Constant of integration

More specifically, if a function ${\displaystyle f(x)}$ is defined on an interval, and ${\displaystyle F(x)}$ is an antiderivative of ${\displaystyle f(x),}$ then the set of all antiderivatives of ${\displaystyle f(x)}$ is given by the functions ${\displaystyle F(x)+C,}$ where ${\displaystyle C}$ is an arbitrary constant (meaning that any value of ${\displaystyle C}$ would make ${\displaystyle F(x)+C}$ a valid antiderivative). For that reason, the indefinite integral is often written as $\int f(x)dx=F(x)+C,$ although the constant of integration might be sometimes omitted in lists of integrals for simplicity.

Origin

The derivative of any constant function is zero. Once one has found one antiderivative ${\displaystyle F(x)}$ for a function ${\displaystyle f(x),}$ adding or subtracting any constant ${\displaystyle C}$ will give us another antiderivative, because $${\displaystyle \frac{d}{dx}(F(x)+C)=\frac{d}{dx}F(x)+\frac{d}{dx}C=F^{\prime }(x)=f(x).}$$The constant is a way of expressing that every function with at least one antiderivative will have an infinite number of them.

Let ${\displaystyle F:\mathbb{R}\to \mathbb{R}}$ and ${\displaystyle G:\mathbb{R}\to \mathbb{R}}$ be two everywhere differentiable functions. Suppose that ${\displaystyle F{}^{\prime }(x)=G{}^{\prime }(x)}$ for every real number x. Then there exists a real number ${\displaystyle C}$ such that ${\displaystyle F(x)-G(x)=C}$ for every real number x.

To prove this, notice that ${\displaystyle [F(x)-G(x){]}^{\prime }=0.}$ So ${\displaystyle F}$ can be replaced by ${\displaystyle F-G,}$ and ${\displaystyle G}$ by the constant function ${\displaystyle 0,}$ making the goal to prove that an everywhere differentiable function whose derivative is always zero must be constant:

Choose a real number ${\displaystyle a,}$ and let ${\displaystyle C=F(a).}$ For any x, the fundamental theorem of calculus, together with the assumption that the derivative of ${\displaystyle F}$ vanishes, implying that$${\displaystyle \begin{array}{cc}& 0={\int }_a^x{F}^{\prime }(t)dt\\ & 0=F(x)-F(a)\\ & 0=F(x)-C\\ & F(x)=C\end{array}}$$thereby showing that ${\displaystyle F}$ is a constant function.

Two facts are crucial in this proof. First, the real line is connected. If the real line were not connected, one would not always be able to integrate from our fixed a to any given x. For example, if one were to ask for functions defined on the union of intervals [0,1] and [2,3], and if a were 0, then it would not be possible to integrate from 0 to 3, because the function is not defined between 1 and 2. Here, there will be two constants, one for each connected component of the domain. In general, by replacing constants with locally constant functions, one can extend this theorem to disconnected domains. For example, there are two constants of integration for $\int dx/x$, and infinitely many for $\int \tan xdx$, so for example, the general form for the integral of 1/x is:$${\displaystyle \int \frac{dx}{x}=\{\begin{array}{ll}\ln \left|x\right|+C^{-}& x<0\\ \ln \left|x\right|+C^{+}& x>0\end{array}}$$Second, ${\displaystyle F}$ and ${\displaystyle G}$ were assumed to be everywhere differentiable. If ${\displaystyle F}$ and ${\displaystyle G}$ are not differentiable at even one point, then the theorem might fail. As an example, let ${\displaystyle F(x)}$ be the Heaviside step function, which is zero for negative values of x and one for non-negative values of x, and let ${\displaystyle G(x)=0.}$ Then the derivative of ${\displaystyle F}$ is zero where it is defined, and the derivative of ${\displaystyle G}$ is always zero. Yet it's clear that ${\displaystyle F}$ and ${\displaystyle G}$ do not differ by a constant, even if it is assumed that ${\displaystyle F}$ and ${\displaystyle G}$ are everywhere continuous and almost everywhere differentiable the theorem still fails. As an example, take ${\displaystyle F}$ to be the Cantor function and again let ${\displaystyle G=0.}$

It turns out that adding and subtracting constants is the only flexibility available in finding different antiderivatives of the same function. That is, all antiderivatives are the same up to a constant. To express this fact for ${\displaystyle \cos (x),}$ one can write: $${\displaystyle \int \cos (x)dx=\sin (x)+C,}$$ where ${\displaystyle C}$ is constant of integration. It is easily determined that all of the following functions are antiderivatives of ${\displaystyle \cos (x)}$: $${\displaystyle \begin{array}{cc}\frac{d}{dx}[\sin (x)+C]& =\frac{d}{dx}\sin (x)+\frac{d}{dx}C\\ & =\cos (x)+0\\ & =\cos (x)\end{array}}$$