If f(x) = f(Pi + e - x) and int_{e}^{Pi} f(x) | vedclass

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(D) Let $I = \int_{e}^{\Pi} x f(x) dx$.Using the property $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a + b - x) dx$,we have:$I = \int_{e}^{\Pi} (e + \Pi -

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$1$

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(D) Let $I = \int_{e}^{\Pi} x f(x) dx$.Using the property $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a + b - x) dx$,we have:$I = \int_{e}^{\Pi} (e + \Pi - x) f(e + \Pi - x) dx$.Since $f(e + \Pi - x) = f(x)$,this becomes:$I = \int_{e}^{\Pi} (e + \Pi - x) f(x) dx = (e + \Pi) \int_{e}^{\Pi} f(x) dx - \int_{e}^{\Pi} x f(x) dx$.$I = (e + \Pi) \left( \frac{2}{e + \Pi} \right) - I$.$2I = 2$.$I = 1$.

If $f(x) = f(\Pi + e - x)$ and $\int_{e}^{\Pi} f(x) dx = \frac{2}{e + \Pi}$,then $\int_{e}^{\Pi} x f(x) dx$ is equal to

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