If f(x) = f(2 - x),then int_{0.5}^{1.5} xf(x) | vedclass

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(B) Let $I = \int_{0.5}^{1.5} xf(x) dx$. Using the property $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx$,we have: $I = \int_{0.5}^{1.5} (0.5 + 1.

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$\int_{0.5}^{1.5} f(x) dx$

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(B) Let $I = \int_{0.5}^{1.5} xf(x) dx$. Using the property $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx$,we have: $I = \int_{0.5}^{1.5} (0.5 + 1.5 - x) f(0.5 + 1.5 - x) dx$ $I = \int_{0.5}^{1.5} (2 - x) f(2 - x) dx$. Since $f(2 - x) = f(x)$,we substitute this into the integral: $I = \int_{0.5}^{1.5} (2 - x) f(x) dx$ $I = 2 \int_{0.5}^{1.5} f(x) dx - \int_{0.5}^{1.5} xf(x) dx$ $I = 2 \int_{0.5}^{1.5} f(x) dx - I$ $2I = 2 \int_{0.5}^{1.5} f(x) dx$ $I = \int_{0.5}^{1.5} f(x) dx$.

If $f(x) = f(2 - x)$,then $\int_{0.5}^{1.5} xf(x) dx$ equals

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