Question

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.

Answer 1

 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. 

Answer 2

Answer:

$\boxed{\boxed{\int\limits^2_0 \int\limits^{x^2}_0 {e^{\frac{y}{x}}} \ dydx=e^2-1}}$

Step-by-step explanation:

The given integral is,

$\int\limits^2_0 \int\limits^{x^2}_0 {e^{\frac{y}{x}}} \ dydx$

Considering the inner integral and treating x as constant,

$=\int\limits^{x^2}_0 {e^{\frac{y}{x}}} \ dy$

$=[{\dfrac{e^{\frac{y}{x}}}{\frac{1}{x}}]_{0}^{x^2}$

$=[xe^{\frac{y}{x}}]_{0}^{x^2}$

$=xe^{\frac{x^2}{x}}-xe^{\frac{0}{x}}$

$=xe^{{x}}-x$

Putting this,

$=\int\limits^2_0 (xe^{{x}}-x)dx$

Computing the indefinite integral first,

$=\int \:(xe^x-x)dx$

$=\int \:xe^xdx-\int \:xdx$

$=xe^x-\int \:e^xdx-\dfrac{x^2}{2}+c$

$=xe^x-e^x-\dfrac{x^2}{2}+c$

Now,

$\int\limits^2_0 (xe^{{x}}-x)dx=[xe^x-e^x-\dfrac{x^2}{2}]_{0}^{2}=e^2-1$