One way to visualize an antiderivative is that the area under the derivative is added to the initial value of the antiderivative to get the final value of the antiderivative over an interval.
The Riemann Series Theorem essentially says that you can basically get any value you want out of a conditionally convergent series by changing the order you add up the terms.
Now consider the function: $$f(x) = 1/x^3$$The function is unbounded over the interval $(-1,1)$, so it not integrable over this interval.
If you break $f(x)$ into Riemann rectangles over the interval $(-1,1)$ and express the area as a Riemann sum, you essentially get a conditionally convergent series. And because of the Riemann Series Theorem, you can make the sum add up to anything. In other words, you can make the rectangles add up to whatever area you want by changing the order in which you add them up. This is in fact why a function needs to be bounded to be integrable - otherwise the area has infinite values/is undefined.
So my question is, in cases like this, how does the antiderivative "choose" an area? I mean in this case the antiderivative, $\frac{1}{-2x^2}$, choose to increase by $\frac{1}{-2(1)^2} - \frac{1}{-2(-1)^2} = 0.$ In other words, the antiderivative "choose" to assign a area of zero to the area under $1/x^3$from $-1$ to $1$ even though the Riemann Series Theorem says the area can be assigned any value.
How did the antiderivative "choose" an area of $0$ from the infinite possible values?
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