Definite Integral of Constant Multiple of Real Function

From ProofWiki
Jump to navigation Jump to search

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 1

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:

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

$\blacksquare$


Proof 2

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

By Linear Combination of Definite Integrals:

$\displaystyle \int_a^b \left({\lambda f \left({t}\right) + \mu g \left({t}\right)}\right) \rd t = \lambda \int_a^b f \left({t}\right) \rd t + \mu \int_a^b g \left({t}\right) \rd t$

for real functions $f$ and $g$ which are integrable on the closed interval $\left[{a \,.\,.\, b}\right]$, where $\lambda$ and $\mu$ be real numbers.

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

$\blacksquare$


Sources