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 $\left[{a \,.\,.\, b}\right]$.

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


Then:

$\displaystyle \int_a^b c f \left({x}\right) \rd x = c \int_a^b f \left({x}\right) \rd x$


Proof

Let $F$ be a primitive of $f$ on $\left[{a \,.\,.\, b}\right]$.

By Primitive of Constant Multiple of Function, $H = c F$ is a primitive of $c f$ on $\left[{a \,.\,.\, b}\right]$.

Hence by the Fundamental Theorem of Calculus:

\(\displaystyle \int_a^b c f \left({x}\right) \rd x\) \(=\) \(\displaystyle \left[{c F \left({x}\right)}\right]_a^b\)
\(\displaystyle \) \(=\) \(\displaystyle c \left[{F \left({x}\right)}\right]_a^b\)
\(\displaystyle \) \(=\) \(\displaystyle c \int_a^b f \left({x}\right) \rd x\)

$\blacksquare$