What is the integral of 0?

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 $0$.

• 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. $$\blacksquare$$ – Inquest Jan 26 '13 at 2:30
• Whether or not the integral is $0$ or $C$ depends on whether you are talking about the indefinite or definite integral. – anon Jan 26 '13 at 2:47
• Maybe it is confusing cause it is "the" integral. Make it clear that "the" integral is not a single function.... – N. S. Jan 26 '13 at 3:14
• Definite integrals $\ne$ Primitive integrals $=$ Antiderivatives. – Did Jan 26 '13 at 3:15
• 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. – d80d2729a352b1366139fc119d3345 Jun 13 '13 at 15:29

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 $\int 0 \,dx = c - c = 0.$

• 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 $0 \int 1 dx = 0x + C$? – Jonathan Hebert Apr 18 '16 at 23:31
• What is the reason to take your position? Why not $0\int 1\, dx = 0\cdot (x + C)$ – Zduff Mar 31 '17 at 20:39

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.

• @downvote ??? Lol. – Rustyn Jan 26 '13 at 6:02
• If $C$ is assumed to be selected randomly, the probability that the friend is right is 0. [tongue in cheek] – Thomas Jan 26 '13 at 7:19

Indefinite integrals (anti-derivatoves) 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 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.

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 $\int 0\, dx= C$

protected by Asaf Karagila♦Mar 20 '15 at 23:13

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