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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(3uyz,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^{3uz} f(3uyz,y,z)h(u) 3\,dy\,dz\,du
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(3uyz,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^{3uz} f(3uyz,y,z)h(u) 3\,dy\,dz\,du

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Whoops I thought the LaTeX would be compiled the first integral is: \[\int\limits f_U(u) h(u)\,du=\int\limits_0^\infty \int\limits_0^\infty\int\limits_{(y+z)/3}^\infty f(3uyz,y,z)h(u) 3\,du\,dy\,dz\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0And the second integral is: \[\int\limits f_U(u) h(u)\,du=\int\limits_0^\infty\int\limits_0^{3u} \int\limits_0^{3uz} f(3uyz,y,z)h(u) 3\,dy\,dz\,du\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I notice the second integration follows the patterns of solving (3uyz) 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!
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