Section 7.3 The Fundamental Theorem of Calculus

We now return to the theme announced at the end of Section 5.4: differentiation and integration are inverse processes, at least under suitable hypotheses. We present the theorem in two standard forms.

Subsection 7.3.1 Indefinite Integrals

Let \(f \in R[a,b]\text{.}\) For each \(c \in [a,b]\text{,}\) the function

\begin{equation*} x \mapsto \int_c^x f(t)\,dt \end{equation*}

is called the indefinite integral of \(f\) with base point\(c\text{.}\)

Theorem 7.3.1. Fundamental Theorem of Calculus (Second Form).

Let \(f \in R[a,b]\text{,}\) and define

\begin{equation*} F(x)=\int_a^x f(t)\,dt. \end{equation*}

Then \(F\) is Lipschitz continuous on \([a,b]\text{.}\) Moreover, if \(f\) is continuous at \(x_0 \in [a,b]\text{,}\) then \(F\) is differentiable at \(x_0\) and

\begin{equation*} F'(x_0)=f(x_0). \end{equation*}

Proof.

Since \(f\) is integrable on \([a,b]\text{,}\) it is bounded. Choose \(M \ge 0\) such that \(|f(x)| \le M\) for every \(x \in [a,b]\text{.}\) Let \(x,x' \in [a,b]\text{.}\) If \(x \lt x'\text{,}\) then part (3) of Basic Properties of the Integral gives

\begin{equation*} F(x')-F(x)=\int_x^{x'} f(t)\,dt. \end{equation*}

Therefore, by part (4) of the same theorem,

\begin{equation*} |F(x')-F(x)| = \left|\int_x^{x'} f(t)\,dt\right| \le M|x'-x|. \end{equation*}

The same inequality is trivial if \(x=x'\text{,}\) and it follows by symmetry when \(x' \lt x\text{.}\) Thus \(F\) is Lipschitz continuous on \([a,b]\text{.}\)

Now suppose that \(f\) is continuous at \(x_0 \in [a,b]\text{.}\) Let

\begin{equation*} \varphi(x)= \begin{cases} \dfrac{1}{x-x_0}\int_{x_0}^{x} f(t)\,dt & \text{if } x \ne x_0,\\ f(x_0) & \text{if } x=x_0. \end{cases} \end{equation*}

Then

\begin{equation*} F(x)-F(x_0)=\varphi(x)(x-x_0) \end{equation*}

for every \(x \in [a,b]\text{.}\) We will prove that \(\varphi\) is continuous at \(x_0\text{.}\) Let \(\varepsilon \gt 0\text{.}\) There exists \(\delta \gt 0\) such that

\begin{equation*} |f(t)-f(x_0)| \lt \varepsilon \end{equation*}

whenever \(t \in [a,b]\) and \(|t-x_0| \lt \delta\text{.}\) If \(x \in [a,b]\) with \(0 \lt |x-x_0| \lt \delta\text{,}\) then

\begin{equation*} \begin{aligned} |\varphi(x)-\varphi(x_0)| &= \left| \frac{1}{x-x_0}\int_{x_0}^{x} f(t)\,dt - \frac{1}{x-x_0}\int_{x_0}^{x} f(x_0)\,dt \right| \\ &\le \frac{1}{|x-x_0|} \int_{\min\{x,x_0\}}^{\max\{x,x_0\}} |f(t)-f(x_0)|\,dt \\ &\lt \frac{1}{|x-x_0|}\varepsilon |x-x_0| = \varepsilon. \end{aligned} \end{equation*}

Thus \(\varphi(x) \to \varphi(x_0)\) as \(x \to x_0\text{,}\) so \(\varphi\) is continuous at \(x_0\text{.}\) By Carathéodory’s Criterion (Proposition 5.1.4), it follows that \(F\) is differentiable at \(x_0\) and \(F'(x_0)=\varphi(x_0)=f(x_0)\text{.}\)

Corollary 7.3.2.

