question_answer

The area (in square unit) enclosed by the curves\[y=4{{x}^{3}}\]and\[y=16x,\]is:

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 \[=2\int_{0}^{2}{(16x-4{{x}^{3}})}dx\]                 \[=2\left[ \frac{16{{x}^{2}}}{2}-\frac{4{{x}^{4}}}{4} \right]_{0}^{2}\] \[=2(32-16)=32\text{ }sq\text{ }unit\]