Let $$f:[-1,2] \rightarrow[0, \infty)$$ 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

$$\int\limits_0^\pi \frac{d x}{4+3 \cos x}=$$

Let $$\mathrm{f}(x)=\int \frac{\sqrt{x}}{(1+x)^2} \mathrm{~d} x, x \geq 0$$, then $$\mathrm{f}(3)-\mathrm{f}(1)$$ is equal to

Let $$\mathrm{f}(x)$$ be positive for all real $$x$$. If $$\mathrm{I}_1=\int_\limits{1-\mathrm{h}}^{\mathrm{h}} x \mathrm{f}(x(1-x)) \mathrm{d} x$$ and $$\mathrm{I}_2=\int_\limits{1-\mathrm{h}}^{\mathrm{h}} \mathrm{f}(x(1-x)) \mathrm{d} x$$, where $$(2 h-1)>0$$, then $$\frac{I_1}{I_2}$$ is