In the book Thomas' Calculus, the Indefinite integral is defined as follows:

we defined the indefinite integral of the function ƒ with respect to x as the set of all antiderivatives of ƒ, symbolized by $\int{ƒ(x)dx}$. Since any two antiderivatives of ƒ differ by a constant, the indefinite integral $\int$ notation means that for any antiderivative F of ƒ , $$\int{f(x)dx=F(x)+C}$$

This directly implies that:

$$\frac{dy}{dx}=f(x) \implies y(x)=\int{f(x)dx} = F(x)+C \tag{1}$$

However, in the book Differential Equations With Applications and Historical Notes, Simmons notes:

We solve it by writing: $$y(x)=\int{f(x)dx} + c \tag{2}$$

The 'it' referring to the differential equation in (1).

Question: Isn't the $c$ redundant? Since the Indefinite integral already contains the collection of antiderivatives, one would think that would be the case, however, from the first fundamental theorem of calculus we get: $$\int_{a}^{x}{f(t)dt}=\int{f(x)dx} \tag{3}$$

and by substituting (3) in (1), $$y(x)=\int_{a}^{x}{f(t)dt}=F(x)-F(a) \tag{4}$$

And now by comparing (1) and (4) we come to the conclusion that: $F(x)+C=F(x)-F(a)$, which clearly has to be False, since a is fixed but C is an arbitrary constant. The only way to make it balanced is to add another arbitrary constant to make: $F(x)+C=F(x)-F(a) + C_1$. But in order to make this happen we have to add the $C_1 to \int{f(x)dx}$ in (1) and hence confirming the validity of (2). So the $c$ in (2) seems necessary again.

How to resolve this contradiction?

  • 2
    $\begingroup$ Why are you comparing equations of two different books, both tell same thing but in different way $\endgroup$ Aug 24, 2021 at 17:42
  • 1
    $\begingroup$ @LalitTolani While they are talking about the solutions of the same differential equation, the two approaches yield different results, and that is a problem. $\endgroup$ Aug 24, 2021 at 17:47

1 Answer 1


Careful, all the second fundamental theorem says is that $$\int_{a}^{x}f(t)\,dt$$ is an antiderivative for $f(x)$. That is not the same as saying that $$\int_{a}^{x}{f(t)\,dt}=\int{f(x)\,dx}$$ which is not true. What you can write is $$\int f(x)\,dx = \int_{a}^{x}f(t)\,dt + C.$$ In any case, yes it is generally considered redundant to write $$\int f(x)\,dx + c,$$ but you would do well to double check how Simmons defines $$\int f(x)\,dx.$$ I would be surprised if their definition varies from the standard, so it may be the case that writing the $+C$ is just their way of making sure that the reader remembers the need for an arbitrary constant. There is certainly nothing incorrect in writing $$\int f(x)\,dx + c$$ because as you have pointed out, if $F(x)$ is an antiderivative of $f(x)$ then $$\int f(x)\,dx = F(x) + C,$$ and since the sum of two arbitrary constants is again an arbitrary constant it is still valid to write $$\int f(x)\,dx + c.$$

  • $\begingroup$ Since $\int_{a}^{x}{f(t)dt}$ is a function of $x$, consider $y(x)=\int_{a}^{x}{f(t)dt}$ and use (1). This will prove that $\int_{a}^{x}{f(t)dt}=\int{f(x)dx}$ $\endgroup$ Aug 24, 2021 at 19:10
  • $\begingroup$ Excuse me I wanted to write and was in fact referring to the first theorem and I have corrected the post $\endgroup$ Aug 24, 2021 at 19:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.