KindiakMath

Defining Integration

Having discussed differentiation, many courses define integration as the reverse of differentiation. While functionally useful in standardised tests, this approach to introduce integration is weak.

Firstly, integration being the reverse of differentiation is more a consequence of integration as area, via the fundamental theorem of calculus, rather than the other way around. Secondly, there are functions such as e^{-x^2} that do not have integrals, by this definition. This poses major challenges, since this function plays a crucial role in probability theory, where integration is, pun intended, integral to the discussion.

We will therefore define integration in the sense of Riemann. To keep the discussion concise, we will define the integral of continuous functions and piecewise-continuous functions.

Definition 1. Let f :[a, b] \to \mathbb R be a continuous function (and thus bounded). Define the Riemann integral of f by

\displaystyle \int_a^b f(x)\,\mathrm dx := \lim_{n \to \infty} \sum_{k=0}^{n-1} f(x_k) \Delta x_k,

whenever the right-hand limit exists, where x_k = a + k \cdot (b-a)/n and \Delta x_k = x_{k+1} - x_k.

Intuitively, this denotes the signed area of the region enclosed by the curve y = f(x), the x-axis, and the lines x=a,x=b.

Interestingly, to establish that this definition actually make sense takes a lot of effort beyond calculus, into real analysis, which takes advantage of the technical properties satisfied by continuous functions. We omit this discussion for now.

It should be straightforward to verify that the process of taking Riemann integrals is linear, that is, for continuous functions f,g :[a,b] \to \mathbb R and any real constant k,

\begin{aligned} \int_a^b(f(x)+g(x))\, \mathrm dx &= \int_a^b f(x)\, \mathrm dx + \int_a^b g(x)\, \mathrm dx, \\ \int_a^b (kf(x))\, \mathrm dx &= k \int_a^b f(x)\, \mathrm dx, \end{aligned}

since taking finite sums \sum_{k=0}^{n-1}(\cdot) and taking limits n \to \infty are linear.

Definition 2. Let f : [a,b] \to \mathbb R be bounded. Suppose f is continuous on except at some point c \in (a, b). Then the integrals

\begin{aligned} \displaystyle \int_a^c f(x)\, \mathrm dx &:= \lim_{t\to 0^+} \int_a^{c-t} f(x)\, \mathrm dx\\ \displaystyle \int_c^b f(x)\, \mathrm dx &:= \lim_{t\to 0^+} \int_{c+t}^b f(x)\, \mathrm dx \\ \int_a^b f(x)\, \mathrm dx &:= \int_a^c f(x)\, \mathrm dx + \int_c^b f(x)\, \mathrm dx. \end{aligned}

can be shown to be well-defined.

Once again to ensure that taking t \to 0^+ yields legitimate integrals requires careful—pun intended—analysis. Assuming that we have formally implemented these notions through real analysis, we just need one lemma to prove the fundamental theorem of calculus.

Lemma (Monotonicity). Let f : [a,b] \to \mathbb R be continuous. Suppose f \geq 0. Then

\displaystyle \int_a^b f(x)\, \mathrm dx \geq 0.

Consequently, if f,g : [a,b] \to \mathbb R are continuous and f \geq g, then

\displaystyle \int_a^b f(x)\, \mathrm dx \geq \int_a^b g(x)\, \mathrm dx .

Proof. By definition of the integral

\displaystyle \displaystyle \int_a^b f(x)\,\mathrm dx = \lim_{n \to \infty} \sum_{k=0}^{n-1} \underbrace{f(x_k)}_{\geq 0} \underbrace{\Delta x_k}_{> 0} \geq \lim_{n \to \infty} \sum_{k=0}^{n-1} 0 = 0.

Now we are ready to prove the fundamental theorem of calculus.

Fundamental Theorem of Calculus. Let f : [a,b] \to \mathbb R be continuous. Then the function F : [a, b] \to \mathbb R defined by

\displaystyle F(x) := \int_a^x f(t)\, \mathrm dt.

is differentiable on (a,b) and F' = f.

Proof. Write the difference quotient as

\displaystyle \begin{aligned} \frac{F(x+h)-F(x)}{h} &= \frac 1h \left(\int_a^{x+h} f(t)\, \mathrm dt - \int_a^x f(t)\, \mathrm dt\right) \\ &= \frac 1h \int_x^{x+h} f(t)\, \mathrm dt. \end{aligned}

Since f is continuous on [x,x+h], use the extreme value theorem to find real numbers c_1(h), c_2(h) \in [x,x+h] such that for any t \in [x,x+h],

f(c_1(h)) \leq f(t) \leq f(c_2(h)).

Integrating all sides with respect to t, monotonicity implies

\displaystyle f(c_1(h))h = \int_x^{x+h} f(c_1(h))\, \mathrm dt \leq \int_x^{x+h} f(t)\, \mathrm dt \leq \int_x^{x+h} f(c_2(h))\mathrm dt = f(c_2(h)) h.

Dividing by h,

\displaystyle f(c_1(h)) \leq \frac 1h \int_x^{x+h} f(t)\, \mathrm dt \leq f(c_2(h)).

Since x \leq c_1(h), c_2(h) \leq x+h, by the squeeze theorem, c_1(h),c_2(h) \to x as h \to 0. By continuity, f(c_1(h)) , f(c_2(h)) \to f(x). By the squeeze theorem again,

\displaystyle F'(x) = \lim_{h \to 0} \frac{F(x+h) - F(x)}{h} = \lim_{h \to 0} \frac 1h \int_x^{x+h} f(t)\, \mathrm dt = f(x),

as required.

This establishes (partly) that differentiation and integration are indeed reverses of each other.

—Joel Kindiak, 19 Oct 24, 1809H

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