Real Analysis: Integrals
The Fundamental Theorem of Calculus
Introduction
The fundamental theorem of calculus connects the two concepts of differentiation and integration to allow for the calculation of closed-form solutions to wide classes of integrals. It is the single most consequential theorem in all mathematics.
The Fundamental Theorem of Calculus, Part 1
If $F$ is a real function defined on $[a, b]$ where
$$F(x) = \int\limits_a^x f(x) \, dx$$
for all $x \in [a, b],$ then $F$ is uniformly continuous on $[a, b].$ Furthermore, if $f$ is continuous at $x,$ then
$$F'(x) = f(x).$$
The Fundamental Theorem of Calculus, Part 2
If $f \in \mathscr{R}([a, b]),$ and there is a differentiable function $F$ defined on $[a, b]$ such that $F' = f,$ then
$$\int\limits_a^b f(x) \, dx = F(b) - F(a).$$
Antiderivatives & Indefinite Integrals
If $F$ is a function defined on $[a, b]$ whose derivative along $[a, b]$ is $f,$ then $F$ is an antiderivative of $f.$ But because the derivative of a constant term is $0,$ there are an infinite number of viable antiderivatives for $f$ over a given interval. For example, if $F,$ $G,$ and $H$ are given by
$$F(x) = x^2 + 3 \quad G(x) = x^2 - 20 \quad H(x) = x^2 + y$$
then each is a valid antiderivative for
$$f(x) = 2x$$
along any interval $[a, b].$
However, the choice of antiderivative does not matter in the evaluation of a definite integral because the constant terms always cancel out. As a result, the only portion of the antiderivative that matters is the portion that varies with respect to the independent variable - i.e., everything but the constant term.
Furthermore, the difficult mathematical work lies in finding an antiderivative for a particular function, rather than evaluating it. The former involves a deeper understanding of the function involved, while the latter is ultimately just algebra that can be performed by a calculator.
To reflect the arbitrariness of the constant term of an antiderivative, as well as the greater value placed on finding antiderivatives than on evaluating them, we introduce the process of indefinite integration. The indefinite integral of a function $f \in \mathscr{R}([a, b])$ is the family of functions given by
$$\int f(x) \, dx = F(x) + c,$$
where $F$ is any antiderivative of $f$ and $c$ is an arbitrary constant called the constant of integration. Note that the integral sign does not have the typical bounds on it of a definite integral. This is because the indefinite integral produces a family of functions, not a real number. The "family of functions" is technically a two-variable function whose inputs are $x$ and $c,$ but it is instead referred to as a family of functions, each member of which differs from every other by only a constant term.
For example, rather than consider $F,$ $G,$ and $H,$ from above, we would instead write out the indefinite integral of $2x$ as
$$\int 2x \, dx = x^2 + c.$$
Problems
The Fundamental Theorem of Calculus - Part 1: Show that if $F$ is a real function defined on $[a, b]$ where
$$F(x) = \int\limits_a^x f(t) \, dt$$
for all $x \in [a, b],$ then $F$ is uniformly continuous on $[a, b],$ and if $f$ is continuous at a point $x \in [a, b],$ then $F$ is differentiable at $x$ and
$$F'(x) = f(x).$$
Part 1: Uniform continuity of $F$: First, note that $f \in \mathbb{R}([a, b]),$ and that therefore $f$ is bounded. Therefore there exists some $M \geq 0$ such that $|f(t)| \leq M$ for all $t \in [a, b].$ By the subdivisibility of domain, for $x, y \in [a, b]$ where $x < y$ we see that
$$\begin{array}{rcl} \left|\displaystyle \int\limits_x^y f(t) \, dt\right| & = & \left|\displaystyle \int\limits_a^y f(t) \, dt - \displaystyle \int\limits_a^x f(t) \, dt \right|\\ \left|\displaystyle \int\limits_x^y f(t) \, dt\right| & = & \left|F(y) - F(x)\right|. \end{array}$$
By the boundedness of $f,$ it follows that
$$\left|F(y) - F(x)\right| \leq M(y - x).$$
For any $\varepsilon > 0,$ we can then see that
$$\left|F(y) - F(x)\right| < \varepsilon \text{ whenever } |y - x| < \dfrac{\varepsilon}{M}.$$
Therefore $F$ is uniformly continuous.
