Find the limit $\lim_{n \to \infty} \frac 1 {n^2} \int_0^{n\pi}x \left|\sin x\right|\,\text dx$

Question

Find the limit $$\lim_{n \to \infty} \frac 1 {n^2}\int_0^{n\pi}x\left|\sin x \right| \, \text dx$$

Tried to eliminate the absolute value by dividing the integral into $n$ parts. But it seems become more complex. What is the right way?

Answer 1

$$(k-1)\pi\le x\le k\pi\;\implies \; (k-1)\pi\int_{(k-1)\pi}^{k\pi}|\sin x|\mathrm dx\le \int_{(k-1)\pi}^{k\pi}x|\sin x|\mathrm dx\le k\pi\int_{(k-1)\pi}^{k\pi}|\sin x|\mathrm dx$$ Thus $$2(k-1)\pi\le \int_{(k-1)\pi}^{k\pi}x|\sin x|\mathrm dx \le2k\pi$$ From $$\int_0^{n\pi}x|\sin x|\mathrm dx =\sum_{k=1}^{n}\int_{(k-1)\pi}^{k\pi}x|\sin x|\mathrm dx$$it follows \begin{align*} \sum_{k=1}^{n}2(k-1)\pi&\le \int_0^{n\pi}x|\sin x|\mathrm dx\le\sum_{k=1}^n2k\pi\\[5pt] \frac{(n-1)n\pi}{n^2}&\le\frac1{n^2}\int_0^{n\pi}x|\sin x|\mathrm dx\le\frac{n(n+1)\pi}{n^2} \end{align*}

Answer 2

Consider $n=2m$. Then: $$\begin{align}\int_{0}^{n\pi}x|\sin x|\text dx&=\sum_{k=0}^{m-1} \left(\int_{2k\pi}^{(2k+1)\pi} x\sin x \text dx-\int_{(2k+1)\pi}^{(2k+2)\pi} x\sin x \text dx\right)=\\ &=\sum_{k=0}^{m-1} \left((-x\cos x+\sin x)|_{2k\pi}^{(2k+1)\pi}-(-x\cos x+\sin x)|_{(2k+1)\pi}^{(2k+2)\pi}\right)=\\ &=\sum_{k=0}^{m-1}\left([(2k+1)\pi+2k\pi]-[-(2k+2)\pi-(2k+1)\pi]\right)=\\ &=4\pi\sum_{k=0}^{m-1}\left(2k+1\right)=4\pi m^2.\\ \end{align}$$ Hence: $$\lim_{n \to \infty} \frac{1}{n^2}\int_{0}^{n\pi}x|\sin x|\text dx=\lim_{m \to \infty} \frac{4\pi m^2}{(2m)^2}=\pi.$$ Similarly, the case $n=2m+1$ is considered.