Definite IntegrationhardFree

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

JEE Maths question with a full step-by-step solution.

Question

The value of the integral

\left|\,\displaystyle\int_{0}^{2\pi}[2\sin x]\,dx\,\right|

, where

[\,\cdot\,]

denotes the greatest integer function, is

\pi

correct

2\pi

3\pi

4\pi

Solution

Step 1: Track the value of

2\sin x

over

[0,2\pi]

. It rises

0\to 2

, falls

2\to 0

[0,\pi]

, then falls

0\to-2

and rises

-2\to 0

[\pi,2\pi]

. The greatest integer

[2\sin x]

changes at the points where

2\sin x

equals

1,2,-1,-2

, i.e. at

\dfrac{\pi}{6},\dfrac{5\pi}{6},\dfrac{7\pi}{6},\dfrac{11\pi}{6}

. Step 2: Read off

[2\sin x]

on each subinterval (single points where it hits the next integer have measure zero):

[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:

\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

\dfrac{7\pi}{6}-\pi=\dfrac{\pi}{6}

and

2\pi-\dfrac{11\pi}{6}=\dfrac{\pi}{6}

. Step 4: Evaluate:

=\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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