If f(x) = f(a-x), then int_0^a x f(x) dx is e | vedclass

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(C) Let $I = \int_0^a x f(x) dx$. Using the property $\int_0^a g(x) dx = \int_0^a g(a-x) dx$, we get: $I = \int_0^a (a-x) f(a-x) dx$. Since $f(a-x) =

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

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(C) Let $I = \int_0^a x f(x) dx$. Using the property $\int_0^a g(x) dx = \int_0^a g(a-x) dx$, we get: $I = \int_0^a (a-x) f(a-x) dx$. Since $f(a-x) = f(x)$, this becomes: $I = \int_0^a (a-x) f(x) dx = \int_0^a a f(x) dx - \int_0^a x f(x) dx$. $I = a \int_0^a f(x) dx - I$. Adding $I$ to both sides, we get: $2I = a \int_0^a f(x) dx$. Therefore, $I = \frac{a}{2} \int_0^a f(x) dx$.

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

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