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! :)