# 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!