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Question

The value of 100π01cos2x dx is

A
2002
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B
400
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C
1002
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D
0
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Solution

The correct option is A 2002
Let I=100π01cos2x dx
I=100π02sin2x dx =2100π0|sinx| dx
Using the property:
a+nTaf(x) dx=nT0f(x) dx[T is the period of f(x)]

Here, the period of |sinx| is π, so
I=1002π0|sinx| dx =1002π0sinx dx =1002(cosx)π0I=2002

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