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

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(B) Given $f(x) = f(1-x)$ and $R_1 = \int_{-1}^2 x f(x) dx$. Using the property $\int_{a}^b g(x) dx = \int_{a}^b g(a+b-x) dx$,where $a = -1$ and $b =

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$2 R_1$

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(B) Given $f(x) = f(1-x)$ and $R_1 = \int_{-1}^2 x f(x) dx$. Using the property $\int_{a}^b g(x) dx = \int_{a}^b g(a+b-x) dx$,where $a = -1$ and $b = 2$,we have $a+b = 1$. Thus,$R_1 = \int_{-1}^2 (1-x) f(1-x) dx$. Since $f(1-x) = f(x)$,we get $R_1 = \int_{-1}^2 (1-x) f(x) dx$. $R_1 = \int_{-1}^2 f(x) dx - \int_{-1}^2 x f(x) dx$. $R_1 = \int_{-1}^2 f(x) dx - R_1$. $2 R_1 = \int_{-1}^2 f(x) dx$. 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,$R_2 = 2 R_1$.

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 $R_2$ is:

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