Definite Integral of Constant Multiple of Real Function

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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 proofs
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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 1

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:

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

$\blacksquare$


Proof 2

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

By Linear Combination of Definite Integrals:

$\ds \int_a^b \paren {\lambda \map f t + \mu \map g t} \rd t = \lambda \int_a^b \map f t \rd t + \mu \int_a^b \map g t \rd t$

for real functions $f$ and $g$ which are integrable on the closed interval $\closedint a b$, where $\lambda$ and $\mu$ be real numbers.

The result follows by setting $\lambda = c$ and $\mu = 0$.

$\blacksquare$


Sources

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