# The Fundamental Theorem of Calculus (Part 1) Proof

\begin{theorem} Suppose that $f$ is continuous on the closed interval $[a,b]$.
\noindent \textbf{Part 1: } If the function $F$ is defined on $[a,b]$ by $$F(x)=\int_a^x f(t)\, dt,$$ then $F$ is an antiderivative of $f$. That is, $F'(x)=f(x)$ for all $x$ in $[a,b]$.
\noindent \textbf{Part 2: } If $G$ is any antiderivative of $f$ on $[a,b]$, then $$\int_a^b f(x)\, dx=\left[G(x)\right]_a^b=G(b)-G(a).$$ \end{theorem} \begin{proof} \textbf{(Part 1)} Let $f$ be a continuous function on the interval $[a,b]$. Define a function $$F(x)=\int_a^x f(t)\, dt$$ for $x$ in $[a,b]$. Then $F(x)$ is the area under the curve $y=f(t)$ between the lines $t=a$ and $t=x$. That is, $F(x)$ is the area of the shaded region in the following diagram. Similarly, $F(x+h)$ is the area under the curve $y=f(t)$ between the lines $t=a$ and $t=x+h$. Hence, $F(x+h)-F(x)$ is the area of the small vertical strip under the curve $y=f(t)$ between the lines $t=x$ and $t=x+h$. This is the area of the pink dotted region in the following diagram. Notice that the base of the small vertical strip has length $h$. So, in a sense we are about to make precise, the height of the small vertical strip is $\frac{F(x+h)-F(x)}{h}$.
Now, $f$ is a continuous function on the closed interval $[x,x+h]$, so $f$ assumes both a maximum value and a minimum value on this interval. Let $m_h=f(u_h)$ be the minimum of $f$ on the interval $[x,x+h]$ and let $M_h=f(v_h)$ be the maximum of $f$ on the interval $[x,x+h]$. Then for all $t$ in the interval $[x,x+h]$, we have $$m_h=f(u_h)\le f(t) \le f(v_h)=M_h.$$ Now, integrate each piece of this inequality, using the fact that integration preserves the order of the inequality: $$m_h\cdot h=\int_x^{x+h} m_h\, dt\le \int_x^{x+h} f(t) \, dt\le \int_x^{x+h} M_h\, dt=M_h\cdot h.$$ This sequence of inequalities has a graphical interpretation as follows. The integral on the left $m_h\cdot h=\int_x^{x+h} m_h\, dt$ is the area of the small shaded rectangle in the following diagram: The integral in the middle $\int_x^{x+h} f(t) \, dt$ is the area of the shaded region under the curve: The integral on the right $\int_x^{x+h} M_h\, dt$ is the area of the large shaded rectangle in the following digram: Dividing the inequalities by $h$ (which we assume is positive), we get $$m_h\le \frac{1}{h}\int_x^{x+h} f(t) \, dt\le M_h.$$ If $h\lt 0$, we get the same inequalities, only reversed. Since $\int_x^{x+h} f(t) \, dt=F(x+h)-F(x)$, we have $$m_h\le \frac{F(x+h)-F(x)}{h}\le M_h,$$ or the same with inequalities reversed. Now, taking the limit as $h$ goes to zero, we note that $u_h$ and $v_h$ are points in the interval $[x,x+h]$, so as $h$ goes to zero, both these points go to $x$. Since $f$ is continuous, $f(u_h)=m_h$ and $f(v_h)=M_h$ go to $f(x)$. Hence, we have $$f(x)\le \lim_{h\to 0} \frac{F(x+h)-F(x)}{h}\le f(x).$$ Hence, by the Squeeze theorem, we have $$F'(x)= \lim_{h\to 0} \frac{F(x+h)-F(x)}{h}= f(x),$$ as desired. \end{proof}