A) \[\frac{8}{3}sq\,unit\]
B) \[\frac{4}{3}sq\,unit\]
C) \[\frac{7}{3}sq\,unit\]
D) \[\frac{2}{3}sq\,unit\]
Correct Answer: B
Solution :
Given curve is \[y=2x-{{x}^{2}}\] or \[{{(x-1)}^{2}}=-(y-1)\] The curve cut the x- axis at \[(0,0)\] and \[(2,0)\]. \[\therefore \] Required area \[=\int_{0}^{2}{y\,dx}\] \[\int_{0}^{2}{(2x-{{x}^{2}})dx=\left[ {{x}^{2}}-\frac{{{x}^{3}}}{3} \right]_{0}^{2}}\] \[=4-\frac{8}{3}=\frac{4}{3}sq\,unit\]You need to login to perform this action.
You will be redirected in
3 sec