The Fundamental Theorem of Riemann Integral Calculus Part 2

# The Fundamental Theorem of Riemann Integral Calculus Part 2

Perhaps one of the most famous and used theorems in elementary calculus is the famous fundamental theorem of calculus. We finally prove the relatively simple-to-prove result.

 Theorem 1: Let $f$ be Riemann integrable on $[a, b]$ and let $g$ be a continuous function on $[a, b]$ such that $f(x) = g'(x)$. Then $\displaystyle{\int_a^b f(x) \: dx = g(b) - g(a)}$.
• Proof: Let $A = \int_a^b f(x) \: dx$. Then for all partitions $P = \{ a = x_0, x_1, ..., x_n = b \} \in \mathscr{P}[a, b]$ we have that:
(1)
\begin{align} \quad g(b) - g(a) = \sum_{k=1}^{n} [g(x_k) - g(x_{k-1})] \end{align}
• Since $g$ is continuous on $[a, b]$ and differentiable such that $g'(x) = f(x)$ we have that $g$ is also continuous and differentiable on each subinterval $[x_{k-1}, x_k]$ and so by the Mean Value theorem for each $k \in \{ 1, 2, ..., n \}$ there exists a $t_k \in (x_{k-1}, x_k)$ such that:
(2)
\begin{align} \quad g'(t_k)[x_k - x_{k-1}] = g(x_k) - g(x_{k-1}) \end{align}
• Substituting this into the sum above gives us:
(3)
\begin{align} \quad g(b) - g(a) &= \sum_{k=1}^{n} g'(t_k)[x_k - x_{k-1}] \\ \quad g(b) - g(a) &= \sum_{k=1}^{n} f(t_k) \Delta x_k \\ \quad g(b) - g(a) &= S(P, f, x) \end{align}
• So for all $\epsilon > 0$ we have that for every partition $P \in \mathscr{P}[a, b]$ that:
(4)
\begin{align} \quad 0 = \mid S(P, f, x) - [g(b) - g(a)] \mid < \epsilon \end{align}
• Therefore we have that:
(5)
\begin{align} \quad \int_a^b f(x) = g(b) - g(a) \quad \blacksquare \end{align}