How ``integration is the inverse of differentiation''

Denote by $D$ the differentiation operator,
and $I_a$ the integration from $a$ operator.

Let $E_a\from \cF \to \bR$ be the ``evaluate at $a$ -funtional-,
and $C\from \bR \to \cF$ the ``constant function map''
  $a \mapsto f \equiv a$

Note that $D$ has kernel, namely the constant functions.
So $D \oplus E_a \from C^1 \to C^0 \oplus \bR$ is invertible,
with inverse $I_a + C$.

IE, $f(x) = f(a) + \int_a^x f'$,
or $\Id = I_a(D) + C(E_a) = (I_a + C)(D \oplus E_a)$.

...and now these are bona fide inverses, the other way too!
