I'm writing a proof for an exercise problem from Thomas' Calculus 14th Edition book on the proof for the Evaluation Theorem. I'm sure I've got the proof nailed down good, but I'd like to have someone take a second look and point out any unseen holes in my proof.
The problem is stated as below:
- Let $a=x_0<x_1<x_2<\cdots<x_n=b$ be any partition of $[a,b]$, and let $F$ be any antiderivative of $f$. Show that $$ F(b)-F(a)=\sum_{i=1}^n\bigg[F(x_i)-F(x_{i-1})\bigg] $$
- Apply Mean Value Theorem to each term to show that $F(x_i)-F(x_{i-1})=f(c_i)(x_i-x_{i-1})$ for some $c_i$ in the interval $(x_{i-1},x_i)$. Then show that $F(b)-F(a)$ is a Riemann sum for $f$ on $[a,b]$
- From part (2) and the definition of the definite integral, show that $$ F(b)-F(a)=\int_a^b f(x)dx $$
My Proof:
Let $f$ be a function that is integrable in $[a,b]$. The Riemann Sum of $f$, $S_n$ in $[a,b]$ is given by
$$
S_n=\sum_{i=1}^nf(x_i)\Delta x_i
$$
where $\Delta x_i$ is the size of the $i$-th of the $n$ partitions of $[a,b]$, $x_i\in (x_{i-1},x_i)$ and $S_n$ is the Riemann Sum of $f(x)$ over $[a,b]$. Since it is given that $F$ is any antiderivative of $f$, we have
$$
F'(x)=f(x) \text{ } \forall \text{ } x\in[a,b]
$$
Consider the term
$$
\frac{F(x_i)-F(x_i-\Delta x_i)}{\Delta x_i}
$$
which is basically the slope of the secant of $F(x)$ between $x_i-\Delta x_i$ and $x_i$. As $\Delta x_i \rightarrow 0$, the value of the above term reaches $F'(x_i)$. That is,
$$
\lim_{\Delta x_i \rightarrow 0} \frac{F(x_i)-F(x_i-\Delta x_i)}{\Delta x_i}=F'(x_i)=f(x_i) \text{ (given)}
$$
Utilizing the above for $S_n$, we can write
$$
\begin{aligned}
S_n &= \lim_{\Delta x_i \rightarrow 0} \sum_{i=1}^n \left[\frac{F(x_i)-F(x_i-\Delta x_i)}{\Delta x_i} \right] \Delta x_i \\
&= \lim_{\Delta x_i \rightarrow 0} \sum_{i=1}^n \big[F(x_i)-F(x_i-\Delta x_i)\big]
\end{aligned}
$$
Notice that, when $\Delta x_i$ is equal, i.e, the partitions are evenly sized,
$$
\begin{aligned}
\Delta x_i &= \frac{b-a}{n} \\
\implies x_i-\Delta x_i &= x_{i-1}
\end{aligned}
$$
Also, as $\Delta x_i \rightarrow 0$, we get $n \rightarrow \infty$. Therefore
$$
\begin{aligned}
S_n &= \lim_{n \rightarrow \infty} \sum_{i=1}^n \big[F(x_i)-F(x_{i-1})\big] \\
&= F(x_1)-F(x_0)+F(x_2)-F(x_1)+\cdots+F(x_n)-F(x_{n-1}) \\
&= F(x_n)-F(x_0) \\
&= F(b)-F(a)
\end{aligned}
$$
Therefore
$$
\boxed{S_n = \lim_{n \rightarrow \infty} \sum_{i=1}^n \big[F(x_i)-F(x_{i-1})\big] = F(b)-F(a)}
$$
Mean Value Theorem for integrals states, $$ \begin{aligned} \exists \text{ } c_i \in (x_i,x_{i-1}) \text{ | } f(c_i) &= \frac{1}{x_i-x_{i-1}}\int_{x_{i-1}}^{x_i}f(x)dx \\ \end{aligned} $$ Hence, $$ (x_i-x_{i-1})f(c_i)=\int_{x_{i-1}}^{x_i}f(x)dx $$
Summing the above from $i=1$ to $n$, we get $$ \sum_{i=1}^n (x_i-x_{i-1})f(c_i) = S_n = \sum_{i=1}^n \int_{x_{i-1}}^{x_i}f(x)dx \\ = \int_a^bf(x)dx $$
Therefore, $$ \boxed{S_n = F(b)-F(a)=\int_a^bf(x)dx} $$