Answer to Question #184355 in Calculus for Jhimwil

Question #184355

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

1
Expert's answer
2021-04-26T05:42:25-0400

"y = -2x^2-4x"

First find x-intersepts:

"-2x^2-4x = 0 \\ |:(-2)\\\\\nx^2+2x = 0\\\\\nx(x+2) = 0\\\\\nx=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{}"


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