We are given that $f(x)$ is an even function defined in the interval $[-2, 2]$, meaning $f(x) = f(-x)$. Based on this, we need to analyze the integrals and select the correct relation from the options.

Key properties of even functions:

For an even function $f(x)$, the following integral property holds: $\int_{-a}^{a} f(x) \, dx = 2 \int_0^{a} f(x) \, dx$ This is because the area on the interval $[0, a]$ is the same as on $[-a, 0]$ due to the symmetry of even functions.

Analysis of the options:

Option a:

$\int_0^2 f(x) \, dx = \frac{1}{2} \int_{-2}^2 f(x) \, dx$ Using the property of even functions, we know: $\int_{-2}^2 f(x) \, dx = 2 \int_0^2 f(x) \, dx$ If we divide both sides by 2, we get: $\int_0^2 f(x) \, dx = \frac{1}{2} \int_{-2}^2 f(x) \, dx$ This matches exactly with option a, making this the correct choice.

Option b:

It states that there is no relationship between the integrals, which is incorrect, as we have already found a clear relationship in option a.

Option c:

$\int_0^2 f(x) \, dx = -\int_{-2}^0 f(x) \, dx$ This does not hold for an even function. In fact, for an even function: $\int_{-2}^0 f(x) \, dx = \int_0^2 f(x) \, dx$ Therefore, option c is incorrect.

Conclusion:

The correct answer is a.

Do you want more details on how the integrals of even functions are derived?

Here are 5 related questions to further your understanding:

1. How would the result change if $f(x)$ were an odd function?
2. What is the relationship between integrals of odd functions over symmetric intervals?
3. Can we generalize this result for any even function on a symmetric interval $[-a, a]$?
4. How can we interpret the integral of an even function geometrically?
5. How do definite integrals behave when the limits of integration are swapped?

Tip: Always check the symmetry properties of functions before integrating over symmetric intervals.