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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?
 one year ago
 one year 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?
 one year ago
 one year 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?
 one year ago

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

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
dw:1352232985691:dw
 one year 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?
 one year 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?
 one year ago

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

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

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
dw:1352233420315:dwyou got it :)
 one year 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
 one year 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!!
 one year 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?
 one year 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
 one year ago

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