## anonymous 5 years ago In reading another site I found the following and I don't understand how the limits of integration are derived when the order of integration is changed: \int f_U(u) h(u)\,du=\int_0^\infty \int_0^\infty\int_{(y+z)/3}^\infty f(3u-y-z,y,z)h(u) 3\,du\,dy\,dz Changing the order of integration we can write: \int f_U(u) h(u)\,du=\int_0^\infty\int_0^{3u} \int_0^{3u-z} f(3u-y-z,y,z)h(u) 3\,dy\,dz\,du

1. anonymous

Whoops I thought the LaTeX would be compiled the first integral is: $\int\limits f_U(u) h(u)\,du=\int\limits_0^\infty \int\limits_0^\infty\int\limits_{(y+z)/3}^\infty f(3u-y-z,y,z)h(u) 3\,du\,dy\,dz$

2. anonymous

And the second integral is: $\int\limits f_U(u) h(u)\,du=\int\limits_0^\infty\int\limits_0^{3u} \int\limits_0^{3u-z} f(3u-y-z,y,z)h(u) 3\,dy\,dz\,du$

3. anonymous

I notice the second integration follows the patterns of solving (3u-y-z) for y, then z, and finally u - if you follow the integration signs from right to left (inside out). I'm confused about the pattern of the first integration though. Its been so long since I've done this that I think my calculator was steam powered!