Definite Integral of Derivative Function

To evaluate the definite integral $\int_{-1}^1 g'(x) \, dx$, we use the Fundamental Theorem of Calculus:

$\int_{-1}^1 g'(x) \, dx = \left[ g(x) \right]_{-1}^1 = g(1) - g(-1)$

From the definition of the function $g(x)$ (which is an odd function, satisfying $g(-x) = -g(x)$ for all $x$ in its domain):

$g(-1) = -g(1)$

Substituting this relation into the evaluated integral:

$\int_{-1}^1 g'(x) \, dx = g(1) - (-g(1)) = g(1) + g(1) = 2g(1)$

Thus, the correct option is (D).