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
∫[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
To evaluate the given double integral ∫R∫ lny/x dA, we need to find the limits of integration for both x and y and then evaluate the integral.
The region R is the rectangle defined by 1<x<e^2 and 1<y<e.
Let's start by evaluating the outer integral with respect to x and then the inner integral with respect to y.
∫R∫ lny/x dA = ∫1< y < e [ ∫1< x < e^2 (lny/x) dx ] dy
Let's evaluate the inner integral first:
∫1< x < e^2 (lny/x) dx
To evaluate this integral, we treat y as a constant and integrate with respect to x:
∫1< x < e^2 (lny/x) dx = [lny * x]1< x < e^2
= ln(y * e^2) - ln(y)
Now, we will substitute the result of the inner integral into the outer integral:
∫1< y < e [ ln(y * e^2) - ln(y) ] dy
Now we can simplify this expression:
∫1< y < e ln(y * e^2) - ∫1< y < e ln(y) dy
Using the logarithmic properties, we can rewrite this as:
∫1< y < e ln(y) + ln(e^2) - ∫1< y < e ln(y) dy
ln(e^2) = 2ln(e) = 2
Simplifying again, we get:
∫1< y < e 2 dy
Evaluating this integral:
[2y]1< y < e = 2 * e - 2
Therefore, the value of the given double integral ∫R∫ lny/x dA is 2e - 2.