# Primitives which Differ by Constant

## 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.

## 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 $f$.

$\blacksquare$

## 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

- 1968: Murray R. Spiegel:
*Mathematical Handbook of Formulas and Tables*... (previous) ... (next): $\S 14$: Definition of an Indefinite Integral - 1977: K.G. Binmore:
*Mathematical Analysis: A Straightforward Approach*... (previous) ... (next): $\S 13.11$