Proving fundamental theorem of calculus
===

Given a function $f : {\Bbb R} \rightarrow {\Bbb R}$ which is continuous over $[0, X]$, we can compute a corresponding function $\Delta f$ given a step length $\delta$

$$
\Delta f(x) = \begin{cases}
f(x + \delta) - f(x) &, \text{if } x\mod \delta = 0 \\
0 &, \text{otherwise}
\end{cases}
$$

We create another function, $A$ (which stands for change in area)

$$
A(n) = \sum_{i = 0}^{n - 1} \delta \cdot \Delta f(i\delta) /\delta
$$

constrained by $n\delta = X$

If $\delta = 1$, $A(n) = (f(1) - f(0)) + (f(2) - f(1)) + \ldots + (f(n) - f(n - 1))$ which gives $A(n) = f(n) - f(0)$, which is the total change in area of the delta function, as the original function ranges in $[0, n]$. We can take $\delta$ equal to any value btw, using $\delta = 1$ to avoid clutter.

As $n \rightarrow \infty$, $\delta \rightarrow 0$. So,

$$
\lim_{\delta \rightarrow 0} \sum \delta \cdot \Delta f(i\delta) / \delta
$$

As $\delta \rightarrow 0$, $\Delta f \rightarrow \delta \text{ } f'$ and thus $\Delta f / \delta \rightarrow f'$ and the sum tends to the riemann integral

$$
f(X) - f(0) = \int_{0}^{X} dx \text{ } \cdot f'
$$

dx dy

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