<p>If <strong>f</strong>(<i>x</i>) = min{<i>x</i><sup>2</sup>, sin(x/2), (<i>x</i> − 2π)<sup>2</sup>}, the area bounded by the curve <i>y</i> = <strong>f</strong>(<i>x</i>), x-axis, <i>x</i> = 0 and <i>x</i> = 2π is given by:</p><p><strong>Note:</strong> <i>x</i><sub>1</sub> is the point of intersection of the curves <i>x</i><sup>2</sup> and sin(<i>x</i>/2); <i>x</i><sub>2</sub> is the point of intersection of the curves sin(<i>x</i>/2) and (<i>x</i> − 2π)<sup>2</sup></p>
<p>(a) \(\int_0^{x_1} x^2 \, dx + \int_{x_1}^{\pi} \sin\left(\frac{x}{2}\right) dx + \int_x^{2\pi} \sin\left(\frac{x}{2}\right) dx + \int_{x_2}^{2\pi} (x - 2\pi)^2 \, dx\)</p>
<p>(b) \(\int_0^{x_1} x^2 \, dx + \int_{x_1}^{x_2} \sin\left(\frac{x}{2}\right) dx + \int_{x_2}^{2\pi} (x - 2\pi)^2 \, dx\), where \(x_1 \in\left(0, \frac{\pi}{3}\right)\) and \(x_2 \in\left(\frac{5\pi}{3}, 2\pi\right)\)</p>
<p>(c) \(\int_0^{x_1} x^2 \, dx + \int_{x_1}^{x_2} \sin\left(\frac{x}{2}\right) dx + \int_{x_2}^{2\pi} (x - 2\pi)^2 \, dx\), where \(x_1 \in\left(\frac{\pi}{3}, \frac{\pi}{2}\right)\) and \(x_2 \in\left(\frac{3\pi}{2}, 2\pi\right)\)</p>
<p>(d) \(\int_0^{x_1} x^2 \, dx + \int_{x_1}^{x_2} \sin\left(\frac{x}{2}\right) dx + \int_{x_2}^{2\pi} (x - 2\pi)^2 \, dx\), where \(x_1 \in\left(\frac{\pi}{2}, \frac{2\pi}{3}\right)\) and \(x_2 \in(\pi, 2\pi)\)</p>
Step-by-Step Solution
Key Concept: Identify the points where the minimum function changes by finding intersections of the three curves, then integrate the minimum function over each interval.
<p>The minimum function changes at the intersection points of the three functions. The area is computed as the sum of integrals over intervals where each function is the minimum.</p><p>∴ Answer is (b).</p>
Correct Answer: B