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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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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
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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
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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
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Definite integrals $\ne$ Primitive integrals $=$ Antiderivatives. –  Did Jan 26 '13 at 3:15
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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. –  Hat Man Jun 13 '13 at 15:29

6 Answers 6

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:

$$\int_a^b\alpha fdx = \alpha \int_a^bf dx$$

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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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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@downvote ??? Lol. –  Rustyn Jan 26 '13 at 6:02
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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

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.

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

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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.

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an arbitrary constant because differentiation of a constant is zero

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You might want to consider adding some extra information to this answer. Also, note how some of the other answers are essentially saying the same thing as you, but are providing a bit more information, so as to be more helpful to the OP. –  Arthur Fischer Jun 12 '13 at 19:04

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