Question

Let \(\int_{-2}^{2} (|\sin x| + |\cos x|) \, dx = 2(3 - \cos 2) + \beta\). Then \(\beta \sin \left( \frac{\beta}{2} \right)\) equals:

• A. 1
• B. 2
• C. 4
• D. 8

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
We evaluate the integral in two parts: \(\int_{-2}^{2} |\sin x| dx\) and \(\int_{-2}^{2} [x \sin x] dx\).
Both \(|\sin x|\) and \(x \sin x\) are even functions, allowing us to halve the interval and double the integral.
Step 2: Key Formula or Approach:
For an even function \(f(x)\), \(\int_{-a}^{a} f(x) dx = 2 \int_0^a f(x) dx\).
For the greatest integer function \([y]\), we must find the intervals where \(y\) lies between consecutive integers.
Step 3: Detailed Explanation:
Let \(I_1 = \int_{-2}^{2} |\sin x| dx\). Since \(\sin x \ge 0\) on \([0, 2]\) (as \(2<\pi\)):
\[ I_1 = 2 \int_0^2 \sin x dx = 2[-\cos x]_0^2 = 2(-\cos 2 - (-1)) = 2(1 - \cos 2) = 2 - 2\cos 2 \] Let \(I_2 = \int_{-2}^{2} [x \sin x] dx = 2 \int_0^2 [x \sin x] dx\).
Analyze \(f(x) = x \sin x\) on \([0, 2]\):
\(f'(x) = \sin x + x \cos x\). For \(x \in (0, 2)\), \(f'(x)>0\), meaning \(f(x)\) is strictly increasing.
At \(x = 0\), \(f(0) = 0\).
At \(x = 2\), \(f(2) = 2 \sin 2 \approx 2(0.909) = 1.818\).
Since the maximum value is strictly between 1 and 2, \(f(x)\) crosses the integer 1 exactly once.
Let this point be \(x_1\), such that \(x_1 \sin x_1 = 1\).
Thus, \([x \sin x] = 0\) for \(x \in [0, x_1)\), and \([x \sin x] = 1\) for \(x \in [x_1, 2]\).
\[ \int_0^2 [x \sin x] dx = \int_0^{x_1} 0 \, dx + \int_{x_1}^2 1 \, dx = 2 - x_1 \] \[ I_2 = 2(2 - x_1) = 4 - 2x_1 \] Combine the two parts:
\[ \text{Total Integral} = I_1 + I_2 = (2 - 2\cos 2) + (4 - 2x_1) = 6 - 2\cos 2 - 2x_1 = 2(3 - \cos 2) - 2x_1 \] According to the problem, this evaluates to \(2(3 - \cos 2) + \beta\).
Equating the two expressions gives \(\beta = -2x_1\).
Step 4: Final Answer:
We are asked to find the value of \(\beta \sin\left(\frac{\beta}{2}\right)\):
\[ \beta \sin\left(\frac{\beta}{2}\right) = (-2x_1) \sin\left(\frac{-2x_1}{2}\right) = -2x_1 \sin(-x_1) = 2x_1 \sin x_1 \] Recall from our earlier deduction that \(x_1\) is precisely defined by \(x_1 \sin x_1 = 1\).
\[ \beta \sin\left(\frac{\beta}{2}\right) = 2(1) = 2 \]