Primitives which Differ by Constant

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Theorem

Let $F$ be a primitive for a real function $f$ on the closed interval $\closedint a b$.

Let $G$ be a real function defined on $\closedint a b$.


Then $G$ is a primitive for $f$ on $\closedint a b$ if and only if:

$\exists c \in \R: \forall x \in \closedint a b: \map G x = \map F x + c$


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


Corollary

Let $f$ be an integrable function on the closed interval $\closedint a b$.

Then there exist an uncountable number of primitives for $f$ on $\closedint a b$.


Proof

Necessary Condition

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

Then $F - G$ is continuous on $\closedint a b$, differentiable on $\openint a b$, and for any $x \in \openint a b$, we have:

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


From Zero Derivative implies Constant Function it follows that $F - G$ is constant on $\closedint a b$.

Hence the result.

$\Box$


Sufficient Condition

Now suppose $\map G x = \map F x + c$.

We compute:

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


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

$\blacksquare$


Sources

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