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Mathematics
Dean Short
8 February, 07:25
F (x) = 8cos (2x) on (0,2π)
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Mark Hess
8 February, 10:11
0
Step-by-step explanation:
First put the lower limit, i. e., x=0,
F (x) = 8cos[2 (0) ]=8cos (0) = 8 (1) = 8
; cos (0) = 1
Now, put the upper limit of given interval, i. e., x = π,
F (x) = 8cos[2 (π) ]=8cos (2π) = 8 (1) = 8
; cos (2π) = 1
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