Definite Integral on Zero Interval

Theorem
Let $$f$$ be a real function which is continuous on the closed interval $$\left[{a \,. \, . \, b}\right]$$.

Let $$\int_a^b f \left({t}\right) dt$$ be the definite integral of $$f$$ on $$\left[{a \,. \, . \, b}\right]$$.

Then $$\forall c \in \left[{a \,. \, . \, b}\right]: \int_c^c f \left({t}\right) dt = 0$$.

Proof
Follows directly from the definition of definite integral.

There is only one subdivision of $$\left[{c \,. \, . \, c}\right]$$ and that is $$\left\{{c}\right\}$$.

Both the lower sum and upper sum of $$f \left({x}\right)$$ on $$\left[{c \,. \, . \, c}\right]$$ belonging to the subdivision $$\left\{{c}\right\}$$ are equal to zero.

Hence the result.