question_answer

The area (in square unit) bounded by the curves \[y={{x}^{3}}\]and \[y=x\] is

A)  \[1/2\text{ }sq\text{ }unit\]     

B)  \[1/4\text{ }sq\text{ }unit\]

C)  \[1/8\text{ }sq\text{ }unit\]      

D)  \[1/16\text{ }sq\text{ }unit\]

Correct Answer: A

Solution :

\[\therefore \]  Required area \[=2\int_{0}^{1}{(x-{{x}^{3}})dx}\] \[=2\left[ \frac{{{x}^{2}}}{2}-\frac{{{x}^{4}}}{4} \right]_{0}^{1}\] \[=2\left[ \frac{1}{2}-\frac{1}{4} \right]=2\times \frac{1}{4}\] \[=\frac{1}{2}sq\,\,unit\]