# Simple Integral

1. Jan 22, 2014

### Radarithm

1. The problem statement, all variables and given/known data
$$F=-mg-m\alpha v$$$$F=m\frac{dv}{dt} =$$
$$\int_0^v \frac{dv'}{g+\alpha v'} = -\int_0^t dt'$$
This equals:
$$\ln \frac{1+\alpha v}{g} = -\alpha t$$

I don't understand why it does; I keep getting an incorrect answer. Can anyone tell me what I'm doing wrong?

2. Relevant equations
3. The attempt at a solution

let $$u=g+\alpha v'$$$$du=g+vdv$$ so $du-g=vdv$
$$-g+\int_0^v udu = \ln (g+\alpha v)-g$$
This has to do with getting the velocity as a function of time; I don't know if it should be in the physics section or not but this is a problem I'm having with calculus.

Last edited: Jan 22, 2014
2. Jan 22, 2014

### Tanya Sharma

Is $F=mg-m\alpha v$ or $F=mg+m\alpha v$ ?

3. Jan 22, 2014

### Radarithm

$$F=-mg-m\alpha$$

4. Jan 22, 2014

### HallsofIvy

Staff Emeritus
This makes no sense. You chose to use v' as the variable of integration. Why does it become v when you find du? Also g is a constant so its derivative is 0. If $u= g+ \alpha v'$ then $du= \alpha dv'$ so that $dv'= du/\alpha$
$$\int_0^v \frac{dv'}{g+ \alpha v'}= \frac{1}{\alpha}\int_g^{g+\alpha v} \frac{du}{u}$$

5. Jan 22, 2014

### Radarithm

Why do the limits of integration change from $0 \to v(t)$ to $g \to g+\alpha v$ ?

6. Jan 22, 2014

### Tanya Sharma

Because the variable of integration has changed from v' to u .

Use the relation between v' and u (u=g+αv') to find the corresponding new limits .

When v' = 0 → u = g (lower limit)
When v' = v → u = g+αv (upper limit)

7. Jan 22, 2014

### HallsofIvy

Staff Emeritus
Because you changed variables from v' to $u= g+ \alpha v'$. When v'= 0, $u= g+ \alpha(0)= g$ and when v'= v, $u= g+ \alpha v$.

8. Jan 22, 2014

### Radarithm

edit: double posted. sorry

9. Jan 22, 2014

### Radarithm

Thanks for the help guys. I appreciate it.

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