Let f:[-1, 2] rightarrow [0, infty) be a cont | vedclass

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(C) Given $R_1 = \int_{-1}^2 x f(x) dx$. Using the property $\int_a^b g(x) dx = \int_a^b g(a+b-x) dx$,we have: $R_1 = \int_{-1}^2 ((-1) + 2 - x) f((-1

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$2 R_1 = R_2$

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(C) Given $R_1 = \int_{-1}^2 x f(x) dx$. Using the property $\int_a^b g(x) dx = \int_a^b g(a+b-x) dx$,we have: $R_1 = \int_{-1}^2 ((-1) + 2 - x) f((-1) + 2 - x) dx$ $R_1 = \int_{-1}^2 (1 - x) f(1 - x) dx$. Since $f(x) = f(1 - x)$,we substitute this into the integral: $R_1 = \int_{-1}^2 (1 - x) f(x) dx = \int_{-1}^2 f(x) dx - \int_{-1}^2 x f(x) dx$. $R_1 = \int_{-1}^2 f(x) dx - R_1$. $2 R_1 = \int_{-1}^2 f(x) dx$. Since $R_2$ is the area bounded by $y = f(x)$ from $x = -1$ to $x = 2$,$R_2 = \int_{-1}^2 f(x) dx$. Therefore,$2 R_1 = R_2$.

Let $f:[-1, 2] \rightarrow [0, \infty)$ be a continuous function such that $f(x) = f(1-x)$ for all $x \in [-1, 2]$. Let $R_1 = \int_{-1}^2 x f(x) dx$,and $R_2$ be the area of the region bounded by $y = f(x)$,$x = -1$,$x = 2$,and the $x$-axis. Then

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