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allsmiles
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Evaluate the following integrals by first reversing the order of integration. ∫(0,8)∫(y^(1/3),2) 8e^x^2 dxdy. Why is the limit when you reverse 0<=x<=2, 0<=y<=x^3 and not x^3<=y<=8?
 2 years ago
 2 years ago
allsmiles Group Title
Evaluate the following integrals by first reversing the order of integration. ∫(0,8)∫(y^(1/3),2) 8e^x^2 dxdy. Why is the limit when you reverse 0<=x<=2, 0<=y<=x^3 and not x^3<=y<=8?
 2 years ago
 2 years ago

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allsmiles Group TitleBest ResponseYou've already chosen the best response.0
I'm confused with finding the limits for each when you reverse the order anyone got any tips?
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
drawing out the region of integration is almost always a good idea
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
dw:1352232985691:dw
 2 years ago

allsmiles Group TitleBest ResponseYou've already chosen the best response.0
Yeah I got that too, but can't you say y^(1/3) to 8 is the limit? Go along the curve y=x^3 and stop at 8?
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
well what would the area between y^(1/3) and y=8 look like?
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
or rather x^3<=y<=8
 2 years ago

allsmiles Group TitleBest ResponseYou've already chosen the best response.0
it'd be the shaded area under y=8 but above y=x^3?
 2 years ago

allsmiles Group TitleBest ResponseYou've already chosen the best response.0
which we're not looking for right
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
dw:1352233420315:dwyou got it :)
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
so in the area we want we want y bounded above by the function, not the line
 2 years ago

allsmiles Group TitleBest ResponseYou've already chosen the best response.0
it's so hard to see sometimes but got it thanks so much for your help!!
 2 years ago

allsmiles Group TitleBest ResponseYou've already chosen the best response.0
so whenever you want it bounded by the curve you'd take it from 0 to the curve?
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
welcome, and never underestimate the power of making sure your drawings are correct
 2 years ago

allsmiles Group TitleBest ResponseYou've already chosen the best response.0
or where ever it starts at
 2 years ago
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