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If the value of 22|xcosπx|dx=k/π then the value of k is

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a
4
b
8
c
12
d
none of these

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detailed solution

Correct option is B

Since the integrand is an even function so required integral is equal to 2∫02 |xcos⁡πx|dx=2∫01/2 xcos⁡πxdx−∫1/23/2 xcos⁡πxdx+∫3/22 xcos⁡πxdx=2xsin⁡πxπ+cos⁡πxπ201/2−xsin⁡πxπ+cos⁡πxπ21/23/2+xsin⁡πxπ+cos⁡πxπ2y22=212π−1π2−−32π−12π+1π2−−32π=8π


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