A) \[32\]
B) \[\frac{32}{3}\]
C) \[\frac{1}{32}\]
D) \[\frac{1}{3}\]
Correct Answer: B
Solution :
Given, \[y=-{{x}^{2}}+2x+3\]and \[y=0\] Therefore, \[x=-1\] and \[x=3\] \ Required area \[=\int_{-1}^{3}{(-{{x}^{2}}+2x+3)dx}\] \[=\left[ -\frac{{{x}^{3}}}{3}+{{x}^{2}}+3x \right]_{-1}^{3}=\frac{32}{3}\].
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