What is the integral of 0?

Solution 1:

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. $

Solution 2:

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.

Solution 3:

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 $$ \frac{\mathrm{d}}{\mathrm{d}t}C=0 $$

Solution 4:

http://mathforum.org/library/drmath/view/65593.html

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.