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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`
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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