# What is an indefinite integral?

2012-06-10

An indefinite integral is the integral whose lower boundary is constant and whose upper boundary is variable and it should be distinguished from a primitive integral that just restores primitive functions. Constants of primitive integration should also be distinguished from constants of indefinite integration that the lower boundary produces. Integration is often said to be the inverse operation of differentiation, but it is mathematically false. Since a set of primitive functions and a set of integrands is not in one-to-one but in many-to-one relationship, integration cannot be an inverse mapping. Differentiation and integration are in the relationship of analysis and synthesis. As a derivative has only partial information on the original function, it cannot restore the whole information.

## 1. The previous explanation of the indefinite integral

What is an indefinite integral? Before I answer the question, let us recognize how it has been explained. There are so many explanations as exponents, but the following is the typical and common one.

An indefinite integral is defined as a set of functions whose derivative is equal to a given function. These functions are named primitive functions or antiderivatives, while the derivative is named an integrand meaning a function to be integrated. An indefinite integral is also called a primitive integral.

As the derivative of a constant is zero, you must add an arbitrary constant C known as the constant of integration so as to construct primitive functions. Let the primitive functions of an integrand f(x) be F(x) + C and we can say:

$f(x)=\frac{d}{dx}\left (F(x)+C \right ) \Leftrightarrow \int f(x) dx=F(x)+C$

Since the integrand f(x) derives from differentiating the primitive functions F(x) + C and integrating the integrand restores the primitive functions, integration is often regarded as an inverse operation of differentiation[1]. The fundamental theorem of calculus seems to demonstrate the inverse relationship between them.

Let f(x) be a continuous function defined on an interval [a, b] and F(x) be its primitive function, continuous on [a, b] and differentiable on the open interval (a, b). For all x in (a, b),

$\frac{d}{dx}\int_a^x f(t)dt = f(x)$

Because of this theorem, some identify an indefinite integral with an integral whose lower boundary is a constant (a) and whose upper boundary is a variable (x) [2], but others disagree on it.

There are four definitions of a primitive function in terms of the integral sign according as the treatment of the constant of integration in addition to this disagreement.

$\; \int f(x) dx$
$\; \int f(x) dx + C$
$\; \int_a^x f(x) dx$
$\; \int_a^x f(x) dx + C$

The lower boundaries a of Eq. 5, 6 is omissible, which however does not mean it has no lower boundaries.

## 2. The indefinite integral should be distinguished from the primitive

The fact of many definitions of a primitive function in terms of the integral sign reflects the confusion of the previous explanations. Although differentiating the four formulae leads to the same derivative, they are not equal. The most questionable of the four is Eq. 5, which Wikipedia (Japanese) adopts[3].

Eq. 5 results in

$\int_a^x f(t)dt =\left [ F(t)+C \right ]_{a}^{x}=\left ( F(x)+C-(F(a)+C) \right )=F(x)-F(a)$

If you regard -F(a) of the last formula as a constant of integration C, it can have the form F(x)+C, but it may not represent all primitive functions. For example, when a cosine function is integrated,

$\int_a^x cos\theta\, d\theta =\left [\, sin\theta+C\, \right ]_{a}^{x}=sinx-cosa$

The domain of -cosa is limited, whatever a may be, as follows

$-1\leq -cosa \leq 1$

So, it cannot represent all constants of the integral.

To prevent such confusion, I propose distinguishing an infinite integral from a primitive integral. That is to say, I define a primitive integral as such that merely constructs primitive functions.

$\int _{P} f(x) dx = F(x) + C$

The subscript “P" is omissible so that this definition is equivalent to the definition of Eq. 3.

On the other hand, I define an indefinite integral as the integral whose lower boundary is constant and whose upper boundary is variable.

$\int_a^x f(t) dt=\left [ \int _{P} f(t) dt \right ]_{a}^{x} = \left [F(t) + C \right ]_{a}^{x}=F(x)-F(a)$

Even if their lower boundary is variable and their upper boundary is constant or both are variables, they can be called indefinite integrals in a broad sense, but the indefinite integral in the narrow sense is the most important owing to the fundamental theorem of calculus (Eq. 2).

Substituting a constant b of Eq. 11 for a variable x makes the indefinite integral a definite integral. It means that the primitive integration is an elemental integration that is used both indefinite and definite integration.

## 3. Integration produces two kinds of constants

What has been called “a constant of integration" is added at the primitive integration and it should be conceptually distinguished from the one that the lower boundary of the indefinite integral produces. So I propose naming the former “a constants of primitive integration" and the latter “a constant of indefinite integration". Constants that differentiation of primitive functions reduces to zero are those of primitive integration, while constants that differentiation at the fundamental theorem of calculus reduces to zero are those of indefinite integration and constants of primitive integration is canceled before differentiation.

