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anonymous
 5 years ago
Double Integral Question
anonymous
 5 years ago
Double Integral Question

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0When one of the boundaries is not constant and is given by a variable lets say: \[\int\limits_{0}^{1} \int\limits_{0}^{x} f(x,y) dA\] why can't I integrate \[\int\limits_{0}^{x} \int\limits_{0}^{1} f(x,y) dA\] I know mathematically I can't but can someone explain why? I know if I did tried to integrate I would end up with x in my answer which doesnt make sense

dumbcow
 5 years ago
Best ResponseYou've already chosen the best response.0the reason has to do with when integrating you integrate over y first then x. If your bounds were from 0 to x when integrating over dx it would be no different that an indefinite integral since you would be plugging in x for x in the upper bounds.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You don't have to integrate with with y first then x. According to fubini you can integrate either way first and the outcome will be the same. Even if you have to integrate with y first I can simply change the bounds (y=x) and I can get \[\int\limits_{0}^{y}\int\limits_{0}^{1}f(x,y) dA\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0This is the same integral as the one above but in this case I can't integrate with y first.

dumbcow
 5 years ago
Best ResponseYou've already chosen the best response.0yes you are correct. My point is still the same though if the variable is part of the bounds of the outside integral, then essentially you are summing the area under the curve for an unknown length. You will never get a real number because the variable will always be part of the result.
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