question_answer

The area enclosed between the curves \[y=x\] and \[y=2x-{{x}^{2}}\]is (in sq. unit)

A) \[\frac{1}{2}\]                   

B)         \[\frac{1}{6}\]                  

C)  \[\frac{1}{3}\]                  

D)         \[\frac{1}{4}\]

Correct Answer: B

Solution :

The equations of curves are \[y=x\]                                 ?(i)                 and        \[y=2x-{{x}^{2}}\]                            ?(iii) On solving Eqs. (i) and (ii), we get \[x=2x-{{x}^{2}}\] \[\Rightarrow \]               \[x=0,1\] Required area\[=\int_{0}^{1}{(2x-{{x}^{2}})-x\,dx}\]                 \[=\int_{0}^{1}{(x-{{x}^{2}})dx=\left[ \frac{{{x}^{2}}}{2}-\frac{{{x}^{3}}}{3} \right]}_{0}^{1}\] \[=\frac{1}{2}-\frac{1}{3}=\frac{1}{6}\text{sq}\,\text{unit}\]