Why do we need $\int f(x) dx$? We already have $\int_a^b f(x) dx$. Is there any sensible reason?

Question

The following sentence is from "Calculus Fourth Edition" by Michael Spivak.

The symbol $$\int f\text{ or }\int f(x)dx$$ means "a primitive of $f$" or, more precisely, "the collection of all primitives of $f$."

Why do mathematicians use and write theorems about $\int f(x)dx$ ?

When we use $\int f(x)dx$, I think we always assume $f$ is continuous.
When $f$ is continuous on $[a,b]$, the collection of all primitives of $f$ is $\{g:[a,b]\to\mathbb{R}:g(x)=\int_c^x f(t)dt+C, C\in\mathbb{R} \}$, where $c$ is an arbitrary fixed point in $[a,b]$.

So, I think we don't need to invent the new symbol $\int f(x) dx$ and we don't need to write theorems about $\int f(x) dx$.
We already have the notion of definite integrals $\int_a^b f(x) dx$ and we already have theorems about $\int_a^b f(x)dx$.

Answer 1

When we use $\int f(x)dx$, I think we always assume $f$ is continuous.

No, not at all. The class of integrable functions is a lot bigger than just continuous functions. And discontinuous functions are considered.

It is correct that if $F,G$ are such that $F'=G'$ then indeed they differ by a constant. And indeed, if $f$ is continuous then by the fundamental theorem of calculus every antiderivative is of the form $x\mapsto \int_0^x f(t)dt + C$. However it is no longer guaranteed to be true if we drop the "continuity" assumption, see this.

And so this:

$$\{g:[a,b]\to\mathbb{R}:g(x)=\int_c^x f(t)dt+C, C\in\mathbb{R} \}$$

is not the collection of antiderivatives in general.

Of course we could write

$$\{g:[a,b]\to\mathbb{R}:g'(x)=f(x)\}$$

but $\int f(x)$ is simply shorter. That's literally what symbols are used for: to make definitions shorter.

It's the same question as: why do we use $\int_{a}^{b} f(t)dt$ if I can simply copy-paste definition here? It is shorter.

Answer 2

Why we need two notations ? Because there are many things that are possible & easy to convey with one notation which are hard or impossible to convey with the other notation.

(1) Consider $Y=\int f(x) dx$ & $Z=\int_a^b f(x) dx$

We have $dY/dx=f(x)$ & $dZ/dx=0$

That is due to $Y$ being a function (indefinite integral) & $Z$ being a Constant (definite integral).

To get the "Same Behaviour" , we have to use the definite integral with the limits $(c,x)$ [ the arbitrary constant $c$ & variable $x$ ] & have to involve a Dummy variable $t$ & then generate the $g$ . . . .

That unnecessary complexity is hidden by the indefinite integral.

(2) What OP gave is somewhat easy with single integration. When we go with 2 (or more) integrations , then adapting the definite integral to indefinite integral via generation of all primitives gets overly cumbersome. Trying to write it out will high-light the Issues involved.

(3) When we go with indefinite integration involving multiple variables , then adapting the definite integral to generate all primitives for all variables (& all orders of integration) gets impossible.

It is very easy to write it like this :
$\iint f(x,y) dxdy$ & $\iiint f(x,y,z) dxdydz$

Trying to write it out with define integrals will high-light the Issues involved here too.

(4) There are ways to consider Deferential Operator $D$ & Integral Operator $I$ where we will have $I=D^{-1}$ : that give a simple notation to write other operators like $D^2$ & $I^2$ & $D^n$ & $I^n$
That is impossible to write with definite integral notation.

(5) One Analogy : We already have notation for addition $(+)$ & we know that $x$ added to itself 5 times is $x+x+x+x+x$. We then have the multiplication $(\times)$ notation that $x+x+x+x+x=5 \times x$ to represent repeated addition.
Similar query here is "why we need multiplication $(\times)$ , when we already have addition $(+)$ ? We can do everything with $+$ without using $\times$"
That is because it is less cumbersome & hides the unnecessary complexities & permits expressing higher concepts [ like Commutativity of multiplication & Distributivity & repeated multiplication giving exponentiation ]

SUMMARY : definite integral & indefinite integral are both very much necessary to simplify the notations & to provide abstractions & to express higher concepts & to reduce the unnecessary complexity. Each notation has its own usage area

Answer 3

Definite integrals and antiderivatives ("primitives" or "indefinite integrals") are not equivalent in construction and there are real differences between them, which is why we continue to distinguish between them. Indeed, while the two are identified with one another by the fundamental theorem of calculus, it is not true that an antiderivative is necessarily equivalent to the class of integrals $\displaystyle \int_c^x f\,\mathrm{d}x$, $c\in\mathbb{R}$.

Recall the definition of each and the operation defining their construction. The definite integral $\displaystyle \int_a^b f\,\mathrm{d}x$ is defined (in the typical sense of a Riemann integral) as the limit of a Riemann sum $\displaystyle \sum f\,\delta x$, where the operation of interest is the integration $f\overset{I}{\longrightarrow} F$. On the other hand, an antiderivative is defined as any function that satisfies $F'=f$, for which we write $\displaystyle F=\int f\,\mathrm{d}x$. The operation of interest is instead the differentiation $F\overset{D}{\longrightarrow} f$, which we seek to invert as $F\overset{D^{-1}}{\longleftarrow}f$. In a certain sense, integration is a "forwards" operation $F=I(f)$ while antidifferentiation is a "backwards" operation $F=D^{-1}(f)$. And this is not a pedantic distinction, as the two may not be equivalent. Indeed for the Volterra function $V$, it has a derivative $v$ so $V=D^{-1}(v)$ but its derivative is not Riemann integrable, so $V\neq I(v)$.