If \(f \in C[a,b]\text{,}\) then the indefinite integral

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

is an antiderivative, or primitive, of \(f\) on \([a,b]\text{.}\)

Proof.

By Continuous Functions Are Integrable, every continuous function on \([a,b]\) is Riemann integrable. Therefore Fundamental Theorem of Calculus (Second Form) applies at every point of \([a,b]\) and yields \(F'=f\text{.}\)

In general, however, an indefinite integral need not be differentiable at every point, and even when it is differentiable at a point, its derivative there need not equal the original integrand.

Example 7.3.3.

Consider the signum function

\begin{equation*} \operatorname{sgn}(x)= \begin{cases} 1 & \text{if } x \gt 0,\\ 0 & \text{if } x=0,\\ -1 & \text{if } x \lt 0. \end{cases} \end{equation*}

Since this is a step function, it is integrable on \([-1,1]\text{.}\) Its indefinite integral with base point \(0\) is

\begin{equation*} F(x)=\int_0^x \operatorname{sgn}(t)\,dt=|x|. \end{equation*}

The function \(|x|\) is not differentiable at \(0\text{,}\) so the indefinite integral of an integrable function need not be differentiable everywhere.

Example 7.3.4.

Let \(h\) be the Thomae function from Example 7.2.8. Since \(h \ge 0\) and \(\int_0^1 h=0\text{,}\) part (2) of Basic Properties of the Integral implies that

\begin{equation*} 0 \le \int_0^x h(t)\,dt \le \int_0^1 h(t)\,dt=0 \end{equation*}

for every \(x \in [0,1]\text{.}\) Hence the function

\begin{equation*} H(x)=\int_0^x h(t)\,dt \end{equation*}

is identically zero on \([0,1]\text{.}\) Therefore \(H'(x)=0\) for every \(x \in [0,1]\text{,}\) while \(h(q) \gt 0\) for every rational \(q \in [0,1]\text{.}\) So it can happen that the indefinite integral is differentiable at a point but its derivative does not equal the integrand there.

Subsection 7.3.2 The Two Forms of the Theorem

Theorem 7.3.5. Fundamental Theorem of Calculus (First Form).

Suppose \(F \colon [a,b] \to \R\) is differentiable on \([a,b]\) and that \(F' \in R[a,b]\text{.}\) Then

\begin{equation*} \int_a^b F'(x)\,dx = F(b)-F(a). \end{equation*}

Proof.

Since differentiability implies continuity, \(F\) is continuous on \([a,b]\text{.}\) Let \(P=\{x_0,\dots,x_n\}\) be any partition of \([a,b]\text{,}\) and let \(m_i\) and \(M_i\) be the infimum and supremum of \(F'\) on \([x_{i-1},x_i]\text{.}\) By the Mean Value Theorem (Theorem 5.3.3), for each \(i=1,\dots,n\) there exists \(t_i \in (x_{i-1},x_i)\) such that

\begin{equation*} F(x_i)-F(x_{i-1})=F'(t_i)(x_i-x_{i-1}). \end{equation*}

Hence

\begin{equation*} m_i(x_i-x_{i-1}) \le F(x_i)-F(x_{i-1}) \le M_i(x_i-x_{i-1}). \end{equation*}

Summing over \(i\) gives

\begin{equation*} L(P,F') \le \sum_{i=1}^n \bigl(F(x_i)-F(x_{i-1})\bigr) \le U(P,F'). \end{equation*}

The sum in the middle telescopes, so

\begin{equation*} L(P,F') \le F(b)-F(a) \le U(P,F'). \end{equation*}

Since \(P\) was arbitrary, we obtain

\begin{equation*} \underline{\int_a^b} F' \le F(b)-F(a) \le \overline{\int_a^b} F'. \end{equation*}

Because \(F'\) is integrable, the upper and lower integrals are equal. Therefore

\begin{equation*} \int_a^b F'(x)\,dx = F(b)-F(a). \end{equation*}

Corollary 7.3.6.

