1. Jun 30, 2015

### Kyle.Nemeth

½m∫ d(v2) = ½m(vf2 - v02) = ΔK

A book on a table is displaced by a net force in the positive x direction, which changes the speed of the book. The integral was taken over the initial speed to the final speed and I'm not quite sure how to incorporate boundaries on my integral with this site (sorry!).

Last edited: Jun 30, 2015
2. Jun 30, 2015

### RUber

$\frac 12 m \int_{v_0}^{v_f} d(v^2)$ is saying that your variable is $v^2$, so your limits of integration should match your variable, which would give you:
$\frac 12 m \int_{v_0^2}^{v_f^2} d(v^2)$
This is essentially just an integral like
$\int_a^b 1 dx = \left. x \right|_a^b = b-a$.

3. Jun 30, 2015

### Kyle.Nemeth

Thank you for the help, I understand now, but there's one more thing that I'm having trouble with also.

I'm looking at this in my book, "Physics for Scientists and Engineers 9th Edition" by Raymond Serway and John Jewett, Jr. In the book, their bounds of integration are not from v02 to vf2, but are just from v0 to vf.

Also, would I not be justified in doing something like this,

½m∫ d(v2) = ½m∫ v dv ?

4. Jun 30, 2015

### RUber

Close. $d(v^2) = 2v dv$, so you could do that too. The bounds were given in terms of the variable v, so sometimes it's easier to change the bounds to match the variable and sometimes its easier to change the variable to match the bounds. In either case, I think that the book's notation can lead to confusion if you just try to do the math without thinking about the physical interpretation.
$\int_{v=v_0}^{v = v_f} d(v^2) = \int_{v^2=v_0^2}^{v^2 = v_f^2} 1 d(v^2) = \int_{v=v_0}^{v = v_f} 2v dv$

5. Jun 30, 2015

### Kyle.Nemeth

So then, for

d(v2) = 2vdv

Are we applying the chain rule somehow?

6. Jun 30, 2015

### RUber

We are taking $\frac{d v^2}{dv} = 2v$ and moving the dv to the right. *edit* Or using the chain rule--both are correct.

7. Jun 30, 2015

### Kyle.Nemeth

Great. Thank you for your help.