$\frac{d}{dx}\int_a^x f(t)dt =\frac{d}{dx} \left [ F(t)+C \right ]_{a}^{x}=\frac{d}{dx}\left ( F(x)+C-(F(a)+C) \right )=\frac{d}{dx}(F(x)-F(a))=\frac{dF(x)}{dx}= f(x)$

Let us recognize the difference of the two constants in a graph. The following figure CI depicts the solutions of a differential equation:

$\frac{dy}{dx}=x^{2}-x-2$

as a direction field and illustrates three sample integral lines.

The blue, brown and turquoise lines represent the following functions respectively:

$y=\frac{1}{3}x^{3}-\frac{1}{2}x^{2}-2x+4$
$y=\frac{1}{3}x^{3}-\frac{1}{2}x^{2}-2x$
$y=\frac{1}{3}x^{3}-\frac{1}{2}x^{2}-2x-4$

They differ in constants of primitive integration by 4. Constants of indefinite integration are the values at lower boundary and three lines have different constants of indefinite integration. Integral (represented by a red line) itself is, however, the same in any line when the lower boundary is 2 and the upper boundary is 3.

## 4. Integration is not the inverse operation of differentiation

Another common mistake is to consider indefinite or primitive integration to be the inverse operation of differentiation. The mathematically defined inverse operation is inverse mapping and we must examine whether integration is inverse mapping or not.

Suppose ƒ is a function mapping elements of a set X to elements of a set Y.

$f\colon X \to Y$

If this mapping is bijection, namely “onto" surjection and “one-to-one" injection, it has a reverse function:

$f^{-1}\colon Y \to X$

Let ƒ be an invertible function and the following universal proposition is true.

$(\forall x) \left (x\in X \wedge f^{-1}\left( \, f(x) \, \right) = x \right )$

For example, multiplication is an inverse operation of division by this standard.

Assuming X is a set of primitive functions and Y is a set of integrands, f corresponds to differentiation and f -1 to integration. Differentiation after integration restores the original function but integration after differentiation does not. So, this universal proposition (Eq. 19) does not apply to differentiation and integration. This is because the set of primitive functions and the set of integrands are not in one-to-one but in many-to-one relationship.

One-to-one mapping is information-preserving, while many-to-one mapping is information-losing. Primitive integration needs the information on a constant of primitive integration and indefinite integration needs the information on a constant of indefinite integration in addition to it. Differentiation loses these two pieces of information, but if we have them, we can restore the original function.

Though differentiation and integration are not in the relationship of inverse operation, they are in the relationship of analysis and synthesis, and therefore we can say integration is the cognition in the opposite direction to differentiation. As Kant’s theory of analytic and synthetic propositions indicates, a product of synthesis has more information than a product of analysis. Generally speaking, integration is more difficult than differentiation. This is because the former requires more information than the latter.

Integral calculus derived from the method of exhaustion of the ancient Greek and was developed independently of differential calculus. Their attempt to reduce the complexity of the whole area by dividing a complex figure into elemental figures whose area was known did not succeeded in exact calculation of the whole area. Old-fashioned systematists might have given up its solution, advocating the holistic proposition, “the whole is more than the sum of the parts." Still reductionism itself is not wrong. They just did not make the most of the determinacy of functions, the invisible part which makes the whole more than the sum of the visible parts.

Integral calculus became easy after James Gregory (1638–1675), Isaac Barrow (1630–1677), Isaac Newton (1643–1727) and Gottfried Leibniz (1646–1716) developed the fundamental theorem of calculus. They succeeded in calculating the area by regarding the function of a boundary line as the rate of area change and integrating the infinitesimal change into the whole area. Integral calculus has been applied to not only area calculating but also any product of variables and their functions. Integral and differential calculus was a successful case of Cartesian reductionism that analyzes the whole into elements and reconstructs them to the whole.

## 5. References

1. 青本 和彦 (著), 砂田 利一 (著).『微分と積分1 － 岩波講座 現代数学への入門 1』 岩波書店 (October 5, 1995). p. 107.
2. 黒田 成俊. 『微分積分 － 共立講座 21世紀の数学 1』 共立出版 (September 1, 2002). p. 145.
3. Wikipedia. “不定積分." 2012年5月24日 (木) 02:39.
4. Pbrks. “Slope field of the equation dy/dx=x^2-x-2." 6 June 2008. Licensed under CC-0.