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:

$\map {F'} x = c$

From Derivative of Function of Constant Multiple:

$\map {\dfrac \d {\d x} } {c x} = c$

From Primitives which Differ by Constant:

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

Hence the result.

$\blacksquare$


Sources

  • 1967: Michael Spivak: Calculus ... (next): Part $\text {III}$: Derivatives and Integrals: Chapter $18$: Integration in Elementary Terms
  • 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 14$: Indefinite Integrals: General Rules of Integration: $14.1$
  • 1976: K. Weltner and W.J. Weber: Mathematics for Engineers and Scientists ... (previous) ... (next): $6$. Integral Calculus: Appendix: Table of Fundamental Standard Integrals
  • 2009: Murray R. Spiegel, Seymour Lipschutz and John Liu: Mathematical Handbook of Formulas and Tables (3rd ed.) ... (previous) ... (next): $\S 16$: Indefinite Integrals: General Rules of Integration: $16.1.$
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