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

On a connected open interval, if $F^{\prime }=g$, every antiderivative of $g$ has the form $F+C$, where $C$ is a constant function. Indeed, two solutions satisfy $(G-F{)}^{\prime }=0$: the zero derivative on a connected open set principle makes their difference constant. In one dimension, the mean value theorem proves this on any closed subinterval. ^[ref]

An initial condition or boundary condition can determine $C$. If a particular antiderivative is $F_0$ and $F(x_0)=y_0$, thenFor $g(x)=x^2$, $F_0(x)=x^3/3$ and $F(1)=2$, this gives $C=5/3$. Without that reference value, choosing $C=0$ simply selects one member of the family.

The freedom has a physical meaning in a simple kinematic model. A velocity $v(t)=at$, with constant acceleration $a$, determines position only as $x(t)=a{t}^2/2+C$. The initial condition $x(0)=x_0$ selects $C=x_0$; changing the starting position does not change the prescribed velocity.

For the differential equation $T^{\prime }=-B/r^2$ on $r>0$, the antiderivative is $T=B/r+C$. The boundary condition $T(R_s)=T_s$ gives $C=T_s-B/R_s$. Discarding the constant of integration therefore needs a specified condition or an approximation that quantifies its effect. On a domain with several connected components, the freedom is larger: antiderivatives on disconnected domains require a separate constant on each component.

Let $U$ be an open subset of the real line, and let $F,G$ be antiderivatives of the same function on $U$. Each connected component $I_j$ is an open interval. On that component, $(G-F{)}^{\prime }=0$, so the mean value theorem gives $G-F=C_j$. The constants on different components are independent: there is no subinterval inside $U$ joining them. Conversely, adding any chosen constant on each component preserves the derivative everywhere on $U$.

For example, the domain of $g(x)=1/x$ is $U=\mathbb{R}\setminus \{0\}$. Every antiderivative on this domain has the formwith arbitrary real $C_{-},C_{+}$. The logarithm of the absolute value differentiates to $1/x$ on each side. The function with $C_{-}=2$ and $C_{+}=-1$ is therefore a valid antiderivative, although it cannot be written as $\log |x|+C$ with one global constant.

An initial condition at $x=1$ fixes $C_{+}$ and leaves $C_{-}$ free. Fixing both constants requires one reference value in each connected component, or additional conditions that couple the components. The locally constant function $F(x)-\log |x|$ captures this remaining freedom.