The value of ∫1−1[x[1+sinπx]+1]dx ([.] denotes the greatest integer)
A
2
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B
1
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C
None of these
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Solution
The correct option is A 2 ∫1−1[x[1+sinπx]+1]dx =∫0−1[x(1+sinπx)+1]dx+∫10[x[1+sinπx]+1]dx Now−1<x<0⇒[1+sinπx]=0 0<x<1⇒[1+sinπx]=1 ⇒[x[1+sinπx]+1]=2 So=∫10[x[1+sinπx]+1]dx=2