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

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(C) Let $I = \int_{a}^{b} x f(x) dx$  --- $(1)$Using the property $\int_{a}^{b} g(x) dx = \int_{a}^{b} g(a+b-x) dx$,we have:$I = \int_{a}^{b} (a+b-x)

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

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(C) Let $I = \int_{a}^{b} x f(x) dx$  --- $(1)$Using the property $\int_{a}^{b} g(x) dx = \int_{a}^{b} g(a+b-x) dx$,we have:$I = \int_{a}^{b} (a+b-x) f(a+b-x) dx$Since $f(a+b-x) = f(x)$,the integral becomes:$I = \int_{a}^{b} (a+b-x) f(x) dx$Expanding the integral:$I = (a+b) \int_{a}^{b} f(x) dx - \int_{a}^{b} x f(x) dx$Substituting $I$ from equation $(1)$:$I = (a+b) \int_{a}^{b} f(x) dx - I$Adding $I$ to both sides:$2I = (a+b) \int_{a}^{b} f(x) dx$Dividing by $2$:$I = \frac{a+b}{2} \int_{a}^{b} f(x) dx$Thus,the correct option is $C$.

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

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