Primitives which Differ by Constant

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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 $f$.

$\blacksquare$


Notes


As there is an uncountable number of possible constants (one for every possible real number), it follows that if a function has a primitive, it has an uncountable number of them.


Sources