Respuesta :
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.
∫(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.
As per the question, the double integer is given as e^y/x dy dx and have outer limits as 0 and 2 and the inner limits are given as 0 and x^2. The integer is a number that functions and tells us about the elements about the data and tells about the process of integration. Thus the answer from the equation of the data is about 1/2.
The double integer problem.
Thus the assumption of the integers are given as e^(y/x): Then ∫(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 Hence the answer is = 1/2.
Find out more information about the double integral limits.
brainly.com/question/2284804.