help with double integral...?

Question

evaluate the double integral (x+4) dxdy when its bounded by y=x and y=x^2

Answer 1

The left intersection point is (0,0) and the right intersection point is (1,1). So you could have:

∫ [from y=0 to y=1] ∫ [from x=y to x=√y of] (x+4) dx dy =
= ∫ [from x=0 to x=1] ∫ [from y=x² to y=x of] (x+4) dy dx

The second one is easier as far as I'm concerned (and clearly the answer must be positive since you're integrating a positive quantity).

= ∫ [from x=0 to x=1] (x+4)y | y=x² to y=x dx
= ∫ [from x=0 to x=1] (x+4)(x-x²) dx
= ∫ [from x=0 to x=1] (4x - 3x² - x³) dx
= 2 - 1 - 1/4
=3/4

Answer 2

your domain here is y = y^1/2, or in short, 0<=y<=1 and y<=x<=sqrt(y)..., so...

your double integral becomes integral from 0 to 1 ofthe integral from y to sqrt(y) of (x +4) dxdy

then integrate with respect to x before integrating with respect to y...

answer, 1/3 + 3/2 - 8/3 = -8/6