How is the integration of displacement performed in this scenario?

[Kyle.Nemeth]

How was this integration performed?

½m∫ d(v²) = ½m(v_f² - 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!).

 

[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
$\int_a^b 1 dx = \left. x \right|_a^b = b-a$.

 

[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 v_f², but are just from v₀ to v_f.

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

½m∫ d(v²) = ½m∫ v dv ?

 

[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.
$\int_{v=v_0}^{v = v_f} d(v^2) = \int_{v^2=v_0^2}^{v^2 = v_f^2} 1 d(v^2) = \int_{v=v_0}^{v = v_f} 2v dv$

 

[Kyle.Nemeth]

So then, for

d(v²) = 2vdv

Are we applying the chain rule somehow?

 

[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.

 

[Kyle.Nemeth]

Great. Thank you for your help.