Definite IntegrationhardFree

|∫[2 sinx] dx| from 0 to 2π (greatest integer) | JEE

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Question
The value of the integral ∣ ∫02π[2sin⁡x] dx ∣\left|\,\displaystyle\int_{0}^{2\pi}[2\sin x]\,dx\,\right|, where [ ⋅ ][\,\cdot\,] denotes the greatest integer function, is
Aπ\picorrect
B2π2\pi
C3π3\pi
D4π4\pi
Solution
Step 1: Track the value of 2sin⁡x2\sin x over [0,2π][0,2\pi]. It rises 0→20\to 2, falls 2→02\to 0 on [0,π][0,\pi], then falls 0→−20\to-2 and rises −2→0-2\to 0 on [π,2π][\pi,2\pi]. The greatest integer [2sin⁡x][2\sin x] changes at the points where 2sin⁡x2\sin x equals 1,2,−1,−21,2,-1,-2, i.e. at π6,5π6,7π6,11π6\dfrac{\pi}{6},\dfrac{5\pi}{6},\dfrac{7\pi}{6},\dfrac{11\pi}{6}. Step 2: Read off [2sin⁡x][2\sin x] on each subinterval (single points where it hits the next integer have measure zero):
[2sin⁡x]={0,(0,π6)∪(5π6,π)1,(π6,5π6)−1,(π,7π6)∪(11π6,2π)−2,(7π6,11π6).[2\sin x]=\begin{cases}0,&\big(0,\tfrac{\pi}{6}\big)\cup\big(\tfrac{5\pi}{6},\pi\big)\\ 1,&\big(\tfrac{\pi}{6},\tfrac{5\pi}{6}\big)\\ -1,&\big(\pi,\tfrac{7\pi}{6}\big)\cup\big(\tfrac{11\pi}{6},2\pi\big)\\ -2,&\big(\tfrac{7\pi}{6},\tfrac{11\pi}{6}\big).\end{cases}
 Step 3: Integrate by summing value ×\times length:
∫02π[2sin⁡x] dx=1 ⁣(5π6−π6)+(−1) ⁣(π6)+(−2) ⁣(2π3)+(−1) ⁣(π6).\int_{0}^{2\pi}[2\sin x]\,dx=1\!\left(\dfrac{5\pi}{6}-\dfrac{\pi}{6}\right)+(-1)\!\left(\dfrac{\pi}{6}\right)+(-2)\!\left(\dfrac{2\pi}{3}\right)+(-1)\!\left(\dfrac{\pi}{6}\right).
 Here 7π6−π=π6\dfrac{7\pi}{6}-\pi=\dfrac{\pi}{6} and 2π−11π6=π62\pi-\dfrac{11\pi}{6}=\dfrac{\pi}{6}. Step 4: Evaluate:
=2π3−π6−4π3−π6=−2π3−π3=−π ⇒ ∣∫02π[2sin⁡x] dx∣=π.=\dfrac{2\pi}{3}-\dfrac{\pi}{6}-\dfrac{4\pi}{3}-\dfrac{\pi}{6}=-\dfrac{2\pi}{3}-\dfrac{\pi}{3}=-\pi\ \Rightarrow\ \left|\int_{0}^{2\pi}[2\sin x]\,dx\right|=\pi.
 Correct answer: (1)
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