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

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


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






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balajia | Student


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



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