If $f:[a,b] \rightarrow \mathbb{R}$ is a constant function, say $f(s) = c$ for every $x \in [a,b]$, show that
$Proof$:
Let $f(x) = c$ for all $x \in [a, b]$. Then,
\[\int_a^b f(x) \, dx = \int_a^b c \, dx\]Since $c$ is a constant, it can be factored out of the integral:
\[\int_a^b c \, dx = c \int_a^b 1 \, dx\]Using the fundamental integral formula:
\[\int_a^b 1 \, dx = b - a\]Thus,
\[c \int_a^b 1 \, dx = c(b - a)\]Therefore,
\[\int_a^b f(x) \, dx = c(b - a)\]$\square$
(I’m not sure whether the explanation above is entirely correct. If you happen to notice any mistakes, I would really appreciate it if you could leave your suggestions in the comments when you have a moment. Thank you very much for your help and corrections.)