# Sum of Integrals on Adjacent Intervals for Continuous Functions

## Theorem

Let $f$ be a real function which is continuous on any closed interval $I$.

Let $a, b, c \in I$.

Then:

$\displaystyle \int_a^c f \left({t}\right) \ \mathrm dt + \int_c^b f \left({t}\right) \ \mathrm dt = \int_a^b f \left({t}\right) \ \mathrm dt$

## Proof

By Continuous Function is Riemann Integrable, $f$ is integrable on $I$.

The result follows by application of Sum of Integrals on Adjacent Intervals for Integrable Functions.

$\blacksquare$

## Comment

This proof would be very simple if we were to use the Fundamental Theorem of Calculus:

 $\displaystyle \int_a^b f \left({t}\right) \ \mathrm dt$ $=$ $\displaystyle F\left({b}\right) - F\left({a}\right)$ $\displaystyle$ $=$ $\displaystyle F\left({b}\right) - F\left({c}\right) + F\left({c}\right) - F\left({a}\right)$ $\displaystyle$ $=$ $\displaystyle \int_c^b f \left({t}\right) \ \mathrm dt + \int_a^c f \left({t}\right) \ \mathrm dt$

... but such a proof would be circular.