To determine which of the following integrals are equal to $\int_2^4 f(x) \, dx$, given that $f$ is an odd function, let's review some properties of odd functions and definite integrals.

1. Odd Function Property: If $f(x)$ is odd, then $f(-x) = -f(x)$.
2. Integral Symmetry Property for Odd Functions: For any odd function $f(x)$, the integral from $-a$ to $a$ is zero: $\int_{-a}^a f(x) \, dx = 0$
3. Integral with Negative Limits: For an odd function, if you reverse the limits of integration (e.g., from $a$ to $-b$), you negate the value of the integral: $\int_{-a}^b f(x) \, dx = -\int_a^b f(x) \, dx$

Given $\int_2^4 f(x) \, dx$, let's analyze the options:

Options Analysis:

1. $\int_2^4 f(x) \, dx$ — This is the original integral, so it is equal to itself.

2. $\int_{-2}^4 f(x) \, dx$ — This integral spans from $-2$ to $4$, and we can split it as: $\int_{-2}^4 f(x) \, dx = \int_{-2}^2 f(x) \, dx + \int_2^4 f(x) \, dx$ Since $f(x)$ is odd, $\int_{-2}^2 f(x) \, dx = 0$. Thus: $\int_{-2}^4 f(x) \, dx = \int_2^4 f(x) \, dx$ So, this option is equal to $\int_2^4 f(x) \, dx$.

3. $\int_{-4}^{-2} f(x) \, dx$ — This integral is in the range from $-4$ to $-2$, so it doesn’t directly relate to the interval $2$ to $4$, and there’s no direct relationship here that would make it equal to $\int_2^4 f(x) \, dx$.

4. $\int_{-4}^2 f(x) \, dx$ — Similar to option 2, we can split it: $\int_{-4}^2 f(x) \, dx = \int_{-4}^{-2} f(x) \, dx + \int_{-2}^2 f(x) \, dx$ Again, ( \int_{-2}^2 f(x) \