`f'(t) = 2t - 3sin(t), f(0) = 5` Find `f`.
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gsarora17
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`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`
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balajia | Student
`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`
Student Answers