TheoremProved

Every antiderivative has the form $F(x)+C$

Statement

If $F$ and $G$ are both antiderivatives of $f$ on an interval $K$ , then there is a constant $C$ such that $G(x) = F(x) + C$ for all $x\in K$ .

Why is it true?

If two cars drive along the same road with identical speed at every instant, they cannot drift apart or come together — the distance between them stays fixed forever. Here $F$ and $G$ are the two cars' positions and $f$ is their shared speed; the theorem says the "gap" $G-F$ must be a constant. Crucially this needs $K$ to be a single interval: on two separate intervals, the gap could jump to a different constant on each piece, since you never have to physically travel from one piece to the other.

Proof sketch

Define $H(x) = G(x) - F(x)$ for $x\in K$ . Since $F$ and $G$ are both antiderivatives of $f$ , $H'(x) = G'(x)-F'(x) = f(x)-f(x)=0$ for every $x\in K$ : $H$ is differentiable on $K$ with derivative identically zero.

Take any two points $x_1 < x_2$ in $K$ (possible since $K$ is an interval, so the whole segment $[x_1,x_2]$ lies in $K$ ). $H$ is differentiable, hence continuous, on $[x_1,x_2]$ , so the Mean Value Theorem (Lagrange) applies: there is $\xi\in(x_1,x_2)$ with $H(x_2)-H(x_1) = H'(\xi)(x_2-x_1) = 0$ , using $H'(\xi)=0$ .

Hence $H(x_2)=H(x_1)$ for every pair $x_1<x_2$ in $K$ : $H$ takes the same value everywhere on $K$ , say $H(x)=C$ for all $x\in K$ . Substituting back, $G(x)-F(x)=C$ , i.e. $G(x) = F(x) + C$ , which is exactly the claim.

Topics that use this theorem

Antiderivative

Step-by-step proofs

No step-by-step proof yet for this theorem.

References

James Stewart (2015). Calculus: Early Transcendentals
Michael Spivak (2008). Calculus
Manuel Bronstein (1998). Symbolic Integration Tutorial