The Fundamental Theorem of Calculus Part 2

Recall that the The Fundamental Theorem of Calculus Part 1 essentially tells us that integration and differentiation are "inverse" operations. We will now look at the second part to the Fundamental Theorem of Calculus which gives us a method for evaluating definite integrals without going through the tedium of evaluating limits.

Theorem 1 (The Fundamental Theorem of Calculus Part 2): If a function $f$ is continuous on an interval $[a, b]$, then it follows that $\int_a^b f(x) \: dx = F(b) - F(a)$, where $F$ is a function such that $F'(x) = f(x)$ ($F$ is any antiderivative of $f$).

• Proof: Let $g(x) = \int_a^x f(t) \: dt$. By the The Fundamental Theorem of Calculus Part 1, we know that $g(x)$ must be an antiderivative of $f(x)$, that is $g'(x) = f(x)$. If $F(x)$ is any antiderivative of $f(x)$, then it follows that $F(x) = g(x) + C$ where $C$ is a constant. Note that $F(x)$ and $g(x)$ differ only a constant. Now suppose we evaluate $g(a)$:

(1)

\begin{align} g(a) = \int_a^a f(t) \: dt \\ g(a) = 0 \end{align}

• Hence if we evaluate $F(b)$ and $F(a)$, we obtain:

(2)

\begin{align} F(b) - F(a) = [g(b) + C] - [g(a) + C] \\ = g(b) - g(a) \\ = g(b) - 0 \\ \end{align}

• But $g(b) = \int_{a}^{b} f(t) \: dt$, so our proof is done, $F(b) - F(a) = \int_{a}^{b} f(t) \: dt$. $\blacksquare$