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21 March, 04:34

F (x) = 8cos (2x) on (0,2π)

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  1. 21 March, 04:56
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    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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