Step-by-Step Solution

1. Option A: The property \(\int_{a}^{b}f(x)dx=\int_{a}^{b}f(a+b-x)dx\) is a standard property of definite integrals. Let's verify it. Let \(I = \int_{a}^{b}f(x)dx\). Substitute \(x = a+b-t\), so \(dx = -dt\). When \(x = a\), \(t = b\), and when \(x = b\), \(t = a\). Therefore, \(I = \int_{b}^{a}f(a+b-t)(-dt) = -\int_{b}^{a}f(a+b-t)dt = \int_{a}^{b}f(a+b-t)dt = \int_{a}^{b}f(a+b-x)dx\). So, option A is correct.
2. Option B: If \(f(x)\) is an even function, then \(f(-x) = f(x)\). The property \(\int_{-a}^{a}f(x)dx=2\int_{0}^{a}f(x)dx\) holds true for even functions. Therefore, \(\int_{-a}^{a}f(x)dx=0\) is incorrect for even functions.
3. Option C: If \(f(x)\) is an odd function, then \(f(-x) = -f(x)\). The property \(\int_{-a}^{a}f(x)dx=0\) holds true for odd functions. Therefore, \(\int_{-a}^{a}f(x)dx=2\int_{0}^{a}f(x)dx\) is incorrect for odd functions.
4. Option D: The given property is \(\int_{0}^{2a}f(x)dx=\int_{0}^{a}f(x)dx-\int_{0}^{a}f(2a+x)dx\). Let's consider the property \(\int_{0}^{2a}f(x)dx = \int_{0}^{a}f(x)dx + \int_{a}^{2a}f(x)dx\). Let \(x = 2a - t\) in the second integral, so \(dx = -dt\). When \(x = a\), \(t = a\), and when \(x = 2a\), \(t = 0\). So, \(\int_{a}^{2a}f(x)dx = \int_{a}^{0}f(2a-t)(-dt) = \int_{0}^{a}f(2a-t)dt = \int_{0}^{a}f(2a-x)dx\). Therefore, \(\int_{0}^{2a}f(x)dx = \int_{0}^{a}f(x)dx + \int_{0}^{a}f(2a-x)dx\). The given option has a minus sign instead of a plus sign and \(f(2a+x)\) instead of \(f(2a-x)\), so it is incorrect.