where D is the region limited by: y=x2, x=2, y=1 (dA = dxdy)
The Attempt at a Solution
So basically i sketched the area, and i get the area under y=x2 0<x<1 and a square at 1<x<2 , 0<y<1
So i divded the integrals; for the square ∫01∫12 of x2+y2dxdy
for the y=x2 part is the part i need help/confirming x goes from 0 to 1(so integral os x2 with those bounds), and i used substitution for y=x2 and got dy=2xdx so integrated from 0 to 1 x42x dx
So i was thinking about this in another way, wich gave me the same answer.
So i integrated x2 and kept the bounds so i got [(x3)/3]10 then integrated y2 and set its bounds to 0 to x2 and substituted so i got [(x3)/3]10 + x^5/3, and moved that inside the bounds.
Am i correct?