Primitives which Differ by Constant

Theorem
Let $$F$$ be a primitive for a real function $$f$$ on the closed interval $$\left[{a \,. \, . \, b}\right]$$.

Let $$G$$ be a real function defined on $$\left[{a \,. \, . \, b}\right]$$.

Then $$G$$ is a primitive for $$f$$ on $$\left[{a \,. \, . \, b}\right]$$ iff $$\exists c \in \R: \forall x \in \left[{a \,. \, . \, b}\right]: G \left({x}\right) = F \left({x}\right) + c$$.

That is, iff $$F$$ and $$G$$ differ by a constant on the whole length of the interval.

Proof
From Differentiation of a Constant and Sum Rule for Derivatives it is clear that $$F \left({x}\right) + c$$ is a primitive for $$f$$.

So, suppose $$G$$ is a primitive for $$f$$.

Then $$F - G$$ is continuous on $$\left[{a \,. \, . \, b}\right]$$, differentiable on $$\left({a \, . \, . \, b}\right)$$, and:

$$\forall x \in \left[{a \,. \, . \, b}\right]: D \left({F \left({x}\right) - G \left({x}\right)}\right) = F^{\prime} \left({x}\right) - G^{\prime} \left({x}\right) = f \left({x}\right) - f \left({x}\right) = 0$$.

From Zero Derivative means Constant Function it follows that $$F - G$$ is constant on $$\left[{a \,. \, . \, b}\right]$$, hence the result.