Question

Sol. [Key Concept: Split the integral; the floor part $\lbrack xsin\ x\rbrack$ is $0$ except where $xsin\ x = 1$.]

$${\int_{- 2}^{2}\mspace{2mu}(|sinx| + \lbrack xsinx\rbrack)dx }{\Rightarrow 2\int_{0}^{2}\mspace{2mu}(|sinx| + \lbrack xsinx\rbrack)dx }{\Rightarrow 2\left\lbrack \int_{0}^{2}\mspace{2mu} sinxdx + \int_{0}^{2}\mspace{2mu}\lbrack xsinx\rbrack dx \right\rbrack }{2\left( \lbrack - cosx\rbrack_{0}^{2} + \int_{0}^{\alpha}\mspace{2mu} 0dx + \int_{\alpha}^{2}\mspace{2mu} 2dx \right) }{2(3 - cos2 - 2 - \alpha) }{2(3 - cos2) - 2\alpha }$$where $\alpha$ is Root of $xsinx = 1$

Now comparing $\beta = - 2\alpha$

$$\begin{matrix} \alpha = - \frac{\beta}{2} \\ - \frac{\beta}{2}sin\left( \frac{- \beta}{2} \right) = 1 \\ \beta sin\left( \frac{\beta}{2} \right) = 2 \end{matrix}$$