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 $\closedint a b$.



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:

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

$\blacksquare$