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Question

The value of the integral 41 π40|cos x|dx is
  1. 2012
  2. 20+12
  3. 19+12
  4. 1912


Solution

The correct option is B 20+12
=41 π40|cos x|dx=10π0|cos x|dx+=41 π410π|cos x|dx
=10π0|cos x|dx+10π+π410π|cos x|dx
[since |cos x| is a periodic cunction of period  π ]
=10π0|cos x|dx+π40|cos x|dx
=10(π20 cos x dxππ2 cos x dx)+sin xπ40
=10(sin xπ20sin xππ2)+12
=10(1+1)+12=20+12
 

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