# Cancelling while integrating

You're allowed to do something like the following, but not because you are actually cancelling dt/dt. Right? Can someone please explain how and why (and if) this works ?

$$\int u_n \frac{\mathrm{d}u_n}{dt} \mathrm{d}t = \int u_n \mathrm{d}u_n$$ $$\int u_n \frac{\mathrm{d}u_{n+1}}{\mathrm{d}t} \mathrm{d}t = \int u_n \mathrm{d}u_{n+1}$$

Note that $u_n$ (position of mass $n$) and $u_{n+1}$ is a function of $t$ (time).

I think I this answer to say that I need to integrate $u_n \frac{\mathrm{d}u_n}{dt}$ w.r.t. $t$, evaluate that answer at the bounds of the integral and use the result as the bounds of the integral $\int \mathrm{d}t$. Right? But I don't have bounds and $u_n \frac{\mathrm{d}u_n}{dt}$ doesn't integrate easily w.r.t. $t$.

(OPTIONAL) I ask because it seems I am applying this rule wrong because I am getting the wrong answer to a bigger problem (see image below). I'm actually trying to find the potential energy by integrating force times velocity with respect to time, like this (correct answer is on top; you can ignore the issues, like that I think I defined $\mathbf{v}$ backwards, not relevant to my question).

• Aug 21, 2013 at 17:05

you can write them as $-\int u_{n}\frac{du_{n+1}}{dt}dt-\int u_{n+1}\frac{du_{n}}{dt}dt=-\int\big(u_{n}\frac{du_{n+1}}{dt}+u_{n+1}\frac{du_{n}}{dt}\big)dt=-\int\frac{d}{dt}(u_nu_{n+1})dt$
$=-u_{n}u_{n+1}+C$ using the product rule
For the answer to your original question, though, the formula $\int u_{n}\frac{du_{n}}{dt}dt=\int u_{n}du_{n}$, which we could also write as $\int f(t)f^{\prime}(t)dt=\int udu$, is correct, since it is just an application of u-substitution.