Part 2: Continuity at $f(x)$ implies $F'(x) = f(x)$: Assume $f$ is continuous at a point $x \in [a, b].$ Then for any $\varepsilon > 0$ there exists a $\delta > 0$ such that
$$|f(x) - f(y)| < \varepsilon \text{ whenever } |x - y| < \delta \text{ and } y \in [a, b].$$
Pick $p$ and $q$ such that $x - \delta \leq p \leq x \leq q \leq x + \delta$ and $p, q \in [a, b].$ Then
$ \left|\dfrac{F(q) - F(p)}{q-p} - f(x)\right| = \left| \displaystyle \int\limits_p^q \dfrac{f(t)}{q-p} \, dt - f(x) \right| \\ \left|\dfrac{F(q) - F(p)}{q-p} - f(x)\right| = \left| \displaystyle \int\limits_p^q \dfrac{f(t)}{q-p} \, dt - \dfrac{1}{q-p}\int\limits_p^q f(x) \, dt \right| \\ \left|\dfrac{F(q) - F(p)}{q-p} - f(x)\right| = \left| \displaystyle \int\limits_p^q \dfrac{f(t) - f(x)}{q-p} \, dt \right| \\ \left|\dfrac{F(q) - F(p)}{q-p} - f(x)\right| = \dfrac{1}{q - p}\left| \displaystyle \int\limits_p^q f(t) - f(x) \, dt \right| \\ \left|\dfrac{F(q) - F(p)}{q-p} - f(x)\right| < \dfrac{1}{q - p}\left| \displaystyle \int\limits_p^q \varepsilon \, dt \right| \\ \left|\dfrac{F(q) - F(p)}{q-p} - f(x)\right| < \dfrac{\varepsilon(q - p)}{q - p} \\ \left|\dfrac{F(q) - F(p)}{q-p} - f(x)\right| < \varepsilon.$
Setting $p=x$ shows that
$$\left|F'(x) - f(x)\right| < \varepsilon.$$
Since $\varepsilon$ is arbitrary, it follows that $F'(x) = f(x).$
The Fundamental Theorem of Calculus - Part 2: Show that if $f \in \mathscr{R}([a, b])$ and there is a differentiable function $F$ defined on $[a, b]$ such that $F' = f,$ then
$$\int\limits_a^b f(x) \, dx = F(b) - F(a).$$
Let $\varepsilon > 0.$ Then there exists a finite interval subdivision $P$ such that $U(f, P) - L(f, P) < \varepsilon.$ By the mean value theorem, for each $p_i \in P$ there is a value $t_i \in p_i$ such that
$$F(\sup(p_i)) - F(\inf(p_i)) = f(t_i)\mu(p_i).$$
It follows that
$$\sum\limits_{i=1}^{|P|} f(t_i) \mu(p_i) = F(b) - F(a),$$
as the lefthand side is a telescoping sum.
Since
$$L(f, P) \leq \sum\limits_{i=1}^{|P|} f(t_i) \mu(p_i) \leq U(f, P),$$
it follows that
$$\left|\sum\limits_{i=1}^{|P|} f(t_i)\mu(p_i) - \int\limits_a^b f(x) \, dx\right| \leq U(f, P) - L(f, P) < \varepsilon.$$
Therefore
$$\left|F(b) - F(a) - \int\limits_a^b f(x) \, dx \right| < \varepsilon.$$
Since $\varepsilon$ is arbitrary, we conclude that
$$\int\limits_a^b f(x) \, dx = F(b) - F(a).$$