question_answer

The area of the region bounded by the curve \[y=4+3x-{{x}^{2}}\]and\[x-\]axis is

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 units