A correction to the definition of ∫f(x)dx

Abstract. This essay points out a widely appeared mistake in some textbooks about the definition of the indefinite integral notation $\int_{}^{}{f\left( x \right)\text{dx}}$.

Some textbooks define $\int_{}^{}{f\left( x \right)\text{dx}}$ as a set of antiderivatives of $f$, such as Thomas’Calculus¹:

The collection of all antiderivatives of $f$ is called the indefinite integral of $f$ with respect to $x$, and is denoted by

$\int_{}^{}{f\left( x \right)\text{dx}}$

The symbol $\int_{}^{}{}$is an integral sign. The function $f$ is the integrand of the integral, and $x$ is the variable of integration.

And also in another widely used book James Stewart’s Calculus²:

…an indefinite integral $\int_{}^{}{f\left( x \right)\text{dx}}$ is a function (or family of functions)…

Be aware of the phrase enclosed in the bracket, a family of functions is a collection of functions whose equations are related³, hence the statement above also indicates that $\int_{}^{}{f\left( x \right)\text{dx}}$ represents a set.

However, $\int_{}^{}{f\left( x \right)\text{dx}}$ is also defined as a primitive function of $f$ in some books, such as Richard Courant & Fritz John’s Introduction to Calculus and Analysis Volume I⁴:

Every primitive function $F\left( x \right)$ of a given function $f\left( x \right)$ continuous on an interval can be represented in the form

$F\left( x \right) = c + \phi(x) = c + \int_{a}^{x}{f\left( u \right)\text{du}}$

where $c$ and $a$ are constants, and conversely, for any constant values of

$a$ and $c$ chosen arbitrarily this expression always represents a primitive function.

And also in book The Fundamentals of Mathematical Analysis Volume 1 by

G. M. Fikhtengol’ts⁵:

Consequently the expression $F\left( x \right) + C$, where $C$ is an arbitrary constant, is the general form of the function which has the derivative $f\left( x \right)$ or the differential $f\left( x \right)\text{dx}$. This expression is called the indefinite integral of $f\left( x \right)$ and is denoted by

$\int_{}^{}{f\left( x \right)\text{dx}}$

which implicitly contains the arbitrary constant.

So which definition of $\int_{}^{}{f\left( x \right)\text{dx}}$ is right?

There is no doubt that if $F\left( x \right)$ is an antiderivative of $f\left( x \right)$ on an interval $I$, then the general antiderivative of $f\left( x \right)$ on an interval $I$ is $F\left( x \right) + C$ where $C$ is an arbitrary constant, and the set of antiderivative of $f\left( x \right)$ on $I$ is $\{ F\left( x \right) + C|C \in R\}$, if one insists that $\int_{}^{}{f\left( x \right)\text{dx}}$ represents the set of antiderivative of $f\left( x \right)$, then there should be

$\int_{}^{}{f\left( x \right)dx = \{ F\left( x \right) + C|C \in R\}}$

But the convention of using $\int_{}^{}{f\left( x \right)\text{dx}}$ in practice is

$\int_{}^{}{f\left( x \right)dx = F\left( x \right) + C}$

where $C$ is any real number, and consequently exists

$\int_{}^{}{e^{x}dx = e^{x} + C}$

$\int_{}^{}{cosxdx = \text{sinx} + C}$

etc., that is, $\int_{}^{}{f\left( x \right)\text{dx}}$ is NOT treated as a set of antiderivatives of $f\left( x \right)$ in using, according to the de facto convention, $\int_{}^{}{\mathbf{f}\left( \mathbf{x} \right)\mathbf{\text{dx}}}$ should be unambiguously defined as the general form of the antiderivative of $\mathbf{f}\left( \mathbf{x} \right)$ rather than a set of antiderivatives of $f\left( x \right)$.

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1. Thomas’ calculus, 14th edition, Joel R. Hass, Christopher E. Heil, Maurice D. Weir , p236 ↩︎

2. Calculus, 8th Edition, James Stewart, p331 ↩︎

3. Calculus, 8th Edition, James Stewart, p29 ↩︎

4. Introduction to Calculus and Analysis Volume I, Reprint of the 1989 edition, Richard Courant, Fritz John, p188 ↩︎

5. The Fundamentals of Mathematical Analysis, Volume 1, 1st Edition, G. M. Fikhtengol’ts, p300 ↩︎