Linear Combination of Integrals/Indefinite

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Theorem

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

Let $\lambda$ and $\mu$ be real numbers.


Then:

$\displaystyle \int \left({\lambda f \left({x}\right) + \mu g \left({x}\right)}\right) \rd x = \lambda \int f \left({x}\right) \rd x + \mu \int g \left({x}\right) \rd x$


Proof

Let $F$ and $G$ be primitives of $f$ and $g$ respectively on $\left[{a \,.\,.\, b}\right]$.

By Linear Combination of Derivatives, $H = \lambda F + \mu G$ is a primitive of $\lambda f + \mu g$ on $\left[{a \,.\,.\, b}\right]$.

Hence:

\(\displaystyle \int \left({\lambda f \left({t}\right) + \mu g \left({t}\right)}\right) \rd t\) \(=\) \(\displaystyle \left[{\lambda F \left({t}\right) + \mu G \left({t}\right)}\right]\)
\(\displaystyle \) \(=\) \(\displaystyle \lambda \int f \left({t}\right) \rd t + \mu \int g \left({t}\right) \rd t\)

$\blacksquare$


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