The proof of the first part of the FTC given here was adapted from Tabrizian (2017Tabrizian, Peyam R. 2017. “Proof of the Fundamental Theorem of Calculus (the One with Differentiation).” 2017. https://youtu.be/4DrCKhCECHo.). The goal is to show that
\[\begin{equation} f(x) = \frac{\mathrm{d}}{\mathrm{d}x}\int\limits_{c}^{x}{f(u)\,\mathrm{d}u} \tag{1.2} \end{equation}\]
where \(c\) is an arbitrary constant. First, expand with the definition of the derivative.6 \[\frac{\mathrm{d}y}{\mathrm{d}x} = \lim_{h \to 0}\frac{y(x+h)-y(x)}{h}\]
\[ f(x)=\lim_{h\to{0}}\left[\frac{\int\limits_c^{x+h}{f(u)\,\mathrm{d}u}-\int\limits_c^{x}{f(u)\,\mathrm{d}u}}{h}\right] \]
Recall that a definite integral of a function on \([a,c]\) is equal to the sum of the definite integral from \([a,b]\) and the definite integral from \([b,c]\) given that \(a < b < c\). Then,
\[ \lim_{h\to{0}}\left[\frac{\int\limits_c^{x+h}{f(u)\,\mathrm{d}u}-\int\limits_c^{x}{f(u)\,\mathrm{d}u}}{h}\right] = \lim_{h \to 0}\left[\frac{\int\limits_{x}^{x+h}{f(u)\,\mathrm{d}u}}{h}\right] \]
Then, we only have to prove that \[\begin{align*} 0&=\lim_{h\to{0}}\left[\frac{\int\limits_x^{x+h}{f(u)\,\mathrm{d}u}}{h}- f(x)\right] \\ &=\lim_{h\to{0}}\left[\frac{\int\limits_x^{x+h}{f(u)\,\mathrm{d}u}}{h}-\frac{f(x)\int_{x}^{x+h}{1\mathrm{d}u}}{h}\right] \\ &=\lim_{h\to{0}}\left[\frac{\int\limits_x^{x+h}{f(u)\,\mathrm{d}u}}{h}-\frac{\int_{x}^{x+h}{f(x)\mathrm{d}u}}{h} \right] \\ &= \lim_{h\to{0}}\left[\frac{\int\limits_x^{x+h}{\left[f(u)-f(x)\right]\,\mathrm{d}u}}{h}\right] \end{align*}\]
Hopefully this form is now intuitive enough that we have “proved” FTC 1.7 You can use a delta-epsilon proof to more rigorously prove the theorem.