I am trying to convince my friend that the integral of $0$ is $C$, where $C$ is an arbitrary constant. He can't seem to grasp this concept. Can you guys help me out here? He keeps saying it is $0$.

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    $\begingroup$ Let $0$ be the only answer to the integral of $\int0\;dx$. Thus, $\dfrac{d}{dx}f(x)=0$ is satisfied ONLY by $f(x)=0$ However, Let $f(x)=0+c,c\in \mathbb{R}$. $$\dfrac{d(0+c)}{dx} =0$$ Thus the assumption that there is only one function satisfying the condition is false. $$\blacksquare$$ $\endgroup$
    – Inquest
    Commented Jan 26, 2013 at 2:30
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    $\begingroup$ Whether or not the integral is $0$ or $C$ depends on whether you are talking about the indefinite or definite integral. $\endgroup$
    – anon
    Commented Jan 26, 2013 at 2:47
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    $\begingroup$ Maybe it is confusing cause it is "the" integral. Make it clear that "the" integral is not a single function.... $\endgroup$
    – N. S.
    Commented Jan 26, 2013 at 3:14
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    $\begingroup$ Definite integrals $\ne$ Primitive integrals $=$ Antiderivatives. $\endgroup$
    – Did
    Commented Jan 26, 2013 at 3:15
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    $\begingroup$ May be he is having the difficulty because: He sees an integral as the area below the curve and the $x$ axis. So under any lower and upper limit it's integral(note: the definition at the beginning ) is zero. $\endgroup$
    – hrkrshnn
    Commented Jun 13, 2013 at 15:29

9 Answers 9


Taking the derivative of any constant function is 0, i.e. $\frac {d}{dx} c = 0$ So the indefinite integral $\int0 \,dx$ produces the class of constant functions, that is $f(x) = c$ for some $c$.

There's something that you have to look at here though, that is "what about the fact $\alpha \int f dx = \int \alpha f dx $?" Can't you say:

$$\int 0\,dx = \int 0 \cdot 1 \,dx = 0 \int 1 \,dx = 0x = 0$$

This gives two conflicting answers. The question is far more complicated that you would first think. But when you say $\int f dx$ and the interval over which you're integrating isn't obvious or defined, what you really mean is "the class of functions that when derived with respect to $x$ produce $f$". The rule stated only applies for definite integrals. That is:

$$\int_a^b\alpha f\,dx = \alpha \int_a^bf \,dx$$

And if you look at textbooks on real analysis (I just looked at Rudin) that's the form in which you will find the theorem.

It should also be noted that the definite integral of $0$ over any interval is $0$, as $\int 0 \,dx = c - c = 0. $

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    $\begingroup$ The answers are only conflicting if you take the position that the arbitrary constant gets added to the integral "before" we multiply by the coefficient $0$, and I see no reason to take that position. IE, why shouldn't I say that $0 \int 1 dx = 0x + C$? $\endgroup$ Commented Apr 18, 2016 at 23:31
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    $\begingroup$ What is the reason to take your position? Why not $0\int 1\, dx = 0\cdot (x + C)$ $\endgroup$
    – Zduff
    Commented Mar 31, 2017 at 20:39
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    $\begingroup$ Don't think this answers the question... I still don't see why $\int 0\,dx = \int 0 \cdot 1 \,dx = 0 \int 1 \,dx = 0(x+c) = 0$ altought we know that the derivative of any constant is $0$ and thus that the integral of $0$ must be a constant. $\endgroup$
    – Quaerendo
    Commented Mar 2, 2020 at 11:49
  • $\begingroup$ Is the indefinite $\int_0^\infty 0 dx $ then also equal to zero? $\endgroup$
    – asmaier
    Commented Apr 22, 2020 at 10:08

You are correct, $\int 0 dx = 0 + C = C$

Your friend is not entirely wrong because $C$ could equal $0$. ie. if
$f(x) = 0$ is one antiderivative. But in general we do not know $C$ unless we are given some initial condition.

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    $\begingroup$ @downvote ??? Lol. $\endgroup$
    – Rustyn
    Commented Jan 26, 2013 at 6:02
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    $\begingroup$ If $C$ is assumed to be selected randomly, the probability that the friend is right is 0. [tongue in cheek] $\endgroup$
    – Thomas
    Commented Jan 26, 2013 at 7:19


There are two types of integrals at play here. Definite integrals are the ones that describe the actual area under a curve. Indefinite integrals are the ones that describe the anti-derivative.

There's no paradox, really. When speaking of indefinite integrals, the integral of $0$ is just $0$ plus the usual arbitrary constant, i.e.,

$\int 0 \, dx = 0 + C = C $

There's no contradiction here. When evaluating the area under a curve $f(x)$, we find the antiderivative $F(x)$ and then evaluate from $a$ to $b$:

$$\int^{b}_{a} f(X) \, dx = F(b) - F(a)$$

So, for $f(x) = 0$, we find $F(x) = C$, and so $ F(b) - F(a) = C - C = 0$. Thus, the total area is zero, as we expected.


