Question

${\int }_0^1|x\sin \pi x|dx$

Answer 1

$\mathrm{For}0<x<1,x>0\mathrm{and}\mathrm{sin\pi }x>0\Rightarrow x\mathrm{sin\pi }x>0$
$$\begin{gathered} \therefore {\int }_0^1\left|x\mathrm{sin\pi }x\right|dx={\int }_0^{1}x\mathrm{sin\pi }xdx \\ \mathrm{Let}I=\int x\mathrm{sin\pi }xdx \\ =x\int \mathrm{sin\pi }xdx-\int \left(\frac{d}{dx}x\int \mathrm{sin\pi }xdx\right)dx \\ =x\left(\frac{-\mathrm{cos\pi }x}{\mathrm{\pi }}\right)-\int \left(\frac{-\mathrm{cos\pi }x}{\mathrm{\pi }}\right)dx \end{gathered}$$
$=\frac{-x\mathrm{cos\pi }x}{\mathrm{\pi }}+\frac{\mathrm{sin\pi }x}{{\mathrm{\pi }}^2}$

Applying the limits, we get

$$\begin{gathered} {\int }_0^1\left|x\mathrm{sin\pi }x\right|dx={\overline{)\left(\frac{-x\mathrm{cos\pi }x}{\mathrm{\pi }}+\frac{\mathrm{sin\pi }x}{{\mathrm{\pi }}^2}\right)}}_0^1 \\ =\left(\frac{-\mathrm{cos\pi }}{\mathrm{\pi }}+\frac{\mathrm{sin\pi }}{{\mathrm{\pi }}^2}\right)-\left(0+0\right) \end{gathered}$$
$$\begin{gathered} =\frac{1}{\mathrm{\pi }}+0-0 \\ =\frac{1}{\mathrm{\pi }} \end{gathered}$$