Question

Find the area bounded by the curve $y=2x-x^2$ and the straight line $y=-x$

Answer 1

Area $={\int }_0^{3}2x-x^2+xdx$

$={\int }_0^{3}3x-x^{2}dx={[\frac{3x^2}{2}-\frac{x^3}{3}]}_0^3$

$=(\frac{3.9}{2}-\frac{27}{3})-0$

$=\frac{27}{3}-\frac{27}{3}$

$=27\left(\frac{3-2}{6}\right)$

$=\frac{27}{6}$