Deltas can be treated algebraically

I took all of AP Calculus BC and got a 5 on the AP exam without truly understanding what an integral was. I get it now though, so here's my best attempt at explaining integrals.

We know from Riemann sums that

\(\displaystyle \int_a^b f(x)dx = \lim_{dx \to 0^+} \sum_a^b f(x)dx\)

I assume that this notation is trivial.

That shared \(f(x)dx\) term implies a very loose definition for an integral, i.e.

\(\displaystyle \int_0^1 f(x) = f(0.0000) + f(0.0001) + f(0.0002) + ...\)

Obviously I'm missing infinitely many values here, just roll with it.

These summations with infinitely many terms would spiral out towards infinity, if it weren't for the \(dx\) culling every term to an infinitessimal amount. That is to say that integration is just an operation and deltas are just variables. Integrals are not tied to specific deltas, and deltas aren't tied to any specific part of an equation.

Let's go through an example problem to show what I mean. Capacitors work with the following equation:

\(\displaystyle q = Cv\)

where \(q\) is charge, \(C\) is capacitance, and \(v\) is voltage. We're trying to solve for voltage with respect to current. We can take the derivative of both sides with respect to time and get this:

\(\displaystyle \frac{dq}{dt}=C\frac{dv}{dt}\)

By definition, \(\frac{dq}{dt}\) is just current, giving us this:

\(\displaystyle i=C\frac{dv}{dt}\)

Now for something weird, we're going to multiply both sides by \(dt\). This works because \(dt\) is just a variable. We can multiply expressions by it to get some other value.

\(\displaystyle i dt=C dv \)

Then we can add up all the possible values (take the integral) of both sides.

\(\displaystyle \int i dt=\int C dv \)

After that it's just basic algebra and calculus

\(\displaystyle vC = \int i dt\)

\(\displaystyle v = 1/C \int i dt\)

\(\displaystyle v(T) = 1/C \int_0^T i dt + v_0\)

This derivation works because deltas are just variables and integration is just an operation. Integrals do not need a specific delta value, and you can multiply deltas to move them around.