Step-by-Step Solution

Given \(f(2a-x) = f(x)\), we need to evaluate \(\int_{0}^{2a} f(x) dx\).

We can use the property of definite integrals: \(\int_{0}^{2a} f(x) dx = \int_{0}^{a} f(x) dx + \int_{a}^{2a} f(x) dx\).

Now, let's substitute \(x = 2a - t\) in the second integral, so \(dx = -dt\). When \(x = a\), \(t = a\), and when \(x = 2a\), \(t = 0\). Thus, \(\int_{a}^{2a} f(x) dx = \int_{a}^{0} f(2a - t) (-dt) = -\int_{a}^{0} f(2a - t) dt = \int_{0}^{a} f(2a - t) dt\).

Since \(f(2a - x) = f(x)\), we have \(f(2a - t) = f(t)\). Therefore, \(\int_{0}^{a} f(2a - t) dt = \int_{0}^{a} f(t) dt = \int_{0}^{a} f(x) dx\).

So, \(\int_{0}^{2a} f(x) dx = \int_{0}^{a} f(x) dx + \int_{0}^{a} f(x) dx = 2\int_{0}^{a} f(x) dx\).