}}{4} \right]_{0}^{2}\] \[=2(32-16)=32\text{ }sq\text{ }unit\]You need to login to perform this action.
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