Question

find the area in the second quadrant bounded by the curve 2x^2+4x+y=0?

Answer 1

$y = -2x^2-4x$

First find x-intersepts:

$-2x^2-4x = 0 \ |:(-2)\\ x^2+2x = 0\\ x(x+2) = 0\\ x=0, x=-2$

So we need to integrate from -2 to 0 of y:

$S = \int_{-2}^0(-2x^2-4x)dx = -\cfrac{2x^3}{3}-2x^2|_{-2}^0=0-0-(\cfrac{16}{3}-8) =8-\cfrac{16}{3} = 2\cfrac{2}3{}$