A) 20
B) 40
C) 60
D) 80
Correct Answer: D
Solution :
[d] \[I=\int\limits_{-20\pi }^{20\pi }{\left| \cos \right|}.dx\] \[I=2\int\limits_{0}^{20\pi }{\left| \cos \right|}.dx\] [\[cosx\] is an even function] \[=2\times 20\int\limits_{0}^{20\pi }{\left| cosx \right|.dx}\] \[=40\int\limits_{0}^{20\pi }{\left| cosx \right|.dx}+40\int\limits_{\pi /2}^{\pi }{\left| cosx \right|.dx}\] [\[\left| cosx \right|\]is periodic with period\[\pi \]] \[=40[sinx]_{0}^{\pi /2}-40[sinx]_{\pi /2}^{\pi }\] \[=40\left( \sin \frac{\pi }{2}-{{\sin }^{{}^\circ }} \right)-40\left( \sin \pi -\sin \frac{\pi }{2} \right)\] \[=40\times 1-40\left( -1 \right)=40+40=80\] Hence option [d] is correct.
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