Given the integral:

$\int_{-2}^{5} g(x) \, dx = 8$

We need to find the value of the integral:

$\int_{-2}^{5} (g(x) + 6) \, dx$

We can separate the integral into two parts:

$\int_{-2}^{5} (g(x) + 6) \, dx = \int_{-2}^{5} g(x) \, dx + \int_{-2}^{5} 6 \, dx$

We already know that:

$\int_{-2}^{5} g(x) \, dx = 8$

Now, let's calculate the second integral:

$\int_{-2}^{5} 6 \, dx$

Since $6$ is a constant, the integral simplifies to:

$6 \int_{-2}^{5} 1 \, dx = 6 \times \left(5 - (-2)\right) = 6 \times 7 = 42$

So, adding the results together:

$\int_{-2}^{5} (g(x) + 6) \, dx = 8 + 42 = 50$

The value of the integral is 50.

Would you like more details or have any questions?

Here are some related questions:

1. How do you find the integral of a constant function over an interval?
2. What properties of integrals allow us to split integrals into sums?
3. How does adding a constant to a function affect its integral?
4. Can you generalize this process to any interval and function?
5. What happens if the constant is negative instead of positive?

Tip: When integrating a sum of functions, you can always split the integral into the sum of integrals for easier calculation.