Question

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

• 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

Answer 1

Answer: (B)

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 $