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

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(A) Given $R_1 = \int_{-1}^2 x f(x) dx$  $(i)$Using the property $\int_a^b f(x) dx = \int_a^b f(a+b-x) dx$,where $a=-1$ and $b=2$,we have $a+b = 1$.So

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$2 R_1=R_2$

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(A) Given $R_1 = \int_{-1}^2 x f(x) dx$  $(i)$Using the property $\int_a^b f(x) dx = \int_a^b f(a+b-x) dx$,where $a=-1$ and $b=2$,we have $a+b = 1$.So,$R_1 = \int_{-1}^2 (1-x) f(1-x) dx$.Since $f(1-x) = f(x)$,this becomes $R_1 = \int_{-1}^2 (1-x) f(x) dx = \int_{-1}^2 f(x) dx - \int_{-1}^2 x f(x) dx$.$R_1 = \int_{-1}^2 f(x) dx - R_1$.$2 R_1 = \int_{-1}^2 f(x) dx$.Since $R_2$ is the area bounded by $y=f(x)$,$x=-1$,$x=2$ and the $X$-axis,$R_2 = \int_{-1}^2 f(x) dx$.Therefore,$2 R_1 = R_2$.

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

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