Indefinite integrals (anti-derivatives) are known modulo a constant function. With definite integrals, the case is different: $$ \int_a^b0\,\mathrm{d}t=0 $$

One way to verify that $C$ is the anti-derivative of $0$ is simply $$ \frac{\mathrm{d}}{\mathrm{d}t}C=0 $$


There is an interesting subtlety here that I think might be tripping up your friend (I've seen it trip up a number of students of my own). None of the existing answers focus on it, although guest196883's answer effectively mentions it in the final sentence, so I'll say a bit about it here.

Suppose $f$ is a continuous function defined on all of $\mathbb{R}$, and consider the following closely-related-but-not-quite-identical classes of functions associated to $f$:

  • Antiderivatives of $f$, that is, functions $F$ satisfying ${dF\over dx}=f$.

  • Area functions of $f$, that is, functions $G$ satisfying $$\int_a^xf(t)dt=G(x)$$ for some "starting point" $a\in\mathbb{R}$.

(Sadly "area function" isn't a standard term; I don't think there is one. It would be more precise to say "definite integral with the upper bound being a variable," but that's a mouthful, and it's misleading to just call these "definite integrals" either in my opinion.)

The fundamental theorem of calculus tells us that every area function of $f$ is an antiderivative of $f$. Usually the converse is true as well. For instance, take as our $f$ the function $$f(x)=x^2$$ and consider the antiderivative $$F(x)={x^3\over 3}+17.$$ By a careful choice of "starting point" $a$, we can construe $F$ as an area function: it's a good exercise to show that $$F(x)=\int_{-\sqrt[3]{3\cdot 17}}^xt^2dt.$$

The right hand side equals ${x^3\over 3}-{(-\sqrt[3]{3\cdot 17})^3\over 3}={x^3\over 3}-17.$

However, sometimes not every antiderivative is an area function. This is exactly what's happening with the constant function $f(x)=0$, and I think this might be what's tripping up your friend: no matter what $a$ I pick, I'll always have $$\int^x_a0dt=0$$ for every $x$. And it's not just the always-zero function which does this; it's a good exercise to show, for example, that $\sin(x)+7$ is not an area function of $\cos(x)$, even though it is an antiderivative. Bump functions provide additional examples, albeit a bit more trivial (= "mostly zero") and less tame (= not analytic).

So there is a sense in which the only "really geometrically meaningful" integral of $0$ is $0$ itself. But your friend is still wrong, since the term "integral" in this context means antiderivative and not area function.


Leaving well enough alone is no fun. At this point we have an interesting follow-up question: which (continuous-on-all-of-$\mathbb{R}$) $f$s have antiderivatives which are not area functions? Call such functions "funky." I've put the complete classification of funky functions - but not its proof - in the spoilered text below.

HINT: What made $x^2$ better than $0$ in this respect? How did I find the crucial starting point $-\sqrt[3]{3\cdot 17}$?

Suppose $f$ is defined and continuous on all of $\mathbb{R}$. Then $f$ is funky iff the set $$\Big\{\int_a^0f(t)dt: a\in\mathbb{R}\Big\}$$ is not all of $\mathbb{R}$, that is, iff there is some number that cannot be achieved as the value of a definite integral with one bound being $0$ (and we can replace $0$ with any other fixed number here).


How about drawing sum upper and lower sums! You won't get very far because you'll be married to the horizontal axis and then, of course, all of the sums are zero and since a definite integral is always sandwiched between any upper and any lower sum. The value is trapped by 0. I.E. 0 <= the integral <= 0. This of course works only for a definite integral. If you are looking for an anti derivative, it shouldn't be too hard to convince your buddy that only constant functions f(x) = C have zero slope.


The integral of 0 is C, because the derivative of C is zero.                                                                  Also, it makes sense logically if you recall the fact that the derivative of the function is the function's slope, because any function    f(x)=C    will have a slope of zero at point on the function. Therefore      ∫0  dx  =   C.   (you can say C+C, which is still just C).


From the fundamental theorem of calculus, we know that $$F(z)-F(w)=\int_w^zf(x)dx$$

Let the integral value of zero be a constant $C$, that is

$$C=\int_w^z0 \quad dx=F(z)-F(w) \quad \forall z\in[a,b]$$

Now choosing $z=w$ gives $C=0$. Consequently, this implies that $F(z)=F(w)$ which is the anti-derivative of the zero function, i.e. $F(z)$ is a constant function. Therefore, the definite integral is always zero.

But, using the formal definition of indefinite integral [see Bartle's book]:

If $f\in \mathcal{R}[a,b]$ , then the function defined by $$F(z)=\int_a^zf(x)dx$$ is called the indefinite integral of f with basepoint a.

This means that the in-diffident integral is a constant function with a possibility to be zero.


Look at this function: F(x)=0. $\frac {d}{dx} F(x) = \frac {d}{dx} 0 = 0$ There is a theorem that says that antiderivatives of any function $f(x)$ has a form of $G(x)+C$, where $G'(x)=f(x)$. If we take $f(x)=0$, then $F'(x)=0=f(x)$, so it's anti derivative has a form of $0+C=C$. So $\int 0\, dx= C$


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