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

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 \(\mathrm{R}_2=\int_{-1}^2 \mathrm{f}(x) \mathrm{d} x=2 \mathrm{R}_1\)