Solve the double integral  `int_1^2 int_0^x e^(y/x) dy dx`  

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lemjay eNotes educator| Certified Educator

`int_1^2 int_0^x e^(y/x) dy dx`

To solve, start with the inner integral.

>> `int_0^x e^(y/x) dy` 

Since we have dy, treat x as a constant. To integrate, use                                 u-substitution method. Let,

`u = y/x`

Differentiate u with respect to y.

`du = 1/x dy`

`xdu =dy`

Then, replace y/x with u and dy with xdu.

`int e^u xdu = x int e^u du = xe^u + C`

Substitute back u=y/x to be able to evaluate limits of the integral. So we have,

>>` int_0^x e^(y/x) dy = xe^(y/x) |_0^x = xe^(x/x) - xe^(0/x)`

     `= xe^1-x = xe-x = x(e-1)`

Then, x(e-1) becomes the integrand of the outer integral.

> `int_1^2 int_0^x e^(y/x) dy dx = int_1^2 x(e-1) dx`

  `= (e-1)int_1^2 x dx = (e-1)x^2/2 |_1^2 = (e-1)/2 x^2|_1^2`

 `= (e-1)/2*(2^2-1^2) = (e-1)/2*3 =(3(e-1))/2`

Hence, `int_1^2 int_0^x e^(y/x) dy dx = (3(e-1))/2` .

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