If f(a + b - x) = f(x),then int_{a}^{b} x cdo | vedclass

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(C) Let $I = \int_{a}^{b} x \cdot f(a + b - x) \, dx$. Using the property $\int_{a}^{b} g(x) \, dx = \int_{a}^{b} g(a + b - x) \, dx$,we get: $I = \in

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$\frac{a + b}{2} \int_{a}^{b} f(x) \, dx$

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(C) Let $I = \int_{a}^{b} x \cdot f(a + b - x) \, dx$. Using the property $\int_{a}^{b} g(x) \, dx = \int_{a}^{b} g(a + b - x) \, dx$,we get: $I = \int_{a}^{b} (a + b - x) \cdot f(a + b - (a + b - x)) \, dx$ $I = \int_{a}^{b} (a + b - x) \cdot f(x) \, dx$ $I = (a + b) \int_{a}^{b} f(x) \, dx - \int_{a}^{b} x \cdot f(x) \, dx$. Since $f(a + b - x) = f(x)$,the original integral is $I = \int_{a}^{b} x \cdot f(x) \, dx$. Substituting this into the equation: $I = (a + b) \int_{a}^{b} f(x) \, dx - I$ $2I = (a + b) \int_{a}^{b} f(x) \, dx$ $I = \frac{a + b}{2} \int_{a}^{b} f(x) \, dx$.

If $f(a + b - x) = f(x)$,then $\int_{a}^{b} x \cdot f(a + b - x) \, dx = $

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