Linear Combination of Integrals/Definite/Proof 1

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


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:

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