(D) Let $I = \int_2^3 x f(x) dx$. Using the property $\int_a^b g(x) dx = \int_a^b g(a+b-x) dx$,we have: $I = \int_2^3 (2+3-x) f(2+3-x) dx = \int_2^3 (

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(D) Let $I = \int_2^3 x f(x) dx$. Using the property $\int_a^b g(x) dx = \int_a^b g(a+b-x) dx$,we have: $I = \int_2^3 (2+3-x) f(2+3-x) dx = \int_2^3 (5-x) f(5-x) dx$. Given $f(5-x) = f(x)$,we substitute this into the integral: $I = \int_2^3 (5-x) f(x) dx = 5 \int_2^3 f(x) dx - \int_2^3 x f(x) dx$. Since $\int_2^3 f(x) dx = 2$,we get: $I = 5(2) - I$. $2I = 10$. $I = 5$.