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\[=\int_{0}^{1}{(2x-{{x}^{2}})-x\,dx}\] \[=\int_{0}^{1}{(x-{{x}^{2}})dx=\left[ \frac{{{x}^{2}}}{2}-\frac{{{x}^{3}}}{3} \right]}_{0}^{1}\] \[=\frac{1}{2}-\frac{1}{3}=\frac{1}{6}\text{sq}\,\text{unit}\]You need to login to perform this action.
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