To solve this problem, let's analyze the given information.

Given

1. The function $f(x)$ is an even function. This means that $f(x) = f(-x)$.
2. We know that $\int_{-5}^{-3} f(x) \, dx = 8.$

Goal

We are asked to determine the value of $\int_{-5}^{5} f(x) \, dx$.

Solution

For an even function, the property of integration over symmetric limits applies: $\int_{-a}^{a} f(x) \, dx = 2 \int_{0}^{a} f(x) \, dx.$ Thus, $\int_{-5}^{5} f(x) \, dx = 2 \int_{0}^{5} f(x) \, dx.$

Since we have $\int_{-5}^{-3} f(x) \, dx = 8$, by symmetry of the even function, the integral over $[3, 5]$ will be the same as the integral over $[-5, -3]$: $\int_{3}^{5} f(x) \, dx = 8.$

Therefore, $\int_{-5}^{5} f(x) \, dx = \int_{-5}^{-3} f(x) \, dx + \int_{-3}^{3} f(x) \, dx + \int_{3}^{5} f(x) \, dx.$

We already know that $\int_{-5}^{-3} f(x) \, dx = 8$ and $\int_{3}^{5} f(x) \, dx = 8$. We need to find $\int_{-3}^{3} f(x) \, dx$.

Since $f(x)$ is even, $\int_{-3}^{3} f(x) \, dx = 2 \int_{0}^{3} f(x) \, dx = 2 \times 4 = 16.$