Correct option is (b) \(-\pi\)
\(\int\limits_0^{2\pi} [2sin x] \,dx = ?\)
We have to subdivide the interval 0 to 2π as under keeping in view that we have to evaluate [2 sinx].
\(\because sin\left( \frac \pi 6\right) = \frac 12 \)
⇒ \(2sin\left(\frac \pi 6\right) = 2 \times \frac 12 = 1\)
\(\therefore [2 sin x] = 0 ;\, \text{if} \;0 \le x < \frac \pi 6\).
\(\because sin\frac \pi2 = 1\)
⇒ \(2 sin\frac \pi2 = 2\)
\(\therefore [2 sin x] = 1;\; \text{if} \;\frac\pi6 \le x < \frac \pi2\)
\(\because sin\left( \pi - \frac \pi6\right) = sin \left(\frac{5\pi}6\right) = \frac 12\)
⇒ \(2sin \left(\frac{5\pi}6\right) = 2\times \frac 12 = 1\)
\(\therefore [2sin x] = 1;\text{if}\; \frac \pi2 < x \le \frac{5 \pi}6\)
\(\because 2sin(\pi) = 0\)
\(\therefore [2sinx] = 0; \text{if}\; \frac{5\pi}6 < x \le \pi\)
\(\therefore sin(\pi + \frac \pi6) = - \frac 12\)
⇒ \(2sin\left(\frac{7\pi}6\right) = -1\)
\(\therefore [2sin x] = -1 ; \text{if}\; \pi < x\le \frac{7\pi}6\)
\(\because sin\left(2\pi - \frac \pi6\right) = - sin\left(\frac \pi 6\right) = \frac {-1}2\)
\(\therefore 2 sin\left(\frac {11\pi}{6}\right) = 1 \) & \( sin 2\pi = 0\)
\(\therefore [2sin x] = -2;\text{if} \; \frac {7\pi}6 < x < \frac{11\pi}6\)
\([2 sin x] = -1;\text{if}\; \frac{11\pi}6 \le x < 2\pi\)
\(\therefore \int\limits_0^{2\pi} (2 sin x) dx ={ \int\limits_0^{\frac\pi6} [2sinx] dx + \int\limits_{\frac\pi6}^{\frac\pi2} [2sinx] dx + \int\limits_\frac\pi3^\frac{5\pi}6 [2sinx] dx + \int\limits_\frac{5\pi}6^\pi [2sinx] dx\\ +\int\limits _\pi^\frac{7\pi}6 [2sinx] dx + \int\limits_\frac{7\pi}6^{\frac{11\pi}6} [2sinx] dx + \int\limits_\frac{11\pi}6^{2\pi} [2sinx] dx}\)
\(={ \int\limits_0^\frac\pi6 0dx + \int\limits_\frac\pi6^\frac \pi2 1dx + \int\limits_{\frac \pi2}^{\frac{5\pi}6} 1dx + \int\limits_{\frac {5\pi}6}^\pi 0dx + \int\limits_\pi ^{\frac{7\pi}6}-1dx \\+ \int\limits_{\frac{7\pi}6}^\frac{11\pi}6 - 2dx + \int\limits_\frac{11\pi}6 ^{2\pi} -1dx}\)
\(= 0+ \left(\frac\pi2 - \frac \pi 6\right) + \left(\frac{5\pi}6 - \frac \pi2\right) - \left(\frac {7\pi}6 - \pi\right) - 2\left(\frac{11\pi}6-\frac{7\pi}6\right) - \left(2\pi - \frac {11\pi}6\right)\)
\(= \frac{8\pi}6-\pi - \frac{8\pi}6 \)
\(= -\pi\)
