Linear Combination of Integrals/Definite/Proof 1

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Theorem

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


Proof

Let $F$ and $G$ be primitives of $f$ and $g$ respectively on $\closedint a b$.

By Linear Combination of Derivatives, $H = \lambda F + \mu G$ is a primitive of $\lambda f + \mu g$ on $\closedint a b$.

Hence by the Fundamental Theorem of Calculus:

\(\ds \int_a^b \paren {\lambda \map f t + \mu \map g t} \rd t\) \(=\) \(\ds \bigintlimits {\lambda \map F t + \mu \map G t} a b\)
\(\ds \) \(=\) \(\ds \lambda \bigintlimits {\map F t} a b + \mu \bigintlimits {\map G t} a b\)
\(\ds \) \(=\) \(\ds \lambda \int_a^b \map f t \rd t + \mu \int_a^b \map g t \rd t\)

$\blacksquare$



There is believed to be a mistake here, possibly a typo.
In particular: This proof relies on $f, g$ being continuous, but they're only given to be integrable
That is originally how this result started out. This page is the proof given in Binmore which has as its premise that $f$ is continuous. The statement was changed by someone back in $2012$ who decided to rewrite a lot of this stuff, and it's only now that I've noticed he changed "continuous" to "integrable". This highlights why we fight hard against editors changing a statement of a result to something different (because it's more general, or whatever) rather than writing a new page with a new statement on it. Nightmare. Going to take ages to get to the bottom of it and work out how to rewrite all this to make it consistent again. I really really hate clever people.
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