I_1+I_2 where _-2^2(| x|+[x x])dx = 2(3- 2)+ ([] greatest integer function).

Question

$I_1+I_2$ where $\displaystyle\int_{-2}^{2}\left(|\sin x|+[x\sin x]\right)dx = 2(3-\cos 2)+\beta$ ($[\cdot]$ greatest integer function). The value of $\beta\sin\left(\dfrac{\beta}{2}\right)$ equals:

Accepted Answer

Split $I = \displaystyle\int_{-2}^2|\sin x|\,dx + \int_{-2}^2[x\sin x]\,dx = I_1+I_2$.

$I_1$: $|\sin x|$ is even, so $I_1 = 2\int_0^2\sin x\,dx = 2(1-\cos 2) = 2-2\cos2$ (since $\sin x>0$ on $[0,2]$ as $2<\pi$).

$I_2$: $f(x)=x\sin x$ is even, so $I_2 = 2\int_0^2[x\sin x]\,dx$. On $[0,2]$, $f$ increases from $0$ to $2\sin2\approx1.82$ (between $1$ and $2$), so there's a unique $\alpha\in(0,2)$ with $\alpha\sin\alpha=1$. For $x\in[0,\alpha)$: $[x\sin x]=0$; for $x\in[\alpha,2]$: $[x\sin x]=1$.

$$I_2 = 2\left(\int_0^\alpha 0\,dx+\int_\alpha^2 1\,dx\right) = 2(2-\alpha) = 4-2\alpha$$

So $I = (2-2\cos2)+(4-2\alpha) = 2(3-\cos2) - 2\alpha$, giving $\beta = -2\alpha$.

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

Solutions · 1