Let int_{-2}^{2} (|sin x| + [x sin x]) dx = 2 | vedclass

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(B) Let $I = \int_{-2}^2 (|\sin x| + [x \sin x]) dx$. Since $|\sin x|$ is an even function, $\int_{-2}^2 |\sin x| dx = 2 \int_0^2 \sin x dx = 2 [-\cos

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$2$

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(B) Let $I = \int_{-2}^2 (|\sin x| + [x \sin x]) dx$. Since $|\sin x|$ is an even function, $\int_{-2}^2 |\sin x| dx = 2 \int_0^2 \sin x dx = 2 [-\cos x]_0^2 = 2(1 - \cos 2)$. Now, consider $f(x) = x \sin x$. Since $f(x)$ is even, $\int_{-2}^2 [x \sin x] dx = 2 \int_0^2 [x \sin x] dx$. For $x \in [0, 2]$, $x \sin x$ increases from $0$ to $2 \sin 2 \approx 1.818$. Thus, $[x \sin x] = 0$ for $x \in [0, x_0)$ where $x_0 \sin x_0 = 1$, and $[x \sin x] = 1$ for $x \in [x_0, 2]$. So, $2 \int_0^2 [x \sin x] dx = 2 \int_{x_0}^2 1 dx = 2(2 - x_0) = 4 - 2x_0$. Thus, $I = 2(1 - \cos 2) + 4 - 2x_0 = 2 - 2\cos 2 + 4 - 2x_0 = 2(3 - \cos 2) - 2x_0$. Comparing this with $2(3 - \cos 2) + \beta$, we get $\beta = -2x_0$. Given $x_0 \sin x_0 = 1$, we find $x_0 \approx 1.11$. However, evaluating the expression $\beta \sin(\frac{\beta}{2})$ with $\beta = -2x_0$ where $x_0 \sin x_0 = 1$, we have $\beta = -2x_0$. Then $\beta \sin(\frac{\beta}{2}) = -2x_0 \sin(-x_0) = 2x_0 \sin x_0 = 2(1) = 2$.

Let $\int_{-2}^{2} (|\sin x| + [x \sin x]) dx = 2(3 - \cos 2) + \beta$, where $[\cdot]$ denotes the greatest integer function. Then $\beta \sin(\frac{\beta}{2})$ equals:

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