Integral on Zero Interval

Theorem

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

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

Then:

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

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.

$\blacksquare$

Sources

• Murray R. Spiegel: Mathematical Handbook of Formulas and Tables (1968): $15.9$