A) \[\frac{125}{6}\]sq units
B) \[\frac{125}{3}\]sq units
C) \[\frac{125}{2}\] sq units
D) None of these
Correct Answer: A
Solution :
Given, \[y=4+3x-{{x}^{2}}\] For the intersection point with\[x-\]axis, put\[y=0\] \[4+3x-{{x}^{2}}=0\] \[\Rightarrow \] \[{{x}^{2}}-3x-4=0\] \[\Rightarrow \] \[(x+1)(x-4)=0\] \[\Rightarrow \] \[x=4,-1\] \[\therefore \]Required area\[=\int_{-1}^{4}{y\,dx}\] \[=\int_{-1}^{4}{(4+3x-{{x}^{2}})dx}\] \[=\left[ 4x+\frac{3{{x}^{2}}}{2}-\frac{{{x}^{3}}}{3} \right]_{-1}^{4}\] \[=16+3.\frac{16}{2}-\frac{64}{3}+4-\frac{3}{2}-\frac{1}{3}\] \[=16+24-\frac{64}{3}+4-\frac{11}{6}\] \[=\frac{125}{6}\]sq unitsYou need to login to perform this action.
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