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

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`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` .**