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Let $\mathrm{f}:[-1,2] \rightarrow[0, \infty)$ be a continuous function such that $\mathrm{f}(x)=\mathrm{f}(1-x), \forall x \in[-1,2]$
Let $\mathrm{R}_1=\int_{-1}^2 x \mathrm{f}(x) \mathrm{d} x$ and $\mathrm{R}_2$ be the area of the region bounded by $y=\mathrm{f}(x), x=-1, x=2$ and the $\mathrm{X}$-axis, then $\mathrm{R}_2$ is
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Let $\mathrm{R}_1=\int_{-1}^2 x \mathrm{f}(x) \mathrm{d} x$ and $\mathrm{R}_2$ be the area of the region bounded by $y=\mathrm{f}(x), x=-1, x=2$ and the $\mathrm{X}$-axis, then $\mathrm{R}_2$ is
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Verified Answer
The correct answer is:
$2 R_1$
Given that $\mathrm{f}(x)=\mathrm{f}(1-x)$ and $\mathrm{R}_1=\int_{-1}^2 x \mathrm{f}(x) \mathrm{d} x$
$\begin{aligned}
& \therefore \quad \mathrm{R}_1=\int_{-1}^2(1-x) \mathrm{f}(1-x) \mathrm{d} x \\
& \quad \cdots\left[\because \int_{\mathrm{a}}^{\mathrm{b}} \mathrm{f}(x) \mathrm{d} x=\int_1^b \mathrm{f}(\mathrm{a}+\mathrm{b}-x) \mathrm{d} x\right]
\end{aligned}$
$\mathrm{R}_1=\int_{-1}^2 \mathrm{f}(x) \mathrm{d} x-\int_{-1}^2 x \mathrm{f}(x) \mathrm{d} x$
... $[\because f(x)=f(1-x)]$
$\therefore \quad \mathrm{R}_1=\int_{-1}^2 \mathrm{f}(x) \overline{\mathrm{d} x}-\mathrm{R}_1$
$\therefore \quad 2 \mathrm{R}_1=\int_{-1}^2 \mathrm{f}(x) \mathrm{d} x$
Note that $\mathrm{R}_2=\int_{-1}^2 \mathrm{f}(x) \mathrm{d} x=2 \mathrm{R}_1$
$\begin{aligned}
& \therefore \quad \mathrm{R}_1=\int_{-1}^2(1-x) \mathrm{f}(1-x) \mathrm{d} x \\
& \quad \cdots\left[\because \int_{\mathrm{a}}^{\mathrm{b}} \mathrm{f}(x) \mathrm{d} x=\int_1^b \mathrm{f}(\mathrm{a}+\mathrm{b}-x) \mathrm{d} x\right]
\end{aligned}$
$\mathrm{R}_1=\int_{-1}^2 \mathrm{f}(x) \mathrm{d} x-\int_{-1}^2 x \mathrm{f}(x) \mathrm{d} x$
... $[\because f(x)=f(1-x)]$
$\therefore \quad \mathrm{R}_1=\int_{-1}^2 \mathrm{f}(x) \overline{\mathrm{d} x}-\mathrm{R}_1$
$\therefore \quad 2 \mathrm{R}_1=\int_{-1}^2 \mathrm{f}(x) \mathrm{d} x$
Note that $\mathrm{R}_2=\int_{-1}^2 \mathrm{f}(x) \mathrm{d} x=2 \mathrm{R}_1$
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