Let \(f \colon [a,b] \to \R\) be continuous, and let \(F\) be an antiderivative of \(f\) on \([a,b]\text{.}\) Then

\begin{equation*} \int_a^b f(x)\,dx = F(b)-F(a). \end{equation*}

Proof.

By Continuous Functions Are Integrable, the function \(f\) is Riemann integrable on \([a,b]\text{.}\) Since \(F'=f\text{,}\) the conclusion follows from Fundamental Theorem of Calculus (First Form).

When \(f\) is continuous, the first form may also be recovered from the second form. Indeed, if

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

then Fundamental Theorem of Calculus (Second Form) shows that \(F_a'=f\text{.}\) Hence \((F-F_a)'=0\) on \((a,b)\text{,}\) so Corollary 5.4.5 yields \(F(x)=F_a(x)+F(a)\text{.}\) Evaluating at \(x=b\) gives \(\int_a^b f = F(b)-F(a)\text{.}\)

Example 7.3.7.

The Thomae function is integrable on \([0,1]\text{,}\) but it has no primitive on \([0,1]\text{.}\) Indeed, if \(H\) were a primitive of \(h\text{,}\) then by Fundamental Theorem of Calculus (First Form),

\begin{equation*} H(x)-H(0)=\int_0^x h(t)\,dt = 0 \end{equation*}

for every \(x \in [0,1]\text{.}\) Thus \(H\) would be constant, which is impossible because then \(H'=0\) everywhere, while \(h(q) \gt 0\) for rational \(q\text{.}\)

Example 7.3.8.

Let

\begin{equation*} H(x)=2\sqrt{x} \qquad (x \in [0,1]). \end{equation*}

Then \(H\) is continuous on \([0,1]\text{,}\) and for \(x \in (0,1]\text{,}\)

\begin{equation*} H'(x)=\frac{1}{\sqrt{x}}. \end{equation*}

This derivative is unbounded on \([0,1]\text{,}\) so it is not Riemann integrable no matter how one defines \(H'(0)\text{.}\) Thus Theorem 7.3.5 does not apply.

Example 7.3.9.

Define

\begin{equation*} F(x)= \begin{cases} x^2\sin(1/x) & \text{if } x \ne 0,\\ 0 & \text{if } x=0 \end{cases} \end{equation*}

on \([-1,1]\text{.}\) Then \(F\) is differentiable on \([-1,1]\text{,}\) and

\begin{equation*} F'(x)= \begin{cases} 2x\sin(1/x)-\cos(1/x) & \text{if } x \ne 0,\\ 0 & \text{if } x=0. \end{cases} \end{equation*}

The derivative is bounded on \([-1,1]\) and is continuous except at \(0\text{,}\) so \(F'\) is integrable. By Theorem 7.3.5,

\begin{equation*} \int_{-1}^1 F'(x)\,dx = F(1)-F(-1) = \sin(1)-(-\sin(1)) = 2\sin(1). \end{equation*}

Example 7.3.10.

Let

\begin{equation*} F(x)= \begin{cases} x^2\cos(1/x^2) & \text{if } x \ne 0,\\ 0 & \text{if } x=0. \end{cases} \end{equation*}

Then \(F\) is differentiable on \([0,1]\text{,}\) and

\begin{equation*} F'(x)= \begin{cases} 2x\cos(1/x^2)+\dfrac{2}{x}\sin(1/x^2) & \text{if } x \ne 0,\\ 0 & \text{if } x=0. \end{cases} \end{equation*}

The term \((2/x)\sin(1/x^2)\) is unbounded near \(0\text{,}\) so \(F'\) is not Riemann integrable on \([0,1]\text{.}\) Again, Theorem 7.3.5 does not apply.

For a much broader characterization of Riemann integrability, see Appendix E. In particular, Theorem E.1.5 explains why bounded functions such as the Thomae function are integrable even though they are not continuous everywhere.

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