A) \[\left( \frac{125}{3} \right)sq\,units\]
B) \[\left( \frac{125}{6} \right)sq\,units\]
C) \[\left( \frac{125}{4} \right)sq\,units\]
D) None of these
Correct Answer: B
Solution :
Given equation of curves are \[y=4+3x-{{x}^{2}}\]and\[y=0\]. On solving these equations, we get \[x=-1,4\] \[\therefore \] Required area\[=\int_{-1}^{4}{y\,dx}\] \[=\int_{-1}^{4}{(4+3x-{{x}^{2}})dx}\] \[=\left[ 4x+\frac{3{{x}^{2}}}{2}-\frac{{{x}^{3}}}{3} \right]_{-1}^{4}\] \[=16+24-\frac{64}{3}+4-\frac{3}{2}-\frac{1}{3}\] \[=44-\frac{65}{3}-\frac{3}{2}\] \[=\frac{264-130-9}{6}\] \[=\frac{125}{6}sq\,unit\]
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