Definite Integral Involving Greatest Integer Function

To find the value of $\beta \sin\left(\frac{\beta}{2}\right)$, we split the given integral into two parts:

$I = \int_{-2}^2 (|\sin x| + [x \sin x]) dx = I_1 + I_2$

where $I_1 = \int_{-2}^2 |\sin x| dx \quad \text{and} \quad I_2 = \int_{-2}^2 [x \sin x] dx$

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Step 1: Evaluate $I_1$

The integrand $|\sin x|$ is an even function. Therefore, $I_1 = 2 \int_0^2 |\sin x| dx$

Since $2 < \pi \approx 3.14$, for $x \in [0, 2]$, $\sin x \ge 0$, which implies $|\sin x| = \sin x$. Thus: $I_1 = 2 \int_0^2 \sin x \, dx = 2 [-\cos x]_0^2 = 2(1 - \cos 2) = 2 - 2 \cos 2$

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Step 2: Evaluate $I_2$

Let $g(x) = x \sin x$. Since $g(-x) = (-x)\sin(-x) = x \sin x = g(x)$, the function $g(x)$ is even, which implies that $[g(x)]$ is also an even function. Hence, $I_2 = 2 \int_0^2 [x \sin x] dx$

Now, let us examine the derivative $g'(x) = \sin x + x \cos x$:

• For $x \in (0, \pi/2]$, both $\sin x > 0$ and $\cos x \ge 0$, so $g'(x) > 0$.
• For $x \in (\pi/2, 2]$, $g'(x) = \cos x (\tan x + x)$. Note that $\cos x < 0$ and $x + \tan x < 0$ (since $2 + \tan 2 \approx 2 - 2.185 < 0$ and $x + \tan x$ is increasing), which yields $g'(x) > 0$.

Thus, $g(x)$ is strictly increasing on $[0, 2]$.

Evaluating $g(x)$ at the boundaries:

• $g(0) = 0$
• $g(2) = 2 \sin 2 \approx 2(0.9093) = 1.8186 < 2$

By the Intermediate Value Theorem, there exists a unique $x_0 \in (0, 2)$ such that: $g(x_0) = x_0 \sin x_0 = 1$

Therefore:

• For $x \in [0, x_0)$, $0 \le x \sin x < 1 \implies [x \sin x] = 0$
• For $x \in [x_0, 2]$, $1 \le x \sin x < 2 \implies [x \sin x] = 1$

Now, integrating $[x \sin x]$ over $[0, 2]$: $\int_0^2 [x \sin x] dx = \int_0^{x_0} 0 \, dx + \int_{x_0}^2 1 \, dx = 2 - x_0$

Thus, $I_2 = 2(2 - x_0) = 4 - 2x_0$

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Step 3: Find $\beta$ and evaluate the final expression

Combining $I_1$ and $I_2$: $I = (2 - 2 \cos 2) + (4 - 2x_0) = 6 - 2 \cos 2 - 2x_0$

We are given that $I = 2(3 - \cos 2) + \beta = 6 - 2 \cos 2 + \beta$.

Equating the two expressions for $I$: $\beta = -2x_0$

Now, we substitute $\beta$ into $\beta \sin\left(\frac{\beta}{2}\right)$: $\beta \sin\left(\frac{\beta}{2}\right) = (-2x_0) \sin\left(\frac{-2x_0}{2}\right) = (-2x_0)(-\sin x_0) = 2x_0 \sin x_0$

Since $x_0 \sin x_0 = 1$: $\beta \sin\left(\frac{\beta}{2}\right) = 2(1) = 2$

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Conclusion

The correct option is B.