# Double integral question

1. Apr 10, 2009

### kieranl

1. The problem statement, all variables and given/known data

Sketch the region of integration and then evaluate the double integral:

2. Relevant equations

$$\int\int$$x2exydA over the region R= {(x,y), y<=x<=1, 0<=y<=1}

3. The attempt at a solution

I have managed to do half of the problem and integrate it respect to x but then have no idea how to finish the problem.

I mangaged to get upto $$\int$$ey(2/y2-2/y3-1/y)+ey2(y-2/y+2/y3) dy

can someone please tell me how to finish it or if i have done something wrong so far??

cheers

Last edited: Apr 10, 2009
2. Apr 10, 2009

### tiny-tim

Welcome to PF!

Hi kieranl! Welcome to PF!

(have a ≤ )
uhh?

start again, and integrate over dy first!

3. Apr 12, 2009

### kieranl

but if i integrated by dy first wouldnt i end up with an answer containing y's and not numbers??

4. Apr 12, 2009

### tiny-tim

If you integrate with respect to dy, then you eliminate y, and only have x's.

5. Apr 12, 2009

### kieranl

but i have to integrate over y<=x<=1 for dx so the final answer would then contain y's?

6. Apr 12, 2009

### Billy Bob

Draw a picture of the region R. Then use the picture to write R in the form

R={(x,y):a<x<b, f(x)<y<g(x)}.

7. Apr 13, 2009

### tiny-tim

Hi kieranl!

(have an integral: ∫ and a ≤ )
No, you have to integrate over 0 ≤ y ≤ x ≤ 1 …

whether you do it over x first or over y first is up to you.