Double integral - e^(x/y) dA D = 1<y<2, y<x<y^3 (all the inequalities have _ under them)
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Seems fairly standard to me. What have you done so far? If you start integrating dx first then it will all work out okay.
that is your integral right it is the same as first we integrate with respect to x and y constant sub now the first one is easy the second let u=y^2 and you continue
Originally Posted by Amer that is your integral right it is the same as first we integrate with respect to x and y constant sub now the first one is easy the second let u=y^2 and you continue Edit: deleted I got it, thanks!
look I integrate not derive it is possible that I make mistakes and typos so feel free as you can see the denominator of this is 1/y I think it is clear
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