Sum of Integrals on Adjacent Intervals for Integrable Functions/Corollary

Corollary to Sum of Integrals on Adjacent Intervals for Integrable Functions
Let $f$ be a real function which is Riemann integrable on any closed interval $\mathbb I$. Let $a_0, a_1, \ldots, a_n$ be real numbers, where $n \in \N$ and $n \ge 2$.

Then:


 * $\displaystyle \int_{a_0}^{a_n} f \left({t}\right) \rd t = \sum_{i \mathop = 0}^{n - 1} \int_{a_i}^{a_{i + 1} } f \left({t}\right) \rd t$

Proof
Proof by induction:

Basis for the Induction
According to Sum of Integrals on Adjacent Intervals for Integrable Functions, $n = 2$ holds.

This is the basis for the induction.

Induction Hypothesis
This is our induction hypothesis:
 * $\displaystyle \int_{a_0}^{a_k} f \left({t}\right) \rd x = \sum_{i \mathop = 0}^{k - 1} \int_{a_i}^{a_{i + 1}} f \left({t}\right) \rd t$

Now we need to show true for $n = k + 1$:
 * $\displaystyle \int_{a_0}^{a_{k + 1} } f \left({t}\right) \rd t = \sum_{i \mathop = 0}^k \int_{a_i}^{a_{i + 1}} f \left({t}\right) \rd t$

Induction Step
This is our induction step:

The result follows by induction.