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The value of int0^(2pi)[2 sin x] dx, w...

The value of `int_0^(2pi)[2 sin x] dx`, where `[.]` represents the greatest integral functions, is

A

`-(5pi)/(3)`

B

`-pi`

C

`(5pi)/(3)`

D

`- 2 pi`

Text Solution

Verified by Experts

The correct Answer is:
A
It is a question of greatest integer function . We have , subdivide the interval `pi` to `2pi`as under keeping in view that we have to evaluate [ 2 sin x]
We know that , `"sin" (pi)/(6)=(1)/(2)`
`:. sin(pi+(pi)/(6))="sin" (7pi)/(6)=-(1)/(2)`
`rArr"sin" (11pi)/(6)=sin(2pi-(pi)/(6))=-"sin"(pi)/(6)=-(1)/(2)`
`rArr"sin" (9pi)/(6)="sin"(3pi)/(6)=-1`
Hence , we divide the interval `pi` to `2pi` as
`(pi,(7pi)/(6)),((7pi)/(6),(11pi)/(6)),((11pi)/(6),2pi)`
`sinx=(0,-(1)/(2)),(-1,-(1)/(2)),(-(1)/(2),0)`
`rArr2sinx=(0,-1),(-2,-1),(-1,0)`
`rArr[2sinx]=-1`
`=int_(pi)^(7pi//6)[2sinx]dx +int_(7pi//6)^(11pi//6)[2sinx]dx+int_(11pi//6)^(2pi)[2sinx]dx`
`=int_(pi)^(7pi//6)(-1)dx+int_(7pi//6)^(11pi//6)(-2)dx+int_(11pi//6)^(2pi)(-1)dx`
`=-(pi)/(6)-2((4pi)/(6))-(pi)/(6)=-(10pi)/(6)=-(5pi)/(3)`
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