Find the value of f(0) when f'(t) = 4 cos(2t), f(pi/4)=4
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You need to find `f(t)` using the integration such that:
`f(t) = int f'(t)dt => f(t) = int 4 cos(2t) dt`
`f(t)= 4 int cos (2t) dt => f(t) = 4*(sin(2t))/2 + c`
`f(t) = 2sin(2t) + c`
Since the problem provides the information that `f(pi/4)=4` , you may find the constant c such that:
`f(pi/4) = 2sin(2pi/4) + c => 4 = 2sin(pi/2) + c`
Substituting 1 for `sin(pi/2)` yields:
`4 = 2 + c => c = 4-2 => c = 2`
Hence, evaluating the function f(t) yields:
`f(t) = 2sin(2t) + 2`
You need to evaluate f(0), hence, you need to substitute 0 for t in equation of the function such that:
`f(0) = 2sin(0) + 2 => f(0) = 2*0 + 2 => f(0) = 2`
Hence, evaluating the value of the function `f(t)` at `t=0` yields `f(0) = 2` .
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