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Question

2nπ0{|sinx|12sinx}dx equals.

A
n
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B
2n
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C
2n
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D
None of the above
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Solution

The correct option is B 2n
Let I=2nπ0{|sinx|12sinx}dx
I=2nπ0{|sinx|12sinx}dx
I=2nπ0{(sinx)12sinx}dx[Tx0ydy=Tx0ydy] (where T is regular period of y)
I=2nπ012sinxdx
I=2n2π0sinxdx
I=n[cosx]π0
I=n[cosπ+cos0]
I=n[(1)+1]
I=2n

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