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*}