Primitive of Constant
Jump to navigation
Jump to search
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: $16.1.$