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 $\closedint a b$.
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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:
\(\ds \int_a^b c \map f x \rd x\) | \(=\) | \(\ds \bigintlimits {c \map F x} a b\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds c \bigintlimits {\map F x} a b\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds c \int_a^b \map f x \rd x\) |
$\blacksquare$