Question

The area bounded by the curve \(y = xsin(\frac{x}{2})\) and the X-axis between the lines \(x = 0\) and \(x = 4π\) (in square units), is....

• A. \(4π\)
• B. \(8π\)
• C. \(12π\)
• D. \(16π\)

Hint:

Split the integral where sin(x/2) changes sign at x = 2pi and take the positive area of each loop.

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
The area between the curve and the X-axis must count regions above and below the axis as positive. So we integrate \(|x\sin(x/2)|\) from 0 to \(4\pi\).

Step 2: Find where the sign changes:
For \(0 \le x \le 2\pi\), \(x/2\) lies in \([0,\pi]\), so \(\sin(x/2) \ge 0\). For \(2\pi \le x \le 4\pi\), \(x/2\) lies in \([\pi, 2\pi]\), so \(\sin(x/2) \le 0\). Hence \[ A = \int_0^{2\pi} x\sin\frac x2\,dx - \int_{2\pi}^{4\pi} x\sin\frac x2\,dx \]

Step 3: Substitute:
Put \(t = x/2\), so \(x = 2t\) and \(dx = 2\,dt\). The limits become 0 to \(2\pi\), and \(x\sin(x/2)\,dx = 4t\sin t\,dt\). So \[ A = 4\int_0^{2\pi} t\,|\sin t|\,dt \]

Step 4: Integration by parts:
\(\int t\sin t\,dt = -t\cos t + \sin t\). On \([0,\pi]\): \([-t\cos t + \sin t]_0^{\pi} = \pi\). On \([\pi, 2\pi]\) we need \(-\int t\sin t\,dt = [t\cos t - \sin t]_\pi^{2\pi} = 2\pi - (-\pi) = 3\pi\).

Step 5: Add up:
\[ A = 4(\pi + 3\pi) = 16\pi \]

Step 6: Why the other options are wrong:
If the sign change is ignored, the integral is \(4[\pi - 3\pi] = -8\pi\), whose size \(8\pi\) is option (B). That counts the lower loop as negative area, which is wrong for an area. Options (A) and (C) do not come from either calculation.

Final Answer:
The area is \(16\pi\) square units, option (D). \[ \boxed{16\pi} \]