Are integration and differentiation really mutually opposite?

Question

Scenario 1:

Suppose, I have a function $f(x)$. Now, let me add 5 to it:

$$F(x):=f(x)+5$$

Now, let me subtract 5 from $F(x)$:

$$F(x)-5$$

$$f(x)+5-5$$

$$f(x)$$

If I add 5 to $f(x)$, I get $F(x)$. Again, if I subtract 5 from $F(x)$, I get $f(x)$. So, we can understand that addition and subtraction are mutually opposite processes.

Scenario 2:

Suppose, I have a function $f(x)$. Now, let me differentiate it with respect to $x$:

$$\frac{d}{dx}f(x)=f'(x)$$

Now, let me find the indefinite integral of $f'(x)$,

$$\int{f'(x)dx}$$

$$f(x)+c$$

If I differentiate $f(x)$, I get $f'(x)$. Now, if I find the indefinite integral of $f'(x)$, I get $f(x)+c$.

Comments:

Integration and differentiation are not quite opposite processes, are they? If I had gotten $f(x)$ instead of $f(x)+c$ after finding the indefinite integral of $f'(x)$, I think we could've said that they are opposite processes. Am I correct?

Answer 1

Since the derivative treats constants as 0, the antiderivative is only unique up to a constant. So the indefinite integral does not output a single function but a family of functions.

We can formalize saying that they are truly opposite by setting up an equivalence class on functions where two functions are in the same equivalence class if they differ by a constant. Then the integral and the derivative truly are inverse functions on those equivalence classes.

It's not meaningful to say they are actual true inverse functions without that, as the derivative maps functions to unique functions whereas the integral maps functions to families of functions.

Answer 2

If you want to look at derivatrive as a map $\mathcal D:\mathcal F\to \mathcal F$ where $\mathcal F$ is the set of all (differentiable) functions, then $\mathcal D$ is neither one-one, nor onto. That's why, it's not okay to talk about inverse of $\mathcal D$ in the same sense as you would talk about inverse of $F$ where $F:\mathcal F\to \mathcal F$ and $F(f)=f+5$.

Since the derivative maps a set of functions (namely those which are separated only by a constant) to one, we need to look at the inverse also as a set of functions. I hope you know that if $f:D\to R$ is not an onto function, then we can (and do) occasionally define $f^{-1}(y)=\{x:f(x)=y\}$. You can think of a similar case here.

Answer 3

I am going to use the symbol $\mathbb{R}^I$ to denote the set of functions $I\rightarrow\mathbb{R},$ and $\mathrm{diff}\left(\mathbb{R}^I\right)$ the set of functions $f\in\mathbb{R}^I$ that are differentiable everywhere, where $I$ is an open interval of $\mathbb{R}.$ Given $\mathrm{diff}\left(\mathbb{R}^I\right),$ the concept of the derivative can be treated in terms of a function $D:\mathrm{diff}\left(\mathbb{R}^I\right)\rightarrow\mathbb{R}^I,$ $D[f]:=f'.$ An antiderivative $F$ of $f$ is a function $F\in\mathrm{diff}\left(\mathbb{R}^I\right)$ such that $D[F]=f.$ Let $\mathrm{Constant}$ denote the set of constant functions in $\mathbb{R}^I.$ These are functions satisfying functional equations $\forall{x,y\in\mathbb{R}},\,f(x)=f(y).$ If $C\in\mathrm{Constant},$ then $D[C]=0,$ where $0$ denotes the zero function.

Everything I have said so far is just a careful restating of things you already know, but I think it is important to restating things this way just so that the point I am about to make is completely clear. If I have a function $F$ such that $D[F]=f,$ then it is also the case $D[F+C]=f,$ where $C\in\mathrm{Constant}.$ As such, $D$ is not an injective function. Therefore, it is not invertible. Since it is not an invertible function, there is no inverse function that will give you a definite answer.

This is why, when we talk about finding an antiderivative, we talk about solving the equation $D[F]=f,$ for which there are many solutions, rather than talking about the inverse of $D$: it is because $D$ is not invertible.

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But also, I should let you know that integrating is not the same thing as antidifferentiating. Integrating is what you do when you evaluate $$\int_a^bf\,\mathrm{d}x.$$ To talk about the integral of a function, we first need to change from working with open intervals to closed intervals. So from now, $I$ will denote a closed interval, not an open interval. The integral of a function $f:I\rightarrow\mathbb{R}$ is equal to some real number, a real number given by the expression $$\int_a^bf\,\mathrm{d}x=\int_If\,\mathrm{d}x,$$ and this is known as the Riemann integral of $f.$ The integral operator can be seen as a function $J:\mathbb{R}^I\rightarrow\mathbb{R},$ with $$J[f]:=\int_If\,\mathrm{d}x=\int_a^bf\,\mathrm{d}x.$$ Well, not quite. Not every function in $\mathbb{R}^I$ is Riemann integrable. So let $R(I)$ denote the set of functions in $\mathbb{R}^I$ that are Riemann integrable. Then $J$ is a function $R(I)\rightarrow\mathbb{R}.$ So this makes it very clear that $J$ is nothing like an inverse function to $D.$ Integration and differentiation are not inverses of each other. However, you can get a sort of "pseudo-inverse," thanks to the fundamental theorem of calculus. Let $I=[a,b],$ and let $f\in{R(I)}.$ Then, for every $x\in[a,b],$ $$\int_a^xf\,\mathrm{d}x$$ exists. Let $F:[a,b]\rightarrow\mathbb{R}$ be defined by $$F(x):=\int_a^xf\,\mathrm{d}x.$$ The first part of the fundamental theorem of calculus states that $D[F]=f.$ $F$n is already defined in terms of integrals of $f,$ so this almost acts like an inverse relationship involving the derivative. You can envision a function $\mathrm{Int}:R(I)\rightarrow\mathbb{R}^I$ defined by $$\mathrm{Int}[f]:=F,\,F(x)=\int_a^xf\,\mathrm{d}.x$$ The first part of the fundamental theorem of calculus can then be restated as simply $D[\mathrm{Int}[f]]=f,$ which is an inverse relationship involving $D$, at least if you restrict the domain of $\mathrm{Int}$ to be the pre-image of $\mathrm{diff}(]a,b[)$ under $\mathrm{Int}$ (remember that $I=[a,b]$).

The second part of the fundamental theorem of calculus states $$\int_If'\,\mathrm{d}x=f(b)-f(a),$$ which can be rewritten as $J[D[f]]=f(b)-f(a),$ at least provided that you restrict the domain of $D$ to be the pre-image of $R(I)$ under $D,$ without the endpoints. This also implies $$\int_a^xf'\,\mathrm{d}t=f(x)-f(a).$$ If you think of the constant function $[f(a)]$ as being the function whose output is $f(a),$ then this allows us to restate the second part of the fundamental theorem of calculus as $$[f(a)]+\mathrm{Int}[D[f]]=f.$$ This is almost like an inverse relationship, again provided the appropriate domain restrictions. So the function $\mathrm{Int}$ is kind of like an inverse function to $D,$ but not quite fully an inverse function, hence why I called it a "pseudo-inverse" earlier. To be clear, let me repeat myself, though: there is no true inverse function for $D$, although you can come close to having something inverse-like.

TL;DR: It is useful to think of integration as being the inverse of differentiation in a very loose way, but they are definitely not inverses in the strict sense of the word "inverse" as mathematicians use it.