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## Homework Statement

∫∫

_{D}x

^{2}+y

^{2}dA

where D is the region limited by: y=x

^{2}, x=2, y=1 (dA = dxdy)

## Homework Equations

## The Attempt at a Solution

So basically i sketched the area, and i get the area under y=x

^{2}0<x<1 and a square at 1<x<2 , 0<y<1

So i divded the integrals; for the square ∫

_{0}

^{1}∫

_{1}

^{2}of x

^{2}+y

^{2}dxdy

for the y=x

^{2}part is the part i need help/confirming x goes from 0 to 1(so integral os x

^{2}with those bounds), and i used substitution for y=x

^{2}and got dy=2xdx so integrated from 0 to 1 x

^{4}2x dx

So i was thinking about this in another way, wich gave me the same answer.

So i integrated x

^{2}and kept the bounds so i got [(x

^{3})/3]

^{1}

_{0}then integrated y

^{2}and set its bounds to 0 to x

^{2}and substituted so i got [(x

^{3})/3]

^{1}

_{0}+ x^5/3, and moved that inside the bounds.

Am i correct?

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