Theorem 6.1.1. The Fundamental Theorem of Calculus (FTC).
If \(f(x)\) is continuous on \([a,b]\) and \(F'(x)=f(x)\) there, then \(\int_a^b f(x)dx = F(b)-F(a)\)
Section 6.1 The Theorem
Even with the basic properties theorem Theorem 5.4.1, there are still very few integrals we can find exactly. This next theorem resolves that problem in a dramatic and fairly comprehensive way.
In other words, all we really need to understand is differentiation. Integrals are just an application of differentiation! So long as we can get good at moving *backwards* through differentiation to *pre* or *anti* derivatives, we can get the value of integrals exactly and quickly. For example, let’s look at \(x^2-1\) from the earlier section. A little thought reveals that
\begin{equation*}
\left( \frac{x^3}{3} - x \right)' = x^2 - 1
\end{equation*}
So, \(\frac{x^3}{3} - x\) is an antiderivative of \(x^2 - 1\) and according to the FTC, we have
\begin{equation*}
\begin{array}{lll}
\int_{-1}^2 (x^2-1)dx \amp = \amp \left(\frac{2^3}{3}-2 \right) - \left(\frac{(-1)^3}{3}-(-1) \right) \\
\amp = \amp \left(\frac{8}{3}-2 \right) - \left(\frac{-1}{3} + 1 \right)\\
\amp = \amp \frac{8}{3}-2 + \frac{1}{3}-1 \\
\amp = \amp \frac{9}{3}-3\\
\amp = \amp 3-3 \\
\amp = \amp 0 \\
\end{array}
\end{equation*}
Just as we found with the limit of Riemann sums earlier!
