Integral of Constant/Definite
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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): $\S 13.5$