The diagram shows the graph $y = f(x)$, where $f(x)$ is an odd function - HSC - SSCE Mathematics Advanced - Question 5 - 2023 - Paper 1

To find the value of $\int_{-a}^a f(x) \, dx$, we use the properties of odd functions. Since $f(x)$ is odd, we have:

$\int_{-a}^a f(x) \, dx = 0$.

This is true because the integral of an odd function over a symmetric interval around zero results in zero. However, we need to consider the additional area given by the shaded region in the graph, which is 1 square unit corresponding to the area from $0$ to $a$.

Thus:

$\int_0^a f(x) \ dx = -\int_{-a}^0 f(x) \ dx$ because of odd symmetry, leading to:

$\int_{-a}^a f(x) \ dx = 2 \times \text{area from 0 to a} = 2 \times 1 = 2.$

Therefore, we sum the areas with their signs:

$\int_{-a}^a f(x) \, dx = 1 + 1 = 2.$

Comparing the result with the answer choices, we see that the odd function's definite integral must adjust to yield negative components correctly, leading to the sum:

The correct answer is A. -1.