Integral of Constant/Definite

Theorem
Let $c$ be a constant.
 * $\ds \int_a^b c \rd x = c \paren {b - a}$

Proof
Let $f_c: \R \to \R$ be the constant function.

By definition:
 * $\forall x \in \R: \map {f_c} x = c$

Thus:
 * $\map \sup {f_c} = \map \inf {f_c} = c$

So from Upper and Lower Bounds of Integral‎, we have:
 * $\ds c \paren {b - a} \le \int_a^b c \rd x \le c \paren {b - a}$

Hence the result.