If $f(a-x) = f(x)$, then $\int_0^a xf(x)dx$ is equal to

Given that $f(a-x) = f(x)$, this indicates the function is symmetric about $x = \frac{a}{2}$.

Let $I = \int_0^a xf(x)dx$

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Using the property $\int_0^a g(x)dx = \int_0^a g(a-x)dx$, replace $x$ with $a-x$:

$I = \int_0^a (a-x)f(a-x)dx$

Since $f(a-x) = f(x)$:

$I = \int_0^a (a-x)f(x)dx$

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Expanding the integral:

$I = \int_0^a af(x)dx - \int_0^a xf(x)dx$

$I = a\int_0^a f(x)dx - I$

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Solving for $I$:

$I + I = a\int_0^a f(x)dx$

$2I = a\int_0^a f(x)dx$

$I = \frac{a}{2}\int_0^a f(x)dx$

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Therefore, $\int_0^a xf(x)dx = \frac{a}{2}\int_0^a f(x)dx$