Question

The area bounded by curve $y=6+x-x^2$ and $x$-axis is ____ sq. units

• A. $\frac{127}{6}$
• B. $\frac{125}{6}$
• C. $\frac{121}{6}$
• D. $\frac{129}{6}$

Answer 1

Answer: (B)

$\frac{125}{6}$

Given, $y=-x^2+x+6$

Solving the above equation we get, $(-x+3),(x+2)$

Area$={\int }_{-2}^{3}6+x-x^{2}dx$

$={[6x+\frac{x^2}{2}-\frac{x^3}{3}]}_{-2}^3$

$=[18+\frac{9}{2}-\frac{27}{3}]-[-12+\frac{4}{2}+\frac{8}{3}]$

$=[30+\frac{5}{2}-\frac{35}{3}]$

$=\frac{125}{6}$sq. units