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anonymous
 3 years ago
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?
anonymous
 3 years ago
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?

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

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

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.1dw:1352232985691:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Yeah 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?

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

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0which we're not looking for right

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

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0it's so hard to see sometimes but got it thanks so much for your help!!

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so whenever you want it bounded by the curve you'd take it from 0 to the curve?

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0or where ever it starts at
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