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I get that $$\int 0 \ {dx}= C$$

but came across this argument:

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

from https://math.stackexchange.com/a/287079/955696

I didn't understand the explanation on why this is false

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

I've always been taught that a constant can be 'pulled' out of the integral regardless of whether it was definite or indefinite. Integrating 0, though, has led to a special case.

Can someone help explain what went wrong?

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    $\begingroup$ The integral given in the answer really should be $\int 0 \;dx= 0 \int 1 \;dx = 0 \cdot x + C= C$. $\endgroup$ Aug 31, 2023 at 21:50
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    $\begingroup$ @mathematics2x2life it is not clear why $0\int 1 dx = 0\cdot x + C$ and not $0\cdot (x+C)$. I believe the question just comes down to notation, but it is very common to view an indefinite integral as "the same kind of object" as any other function, which leads one to think the latter interpretation is okay $\endgroup$
    – Carlyle
    Aug 31, 2023 at 21:54
  • $\begingroup$ @Carlyle I would think it should be clear why $0 \int 1 dx= 0 \cdot x + C$ and not $0 \cdot (x + C)$ because the indefinite integral is unique only up to constant. The output of the indefinite integral $0 \int 1 dx$ is $0 \cdot x$ and not $x + C$. But that could be due to my education and how I instruct - placing pause and emphasis on this before the addition of '$+C$'. I am not sure I agree that your suggestion is the precise reason why the latter misinterpretation comes up, but I certainly can understand how it might arise after your comment. $\endgroup$ Aug 31, 2023 at 22:01
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    $\begingroup$ Just take the derivative of $C$ and you will find $0$ ... $\endgroup$
    – LL 3.14
    Aug 31, 2023 at 22:35

1 Answer 1

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Let's be very clear. The notation

$$\int f(x) dx = F(x) + C$$

represents an entire class of functions, in other words it represents the set:

$$\{G(x) : G'(x) = f(x)\}$$

and, as it so happens, if $F(x)$ is one function that belongs to this set, then we can equivalently write the set as

$$\{F(x) + C : C \in \mathbb{R}\}$$

because we know that all of the functions that represent possible antiderivatives of $f$ will differ by some constant $C$.

The sneaky thing is that when we get into expressions like $a \int f(x) dx$, we actually wind up matching the second definition - i.e. this is actually the set $\{a F(x) + C \}$, rather than $\{a(F(x) + C)\}$. And this only makes a difference when $a = 0$ - in other words, $0 \int 1 dx$ represents the set $\{ 0 F(x) + C : F'(x) = 1\}$, which turns out to just be the set of all possible constants $C$.

Why do we do this? Mostly because it lets us write the identity $\int a f(x) dx = a \int f(x) dx$ without needing to specify that $a$ cannot be zero.

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