# calculus

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evaluate the double integral
∫R∫ lny/x dA
for the region R is the rectangle defined by 1<x<e^2 and 1<y<e

• calculus -

∫[1,e^2]∫[1,e] lny/x dy dx
= ∫[1,e^2](y/x (lny - 1))[1,e] dx
= ∫[1,e^2] [(e/x (1-1))-(1/x (0-1))] dx
= ∫[1,e^2] 1/x dx
= lnx [1,e^2]
= 2

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