A) \[\frac{31}{3}\]
B) \[\frac{29}{3}\]
C) \[\frac{34}{3}\]
D) \[\frac{32}{3}\]
Correct Answer: D
Solution :
\[{{x}^{2}}-y\le 0\] and \[2x+y-3\le 0\] For Point of intersection we have \[{{x}^{2}}+2x-3=0~~\] \[\Rightarrow \,\,\,\,\,\,\,\,x=1,x=-3\] \[\therefore \] P (1, 1) and Q (-3, 9) are point of intersection \[\therefore \] Required area \[\text{=}\int\limits_{-3}^{1}{\left( 3-2x-{{x}^{2}} \right)dx}\] \[\text{=12}-\left( {{x}^{2}} \right)_{-3}^{1}-\frac{1}{3}\left( {{x}^{3}} \right)_{-3}^{1}\] \[\text{=12}-\left( 1-9 \right)-\frac{1}{3}\left( 1+27 \right)\] \[\text{=20}-\frac{28}{3}=11-\frac{1}{3}=\frac{32}{3}\]
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