Why no constant $C$ when integrating $\int_{0}^{x}f(t)dt$?

Question

Why is there no constant $C$ added when integrating from $0$ to $x$ in $$ \int_{0}^{x}f(t)dt$$

Is this the only case when $C$ can be omitted?

Answer 1

if you assume the integral of function f(t) is g(t)+C, so $$\int_{0}^{x}f(t)dt=[g(x)+C]-[g(0)+C]=g(x)-g(0)$$

Answer 2

$\newcommand{\eps}{\varepsilon}$In calculus, the integral sign is used to denote two substantially distinct concepts, antidifferentiation and integration (a.k.a. indefinite integration and definite integration), that turn out to be closely related (for the class of continuous functions) by the fundamental theorems.

Let $I = [a, b]$ be a closed, bounded interval of real numbers.

• If $f$ is a derivative on $I$, i.e., if there exists a function $F$ on $I$ such that $F' = f$,[1] then we write $$ \int f(t)\, dt = F(t) + C $$ to signify that the general antiderivative of $f$ on $I$ is $F$ plus a constant.[2]

• If $f$ is (Riemann) integrable on $I$, i.e., if $f$ is bounded, and the lower and upper sums of $f$ with respect to a partition $P$ of $I$ can be made arbitrarily close,[3] then $$ \int_{a}^{b} f(t)\, dt $$ denotes the least upper bound of the lower sums of $f$ on $I$, which by hypothesis is equal to the greatest lower bound of the upper sums of $f$ on $I$.

Clearly these usages of the integral sign are logically distinct. In fact, they apply to slightly different classes of function; not every derivative is Riemann integrable, and not every integrable function is a derivative.

In many elementary calculus courses, the "common ground" between these usages is to assume $f$ is continuous on $I$. Every continuous function turns out to be both a derivative and to be Riemann integrable, and one fundamental theorem of calculus asserts that $$ F(x) = \int_{a}^{x} f(t)\, dt $$ is an antiderivative of $f$ on $[a, b]$. Moreover, if $G$ is an arbitrary antiderivative of $f$ on $[a, b]$, the other fundamental theorem of calculus asserts $$ \int_{a}^{b} f(t)\, dt = G(t)\Big|_{t=a}^{t=b} = G(b) - G(a). $$ That is, in the appropriate setting we may regard the symbol $$ \int_{a}^{b} f(t)\, dt $$ as either a definite integral (in which case there is no "$+C$"), or as the difference between two values of an antiderivative of $f$ (in which case the "$+C$"s cancel), and these interpretations agree.

Finally, a coda to your question: If you encounter an integral sign with limits (even if one or both limits are "variables"), such as $$ \int_{0}^{x} f(t)\, dt, $$ context dictates definite integration.

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Notes:

1. To split hairs, $F' = f$ on $[a, b]$ means: There exists an open interval $(c, d) \supset [a, b]$ and a differentiable function $F$ on $(c, d)$ such that $F'(t) = f(t)$ for all $t$ in $[a, b]$.

2. Formally, if $G' = f$ on $I$, there exists a real number $C$ such that $G(t) = F(t) + C$ for all $t$ in $I$.

3. Precisely, for every $\eps > 0$, there exists a partition $P$ of $[a, b]$ such that $$ U(f, P) - L(f, P) < \eps. $$

Answer 3

If You write $\int f(t)dt=F(x)+C$ for the indefinite integral You actually mean the set of all antiderivatives:

$\int f(t)dt=\{F:D\rightarrow \mathbb{R}:F´=f\}$. Since two functions $F_1$ and $F_2$ in this set satisfy $(F_1-F_2)´\equiv 0$ they can only differ by a constant $C\in \mathbb{R}$. So

$\int f(t)dt=\{F+C:C\in\mathbb{R}\}$ where $F´=f$

This is why one writes ( a bit sloppy)$\int f(t)dt=F(x)+C$.

The link between this indefinite integral and the limits of the Riemann sums (the definite integrals) is then given by the fundamental theorem of calculus:

$\int_a^b f(t)dt=F(a)-F(b)$ for some $F\in \int f(t)dt$.