Evaluate.
\[
\int_{0}^{5} \int_{-3}^{-1}(4 x+y) d x d y
\]
$\int_{0}^{5} \int_{-3}^{-1}(4 x+y) d x d y=\square$ (Simplify your answer.)

Step 1 ：First, we solve the inner integral by treating y as a constant. The integral of \(4x+y\) with respect to x is \(2x^2+xy\).

Step 2 ：Next, we substitute the limits of x which are -3 and -1 into the expression \(2x^2+xy\). This gives us \(2*(-1)^2+(-1)*y - (2*(-3)^2+(-3)*y) = 2*y - 16\).

Step 3 ：Then, we solve the outer integral by integrating the expression \(2*y - 16\) with respect to y. The integral of \(2*y - 16\) with respect to y is \(y^2 - 16y\).

Step 4 ：Finally, we substitute the limits of y which are 0 and 5 into the expression \(y^2 - 16y\). This gives us \(5^2 - 16*5 - (0^2 - 16*0) = -55\).

Step 5 ：So, the value of the double integral \(\int_{0}^{5} \int_{-3}^{-1}(4 x+y) d x d y\) is \(\boxed{-55}\).