$\int_{1}^{5} \int_{x}^{0}(2 y-8 x) d y d x$

Step 1 ：First, we integrate with respect to y: \(\int_{1}^{5} \int_{x}^{0}(2 y-8 x) d y d x = \int_{1}^{5} [y^2 - 8xy]_{x}^{0} dx\)

Step 2 ：This simplifies to: \(\int_{1}^{5} [0 - x^2 + 8x^2] dx = \int_{1}^{5} 7x^2 dx\)

Step 3 ：Next, we integrate with respect to x: \(\int_{1}^{5} 7x^2 dx = [ \frac{7}{3}x^3]_{1}^{5}\)

Step 4 ：This simplifies to: \( \frac{7}{3}(5^3) - \frac{7}{3}(1^3) = \frac{7}{3}(125 - 1) = \frac{7}{3}(124) = 868/3\)

Step 5 ：So, the value of the double integral \(\int_{1}^{5} \int_{x}^{0}(2 y-8 x) d y d x\) is \(868/3\)

Step 6 ：Finally, we check the result. The result is a number, so it meets the requirements of the problem. The final answer is \(\boxed{\frac{868}{3}}\)