Question

Area bounded by the curve y = 2x - x^2 and the straight line y =-x is

Answer 1

So lets draw graph for both curves
Now find the points of intersection
substitute
 $y = - x$
in
$y = 2x - x^2$
like
$-x = 2x - x^2$
$x^2 - 3x = 0 \\ x(x-3) = 0$
$x = 0, 3$
the intersection points are (0,0) and (3,-3)
So for area
$\int\limits^{3}_{0} {f(x) - g(x)} \, dx = \int\limits^{3}_{0} {3x-x^2} \, dx = \\ |\frac{3x^2}{2} - \frac{x^3}{3}|^{3}_{0} = \frac{9}{2}$