# Double Integral

• Nov 10th 2010, 11:54 AM
ineedyourhelp
Double Integral
Let
f be continuous on [0; 1] and let R be the triangular region with vertices (0; 0); (1; 0) and (0; 1). Show that

$\displaystyle \int\int_{R}f(X+Y)=\int_{0}^{1}uf(u)\, du$

Hi All,

I got stuck with the above question. Basically I got LHS:
$\displaystyle \int\int_{R}f(X+Y)=\int_{0}^{1}\int_{0}^{1-x}f(x+y)\, dydx$

How do I continue from here?
• Nov 10th 2010, 12:04 PM
metallicsatan
Hi QF203 classmate. Use the change of variable method taught in Lesson 9. Haha. Use u=x+y, and v=y. :D
• Nov 10th 2010, 01:20 PM
Prove It
Quote:

Originally Posted by metallicsatan
Hi QF203 classmate. Use the change of variable method taught in Lesson 9. Haha. Use u=x+y, and v=y. :D

You will also need to compute the Jacobian so that $\displaystyle \displaystyle dy\,dx$ is transformed to $\displaystyle \displaystyle |J|\,du\,dv$ or $\displaystyle \displaystyle |J|\,dv\,du$ (whichever is easier).
• Nov 10th 2010, 01:34 PM
stacyk
Thanks for the help. I had a similar problem too.