Double integral question

[kieranl]

Homework Statement

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

Homework Equations

$$\int\int$$x²e^xydA over the region R= {(x,y), y<=x<=1, 0<=y<=1}

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 $$\int$$e^y(2/y²-2/y³-1/y)+e^y2(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

 

[tiny-tim]

Welcome to PF!

Hi kieranl! Welcome to PF!

(have a ≤ )

$$\int\int$$x²e^xydA over the region R= {(x,y), y<=x<=1, 0<=y<=1}
…
I mangaged to get upto $$\int$$e^y(2/y²-2/y³-1/y)+e^y2(y-2/y+2/y³) dy

uhh?

start again, and integrate over dy first!

 

[kieranl]

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

 

[tiny-tim]

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

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

 

[kieranl]

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

 

[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)}.

 

[tiny-tim]

Hi kieranl!

(have an integral: ∫ and a ≤ )

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

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. ​