Linear Combination of Integrals

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:
 * $\displaystyle \int \left({\lambda f \left({x}\right) + \mu g \left({x}\right)}\right) \ \mathrm dx = \lambda \int f \left({x}\right) \ \mathrm dx + \mu \int g \left({x}\right) \ \mathrm dx$

and:
 * $\displaystyle \int_a^b \left({\lambda f \left({t}\right) + \mu g \left({t}\right)}\right) \ \mathrm dt = \lambda \int_a^b f \left({t}\right) \ \mathrm dt + \mu \int_a^b g \left({t}\right) \ \mathrm dt$

Proof for Indefinite Integrals
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,

Proof for Definite Integrals
It is clear that for step functions $s$ and $t$ that


 * $\displaystyle \int_a^b \lambda s(x) + \mu t(x)dx = \lambda \int_a^b s(x)dx + \mu \int_a^b t(x) dx$

Under any partition, the lower sums and upper sums of $f$ and $g$ are step functions, so the above formula relates the lower and upper sums of $f$ and $g$ to the lower and upper sums of the linear combinations of $f$ and $g$.

Because this identity is preserved for all possible partitions of $[a,b]$, it is preserved for the supremum and infimum of all possible lower and upper sums, so the linear combinations of $f$ and $g$ are integrable.