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While I was first learning about integrals and antiderivatives, I took for granted that, for example, $\int 2xdx = x^2 + c$. But if the constant can be any number, that means that $\int 2xdx = x^2 = x^2 + 1$ which doesn't make any sense. I understand that the constant is usually just left as $c$, but it is implied that it can be any number, right? We don't define the square root of $4$ to be $2$ and $-2$ and we don't define $\arcsin(0)$ to be $0$, $2\pi$, $4\pi$...

If you try to define $\int\frac{d}{dx}(f(x))dx = f(x)$, a problem arises. Mainly, if $f(x)$ is something like $x^3 + 4$. The $4$ gets lost when taking the derivative so we can't be sure of what a function is if we only know its derivative.

A potential solution to this problem would be to define a function $S(f)$ such that it takes a function as its input and outputs a function stripped of its constant e.g. $S(x^3 + 4) = x^3$. If this is done, the antiderivative can be defined as follows: $$\int \frac{d}{dx}(f(x))dx = S(f(x))$$ $$\frac{d}{dx}(\int f(x)dx)=f(x)$$ Is it possible to define $S(f)$? If not, is there any other way to define the antiderivative so that it doesn't have a constant?

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    $\begingroup$ It is not true that $\int 2x\,dx=x^2$. The left hand is only defined up to an additive constant. $\endgroup$
    – lulu
    Jul 19, 2020 at 15:45
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    $\begingroup$ $C$ is an unknown constant. It can not be determined without additional information. If, say, you want the function $F(x)$ such that $F'(x)=2x$ and $F(0)=0$ then you can use the indefinite integral and conclude that $F(x)=x^2$. But you need some additional information or $C$ remains undetermined. $\endgroup$
    – lulu
    Jul 19, 2020 at 15:56
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    $\begingroup$ @NZQRC The number of functions with a given derivative is infinite. Why would there be a particular one which is preferred over any other? $\endgroup$ Jul 19, 2020 at 16:23
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    $\begingroup$ You're effectively defining the antiderivative to be$$\int f(x)\mathrm{d}x\equiv\int_0^x f(t)\mathrm{d}t$$but this definition isn't consistent with the definition of the antiderivative which is the class of functions whose derivative is the provided function. $\endgroup$ Jul 19, 2020 at 16:39
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    $\begingroup$ You can define $F(x)$ to be $F'(x) = f(x)$ and $F(0) = 0$. You could but we don't. $\endgroup$
    – fleablood
    Jul 19, 2020 at 16:44

1 Answer 1

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The integral, while commonly thought of as an inverse to the differentiation operator, is not really a "true" inverse.

Just like ${f(x)=x^2}$ is, without any extra constraints, technically is not invertible. It lacks injectivity. ${f(x)=f(y)}$ does not imply that ${x=y}$. If you want an inverse - you must either resort to multi-valued functions, or you must restrict the domain you are interested in. If we do the latter, then indeed

$${f^{-1}(x) = \sqrt{x}}$$

In exactly the same way,

$${\frac{df}{dx}=\frac{dg}{dx}}$$

does not imply ${f(x)=g(x)}$. The differential operator clearly lacks injectivity, and you gave concrete examples

$${\frac{d}{dx}\left(x^2 + 1\right)=\frac{d}{dx}\left(x^2 \right)}$$

but obviously ${x^2 + 1\neq x^2\ \forall\ x \in \mathbb{R}}$.

Making the argument that you can set ${c}$ to be whatever you want in two different contexts and claiming the results are still equal is an invalid step. It's like saying ${2^2 = (-2)^2 = 4}$, so ${2=-2}$ - injectivity does not hold.

Now, when it comes to evaluating definite integrals - it doesn't matter which anti-derivative you pick. The Fundamental Theorem of Calculus simply states that if ${f(x)}$ is continuous on ${[a,b]}$ then

$${\int_{a}^{b}f(x)dx=F(b)-F(a)}$$

where ${F(x)}$ is any function satisfying ${\frac{d}{dx}\left(F(x)\right)=f(x)}$. So it doesn't matter what constant you specify. And it's easy to see why:

$${\left(F(b) + c\right) - \left(F(a) + c\right)=F(b)-F(a)}$$

To summarise: you cannot do anything about the missing information (unless in context you can Algebraically find out the value of the ${+c}$ constant) and it's not a bug. It's a feature! (Sorry for the terrible overused Computer Science reference).

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