1
$\begingroup$

Please help in understanding the following integrals.

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).

Potential Energy

$\endgroup$
  • 1
    $\begingroup$ Related. $\endgroup$ – Git Gud Aug 21 '13 at 17:05
1
$\begingroup$

The problem occurs in dealing with the two terms in the middle;

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.