# `f'(t) = 2t - 3sin(t), f(0) = 5` Find `f`.

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### 2 Answers

`f'(t)=2t-3sin(t)`

`f(t)=int(2t-3sin(t))dt`

`f(t)=2(t^2/2)-3(-cos(t))+C` ,C is constant

`f(t)=t^2+3cos(t)+C`

Now , evaluate C , given f(0)=5

`f(0)=5=0^2+3cos(0)+C`

`5=3+C`

`C=2`

`:.f(t)=t^2+3cos(t)+2`

`f'(t)=2t-3sint`

Integrating both sides with respect to x,we get

`int f'(t) =int(2t-3sint)dt`

`f(t) = 2(t^2/2)-3(-cost)+c`

`= t^2+3cost+c`

Given f(0) = 5

`f(0)=0+3+c`

c=2

Therefore `f(t)= t^2+3cost+2`