A) \[\frac{9}{2}\] sq units
B) \[\frac{43}{6}\] sq units
C) \[\frac{35}{6}\] sq units
D) None of these
Correct Answer: A
Solution :
The equations of given curve and a line are \[y=2x-{{x}^{2}}\] ... (i) and \[y=-x\] ... (ii) On solving Eqs. (i) and (ii), we get the points of intersection of curves which are (0, 0) and (3, - 3). \[\therefore \] Required area \[=\int_{0}^{3}{\{(2x-{{x}^{2}})-(-x)\}dx}\] \[=\int_{0}^{3}{(3x-{{x}^{2}})dx}\] \[=\left[ \frac{3{{x}^{2}}}{2}-\frac{{{x}^{3}}}{3} \right]_{0}^{3}\] \[=\frac{27}{2}-\frac{27}{3}=\frac{27}{2}-9\] \[=\frac{9}{2}\] sq unitsYou need to login to perform this action.
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