Why is the integral of dt = t + C1

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SUMMARY

The integral of dt is expressed as t + C1, where C1 represents a constant of integration. This conclusion arises from the fundamental theorem of calculus, which states that if df = dt, then f must equal t plus a constant. The discussion highlights that while f = t is one solution, there are infinitely many solutions differing by a constant, reinforcing the concept of indefinite integrals.

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Why is the integral of dt = t + C1

Can someone explain that to me?

Thanks.
 
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It seem the integral you are interested in is the indefinite type. So you are looking for some function f so that
df=dt
clearly f=t is one such function, but there are others
what if we are intersted in g so that dg=0
but for any c c*dg=c*0=0 also
d(1)=0
but also
d(c)=0

so if du=dv
we can say
d(u-v)=0
but we cannot conclude u=v
since d(u-v)=0
we can conclude
u-v=c for some c

so if df=dt
f=t+c
 
thanks, i understand now.
 

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