KindiakMath

What with the +C?

Students in integral calculus are often penalised for omitting the arbitrary constant of integration, namely,

\displaystyle \int x^2\, \mathrm dx = \frac{x^3}{3}\quad \text{vs.} \quad \displaystyle \int x^2\, \mathrm dx = \frac{x^3}{3} + C.

Here, the answer on the left would be penalised for incompleteness. But why?

There are several layers of explanations for this, but all of them are re-packagings of the following result:

Constancy Lemma. Let f : [a, b] \to \mathbb R be continuous and differentiable on (a, b). Suppose f' = 0. Then for any x \in [a, b], f(x) = f(a).

Proof. Fix x \in (a, b]. By the mean value theorem, find c \in (a, x) such that

f(x) = f(a) + f'(c) (x - a) = f(a) + 0 \cdot (x - a) = f(a).

In other words, a differentiable function with identically zero gradient is necessarily constant.

Therefore, for any constant +C, it is no surprise that differentiating that yields 0. By the fundamental theorem of calculus, integrating 0 should yield any constant. We don’t even know what this constant is!

We can formalise this finding using this result, known as the second part of the fundamental theorem of calculus.

Fundamental Theorem of Calculus II (FTC II). Let f : [a, b] \to \mathbb R be continuously differentiable (i.e. f is differentiable on (a, b) and f' be continuous on [a, b]). Then for any x \in [a, b] and any choice of f(a),

\displaystyle f(x) = f(a) + \int_a^x f'(t)\, \mathrm dt.

Denoting f(a) = C,

\displaystyle f(x) = C + \int_a^x f'(t)\, \mathrm dt.

Proof. Define g : [a, b] \to \mathbb R by

\displaystyle g(x) := f(x) - \int_a^x f'(t)\, \mathrm dt.

By the first fundamental theorem of calculus, g is differentiable and

\displaystyle g'(x) := f'(x) - f'(x) = 0.

By the constancy lemma,

\displaystyle f(x) - \int_a^x f'(t)\, \mathrm dt = g(x) = g(a) = f(a).

By algebra,

\displaystyle f(x) = f(a) + \int_a^x f'(t)\, \mathrm dt.

The existence of the f(a) is the reason for the +C. The quantity f(a) could have been any number, and the derivative of f would still be f'. In fact, once f(a) is specified, we necessarily obtain a unique f(x). This plays a crucial role in studying differential equations. This is also why we use the notation

\displaystyle \int f'(x)\, \mathrm dx = f(x) + C,

where f(x) can be computed quickly, and the +C allows us to abbreviate the infinitely many possible choices for f(a).

For all intents and purposes, we write f' to mean any function where we know what f is. Usually, we choose a convenient expression for f(x). We will complete a quick calculation to illustrate this calculation point.

Theorem. We have for nonnegative integers n,

\displaystyle \int x^n\, \mathrm dx = \frac{x^{n+1}}{n+1} + C.

Proof. Define f(x) = x^{n+1}/(n+1). By the power rule

f'(x) = \displaystyle \frac{\mathrm d}{\mathrm dx} \left( \frac{x^{n+1}}{n+1} \right) = \frac{(n+1)x^n}{n+1} = x^n.

By FTC II,

\displaystyle \int f'(x)\, \mathrm dx = f(x) + C.

Substituting,

\displaystyle \int x^n\, \mathrm dx = \frac{x^{n+1}}{n+1} + C.

The final observation that

\displaystyle \int \frac{\mathrm d}{\mathrm d x}(f(x))\, \mathrm dx = \int f'(x)\, \mathrm dx = f(x) + C,

justifies the popular heuristic of integration as “reverse differentiation”, plus C.

—Joel Kindiak, 23 Oct 24, 0908H

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