Expressing x(t) of a particle (air resistance)

[Cooojan]

1.
2D -problem
A particle is moving on a frictionless, horizontal surface from ##O_{rigo}## with initial speed ## v_0 ##, as it faces the air resistance force ## F_R ##.2.
## F_R = -mαv ## (where v is a particle velocity)

## \frac {dv}{dt} + αv = 0 ##Show that:

## x(t) = \frac 1 α (v_0)(1-e^{-αt}) ##3.
So I menaged, somewhat, to complete this exercise, but I only come to this solution:

## x(t)=\frac1α(v_0)(-e^{-αt}) ##

Can somebody explain me, how does this missing ## "1" ## get there?

## a=-αv ##

## \frac{dv}{-αv}=dt ##

## -\frac1α\int{\frac1v}\,dv=∫dt ##

## e^{ln(v)} = e^{-αt+C_1} ##

## v=Ce^{-αt} ##

## v(t) = v_0e^{-αt} ##

## v(t)=\frac{dx}{dt} ##

## ∫dx=\int v \,dt ##

## x(t)=v_0 \int e^{-αt}\,dt ##

## x(t)=\frac1α(v_0)(-e^{-αt}) ##

But it should be

## x(t) = \frac 1 α (v_0)(1-e^{-αt}) ##

Tnx a lot. Any help would be very useful and much appriciated :)

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[kuruman]

∫dx=∫vdt

You need limits of integration here. When t = 0 x = 0 and v = v₀; when t = t, x = x(t) and v = v(t).

 

[Cooojan]

You need limits of integration here. When t = 0 x = 0 and v = v₀; when t = t, x = x(t) and v = v(t).

Worked perfectly. Thank you 4 your time! :)