Definite Double Integral of a single variable

1. Aug 18, 2014

Narroo

1. The problem statement, all variables and given/known data
This isn't actually homework. I was messing around in my notebook trying something when I ended up writing something to the effect of this:

$dT = \frac{V^{2}}{R(1+α dT)}dQ$
$R(1+α dT) dT = V^{2}dQ$
Where α and V are constants.

Now, I'm fairly sure what I had done made no sense physical and I screwed up, which is fine, but as I wrote this, I realized something: I'm not sure how to solve this, or if this is even a valid equation

3. The attempt at a solution

So, I tried integrating with a naive and hopeful glimmer in my eye:
$RΔT+\frac{a}{2}ΔT^{2}= V^{2}Q$

Where $ΔT=T_{2}-T_{1}$
You see, I'm interested in a definite integral, I'm interested in the change between $T_{1}$ and $T_{2}$.

Now, I'm pretty sure this is wrong enough to make someone cry, so I'll just leave it at that as ask where I went wrong and why?

2. Aug 18, 2014

gopher_p

Having no reference for what the constants $\alpha$ and $V$ and the variables (?) $Q$, $R$, and $T$ are intended to represent, there is no reason why the equation $$\Delta T = \frac{V^{2}}{R(1+α \Delta T)} \Delta Q$$ or its "differential brother" $$dT = \frac{V^{2}}{R(1+α dT)}dQ$$ can't model some physical system.

If your end goal it to integrate something (and it looks like it is) then the hand-waver would look at $$RdT+\alpha R dT^2=V^2 dQ$$ and say something along the lines of,

The key reason why the "negligible term" business works is because, in the case where $f$ is an integrable function, we have

$$\lim_{n\rightarrow\infty}\sum_{i=1}^n\left(f(x_i)(\Delta x)^2\right) =\lim_{n\rightarrow\infty}\left(\Delta x\sum_{i=1}^nf(x_i)\Delta x\right) =\lim_{n\rightarrow\infty}\Delta x\lim_{n\rightarrow\infty}\sum_{i=1}^nf(x_i)\Delta x =0\cdot\int_a^bf(x)\ dx=0$$ for $\Delta x=\frac{b-a}{n}$ a regular partition.

3. Aug 19, 2014

Narroo

Thanks. But, I'm not really interested in handing waving the odd term away. I'm interested in how I'd actually solve the equation exactly.

4. Aug 19, 2014

HallsofIvy

Staff Emeritus
What gopher_p did is NOT "hand waving" he showed why $\int\int dx dx= 0$ for any variable x.

5. Aug 19, 2014

Narroo

Sorry, I miss-read it. That said, there are two things I do not understand:
In the following:
$$\lim_{n\rightarrow\infty}\sum_{i=1}^n\left(f(x_i)(\Delta x)^2\right) =\lim_{n\rightarrow\infty}\left(\Delta x\sum_{i=1}^nf(x_i)\Delta x\right) =\lim_{n\rightarrow\infty}\Delta x\lim_{n\rightarrow\infty}\sum_{i=1}^nf(x_i)\Delta x =0\cdot\int_a^bf(x)\ dx=0$$ for $\Delta x=\frac{b-a}{n}$ a regular partition

Why does ΔX come out of the sum? Why can I write
$\lim_{n\rightarrow\infty}\Delta x\lim_{n\rightarrow\infty}\sum_{i=1}^nf(x_i)\Delta x$

I can't write $\lim_{n\rightarrow\infty}\Delta x^{2}\lim_{n\rightarrow\infty}\sum_{i=1}^nf(x_i)$ can I?

And secondly, wouldn't $\int\int dx dx= \frac{1}{2}x^{2}?$

6. Aug 19, 2014

gopher_p

If you have a regular partition, the $\Delta x$ is constant with respect to the sum; $\frac{b-a}{n}$ is independent of $i$.

You can write it the second way, but that doesn't help you evaluate the limit. It's likely that $\sum\limits_{i=1}^nf(x_i)$ is not a convergent sequence.

7. Aug 19, 2014

Zondrina

If $\Delta x = \frac{b-a}{n}$, then as $n → ∞$, $\Delta x → 0$.

Notice $\Delta x$ is independent of the summation index $i$, and can therefore be pulled outside of the sum.

The other part is a Riemann sum, which converges to an integral:

$$=(\lim_{n\rightarrow\infty}\Delta x) (\lim_{n\rightarrow\infty}\sum_{i=1}^nf(x_i)\Delta x) =0\cdot\int_a^bf(x)\ dx=0$$

More to the point though, without knowing what $T$ is exactly, the problem hasn't been stated clearly.

8. Aug 19, 2014

Narroo

T is an independent variable of the system in question. Actually, the equation was dependent on the change in T, ΔT, not T itself, which is why the integral was definite.

And thanks for clearing everything up everyone!