Solve the double integral of e^y/x dy dx with outer limits as 0 and 2 and inner limits as 0 and x^2.

lets say integrand is e^ (y/x):

∫ (x = 0 to 1) ∫ (y

= 0 to x^2) e^ (y/x) dy dx

= ∫ (x = 0 to 1) xe^ (y/x)

= ∫ (x = 0 to 1) x (e^x - 1) dx

= ∫ (x = 0 to 1) (xe^x - x) dx

= (xe^x - e^x) - (1/2) x^2

= 1/2.