What is the simplified integral for evaluating double integrals with IBP?

[rmiller70015]

Homework Statement

∫∫[ye^(-xy)]dA R=[0,2]×[0,3] evaluate the integral.

Homework Equations

The Attempt at a Solution

So I started with some algebra changing the integral to ∫(e^-x)[∫ye^-ydy]dx
I evaluated the y portion first because its more difficult to deal with and wanted to get it out of the way.
I ended up integrating by parts with:
U=y dv=e^-x
Dy=dy v=-e^-x and got
-ye^(-y) - ∫-e^(-y)dy on the interval [0,3] and got -4e^(-3) + 1

This is now a constant, pulled it out of the x integral leaving:
(-4e^(-3) +1)∫e^(-x)dx
The final integral I evaluated as:
-e^(-x) on [0,2] gives
-e^(-2) + e^0 and this is multiplied by the previous number to give:
4e^(-6)-4e^(-3)-e^(-2)+1
Using a calculator to approximate I get a value of 0.692468
The answer in the book is .5e^(-6) + (5\2) which is approximated by a calculator as 2.5012394

I'm doing something wrong and I am wondering if there is a rule I forgot, do I need to integrate by x first and then y?

 

[ehild]

Homework Statement

∫∫[ye^(-xy)]dA R=[0,2]×[0,3] evaluate the integral.

Homework Equations

The Attempt at a Solution

So I started with some algebra changing the integral to ∫(e^-x)[∫ye^-ydy]dx

It is wrong: $(e^{-x})(ye^{-y})=ye^{-x-y}$ instead of $ye^{-xy}$

ehild

 

[rmiller70015]

I was walking back from the corner store when I realized this, but thank you for confirming it. I had, as another helper once told me "abused the chain rule."

 

[LCKurtz]

I was walking back from the corner store when I realized this, but thank you for confirming it. I had, as another helper once told me "abused the chain rule."

Oh my! Chain rule abuse! Oh Guard -- Guard! Come quickly before he gets away!

 

[verty]

Hint: if y was constant, the integrand would be quite simple.

 

[HallsofIvy]

Verty's hint suggest that you do the integration in the other order:
\int_{y=0}^3\left[\int_{x= 0}^2 ye^{-xy}dx\right] dy

You should find that much simpler.