Integral of Constant/Definite

Theorem

Let $c$ be a constant.

$\ds \int_a^b c \rd x = c \paren {b - a}$

Corollary

$\ds \int_a^b \rd x = b - a$

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 Darboux's Theorem, we have:

$\ds c \paren {b - a} \le \int_a^b c \rd x \le c \paren {b - a}$

Hence the result.

$\blacksquare$

Sources

• 1977: K.G. Binmore: Mathematical Analysis: A Straightforward Approach ... (previous) ... (next): $\S 13.5$