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Q.

∫02π[2sinx]dx=

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a

−5π3

b

−π

c

5π3

d

−2π

answer is B.

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Detailed Solution

I=∫0π60.dx+∫π65π61.dx+∫5π6π0.dx+∫π7π6(−1)dx+∫7π611π6(−2)dx+∫11π62π(−1)dx    =4π6−π6−8π6−π6=−π
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