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. 

Im sorry i cant answer it : ( im so sorry