# 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 $\left[{a \,.\,.\, b}\right]$ if and only if:

$\exists c \in \R: \forall x \in \left[{a \,.\,.\, b}\right]: G \left({x}\right) = F \left({x}\right) + c$

That is, if and only if $F$ and $G$ differ by a constant on the whole interval.

## Proof

### Necessary Condition

Suppose $G$ is a primitive for $f$.

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

 $\displaystyle D_x \left({ F \left({x}\right) - G \left({x}\right) }\right)$ $=$ $\displaystyle D_x \left({ F \left({x}\right) }\right) - D_x \left({ G \left({x}\right) }\right)$ Sum Rule for Derivatives $\displaystyle$ $=$ $\displaystyle f \left({x}\right) - f \left({x}\right)$ $F, G$ are a primitives for $f$ $\displaystyle$ $=$ $\displaystyle 0$

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

$\Box$

### Sufficient Condition

Now suppose $G \left({x}\right) = F \left({x}\right) + c$.

We compute:

 $\displaystyle D_x G \left({x}\right)$ $=$ $\displaystyle D_x \left({ F \left({x}\right) + c }\right)$ $\displaystyle$ $=$ $\displaystyle D_x \left({ F \left({x}\right) }\right) + 0$ Sum Rule for Derivatives and Derivative of Constant $\displaystyle$ $=$ $\displaystyle f \left({x}\right)$ $F$ is a primitive for $f$

Hence $G$ is also a primitive for $f$.

$\blacksquare$