Final answer:
The integral of the odd function f(x) from -2 to 2 cancels out, leaving only the integral of the constant 6, which when evaluated from -2 to 2 is 24. Therefore, the integral sought is 24.
Step-by-step explanation:
The student's question pertains to properties of odd and even functions and how they apply to definite integrals. Specifically, the function f mentioned in the question is an odd function because it satisfies the condition f(-x) = -f(x) for all x.
Firstly, it's given that ∫[0 to 2] f(x) dx = 5. This integral represents the area under the curve of f(x) from 0 to 2. However, since f(x) is an odd function, the integral from -2 to 0 of f(x) would yield an area of -5, because the areas on the negative side of the y-axis are the equal in magnitude but opposite in sign to the areas on the positive side.
Next, we are asked to determine the value of ∫[-2 to 2] (f(x) + 6) dx. We can split this integral into two parts: ∫[-2 to 0] (f(x) + 6) dx + ∫[0 to 2] (f(x) + 6) dx. The integral of f(x) from -2 to 0 is -5 (from the odd function property), and the integral from 0 to 2 of f(x) is 5 (as given). Integrating 6 over the interval [-2, 2] simply yields 6 times the length of the interval, which is 24.
Combining these results, the first part (-5) cancels out with the f(x) from 0 to 2 (5), and we are left with the integral of 6, which is 24. Hence, ∫[-2 to 2] (f(x) + 6) dx = 24, or answer choice (C).