$\int_{3}^{5} \int_{1}^{\sqrt{x}} 2 y e^{-x} d y d x$

Step 1 ：Given the double integral problem: \(\int_{3}^{5} \int_{1}^{\sqrt{x}} 2 y e^{-x} d y d x\). The inner integral is with respect to y and the outer integral is with respect to x. The function to be integrated is \(2 y e^{-x}\). The limits of the inner integral are from 1 to \(\sqrt{x}\) and the limits of the outer integral are from 3 to 5.

Step 2 ：First, solve the inner integral, treating x as a constant. The function to be integrated becomes \(2 y e^{-x}\).

Step 3 ：The result of the inner integral is \(x e^{-x} - e^{-x}\).

Step 4 ：Next, substitute the limits of the inner integral into the result. This gives us the function to be integrated for the outer integral.

Step 5 ：Then, solve the outer integral, treating the result of the inner integral as the function to be integrated.

Step 6 ：The result of the outer integral is \(-5 e^{-5} + 3 e^{-3}\).

Step 7 ：Finally, substitute the limits of the outer integral into the result to get the final answer.

Step 8 ：The final answer is \(\boxed{-5 e^{-5} + 3 e^{-3}}\)