# Primitive of Constant

## Theorem

Let $c$ be a constant.

$\ds \int c \rd x = c x + C$ where $C$ is an arbitrary constant.

## Proof

Let:

$\ds \map F x = \int c \rd x$

From the definition of primitive:

$\map {F'} x = c$
$\map {\dfrac \d {\d x} } {c x} = c$
$\map {\dfrac \d {\d x} } {c x + C} = c$

Hence the result.

$\blacksquare$