# Definite Integral of Constant Multiple of Real Function/Proof 1

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## Theorem

Let $f$ be a real function which is integrable on the closed interval $\closedint a b$.

There is believed to be a mistake here, possibly a typo.In particular: $f$ is only integrable here, but theorems used requiring that $f$ is continuous are used in both proofsYou can help ProofWiki by reviewing it, and either correcting it or adding some explanatory material as to why you believe it is actually correct after all.To discuss this page in more detail, feel free to use the talk page.When this work has been completed, you may remove this instance of `{{Mistake}}` from the code.If you would welcome a second opinion as to whether your work is correct, add a call to `{{Proofread}}` the page. |

Let $c \in \R$ be a real number.

Then:

- $\ds \int_a^b c \map f x \rd x = c \int_a^b \map f x \rd x$

## Proof

Let $F$ be a primitive of $f$ on $\closedint a b$.

By Primitive of Constant Multiple of Function, $H = c F$ is a primitive of $c f$ on $\closedint a b$.

Hence by the Fundamental Theorem of Calculus:

\(\ds \int_a^b c \map f x \rd x\) | \(=\) | \(\ds \bigintlimits {c \map F x} a b\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds c \bigintlimits {\map F x} a b\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds c \int_a^b \map f x \rd x\) |

$\blacksquare$