Fundamental Theorem of Calculus

Section 2.3 Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus (FTC) forges the bridge between differentiation and integration. Given a function \(f(x)\) on a closed interval \(I\) and a point \(a \in I\text{.}\) One can consider the function \(F_a(x) = \int_a^x f(t) dt\text{,}\) i.e. the "signed area" from \(a\) to \(x\) underneath the graph \(y=f(x)\text{.}\) Even though \(F_a(x)\) may not exist for an arbitrary \(f(x)\text{,}\) it does exist, for example, when \(f(x)\) is piecewise continuous. A version of the FTC asserts that \(F' = f\) at the continuous points of \(f\text{.}\) In particular, if \(f\) is continuous on \(I\text{,}\) then the \(F(x)\) thus defined is an anti-derivative of \(f(x)\) (the one that vanishes at \(a\)). In addition, if \(F\) is any antiderivative of a continuous \(f\) on \(I\text{,}\) then the definite integral is evaluated by \(\int_a^b f(x)\,dx = F(b)-F(a)\text{.}\) Together, these statements tell us that “differentiation undoes integration”, and they provide a practical way to compute integrals.

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Example 2.11. Integrating a step function.

Consider the step function

\begin{equation*} f(t)= \begin{cases} 2, & 0 \le t \lt 1,\\[4pt] -1, & 1 \le t \lt 3,\\[4pt] 1, & 3 \le t \le 4. \end{cases} \end{equation*}

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By finding the area of rectangles, one checks easily that the function \(F_0(x) = \int_0^x f(t)\, dt\) is given by

\begin{equation*} F_0(x) = \begin{cases} 2x, & 0 \le x \lt 1,\\ 3 - x, & 1 \le x \lt 3,\\ x - 3, & 3 \le x \le 4,\\ \end{cases} \end{equation*}

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We now give a proof of the fundamental theorem of calculus for bounded functions on closed interval that are continuous except at finitely many points.

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Theorem 2.14. Fundamental Theorem of Calculus.

Let \(f\) be a bounded function on a closed interval \(I\text{.}\) Suppose \(f\) is continuous at all but finitely many points in \(I\text{.}\) Then for any fixed point \(a \in I\text{,}\) the function

\begin{equation*} F_a(x) = \int_a^x f(t)\; dt \end{equation*}

is continuous. Moreover, \(F_a'(x_0) = f(x_0)\) whenever \(f(x)\) is continuous at \(x_0\text{.}\)

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Proof.

By assumption \(|f(x)| \le B\) on \(I\) for some \(B\text{.}\) So, for any \(x_0 \in I\text{,}\)

\begin{equation*} |F_a(x)-F_a(x_0)| = \left| \int_{x_0}^{x} f(t)\ dt\right| \le B|x-x_0| \to 0 \end{equation*}

as \(|x-x_0| \to 0\text{.}\) Therefore, \(F_a(x)\) is continuous at \(x_0\text{.}\) Furthermore, if \(f\) is continuous at \(x_0\text{,}\) then for any \(\varepsilon \gt 0\text{,}\) there exists \(\delta \gt 0\) such that \(|f(t)-f(x_0)| \lt \varepsilon\) whenever \(|t-x_0| \lt \delta\text{.}\) Thus, for \(x \in I\) with \(|x-x_0| \lt \delta\text{,}\)

\begin{equation*} \left| \int_{x_0}^x f(t)-f(x_0)\; dt \right| \lt \varepsilon|x-x_0|. \end{equation*}

Consequently,

\begin{align*} \left| \frac{F_a(x) - F_a(x_0)}{x-x_0} - f(x_0)\right| & = \left|\frac{\int_{x_0}^{x} f(t) - f(x_0) \ dt}{x-x_0}\right| \lt \varepsilon. \end{align*}

This shows that \(F'(x_0)\) exists and is \(f(x_0)\text{.}\)

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In particular, if \(f\) is continuous on \(I\text{,}\) then \(F_a(x)\) is the antiderivative of \(f(x)\) that vanishes at \(a\text{.}\) Moreover, if \(F(x)\) is an antiderivative of \(f(x)\) then by the MVT, \(F(x)-F_a(x)\) is constant on \(I\) and that constant must be \(F(a)-F_a(a) = F(a)\text{.}\) Thus, for any \(b \in I\text{,}\) \(F(b)-F_a(b) = F(a)\text{,}\) that means

\begin{equation*} \int_a^b f(t)\ dt = F_a(b) = F(b)-F(a). \end{equation*}

This last equation turns the problem computing definite integrals of continuous functions into the one of finding antiderivatives.

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