# 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 Darboux 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:

$\displaystyle \int_{a_0}^{a_n} \map f t \rd t = \sum_{i \mathop = 0}^{n - 1} \int_{a_i}^{a_{i + 1} } \map f t \rd t$

## 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:

$\displaystyle \int_{a_0}^{a_k} \map f t \rd x = \sum_{i \mathop = 0}^{k - 1} \int_{a_i}^{a_{i + 1} } \map f t \rd t$

Now we need to show true for $n = k + 1$:

$\displaystyle \int_{a_0}^{a_{k + 1} } \map f t \rd t = \sum_{i \mathop = 0}^k \int_{a_i}^{a_{i + 1} } \map f t \rd t$

### Induction Step

This is our induction step:

 $\displaystyle \int_{a_0}^{a_{k + 1} } \map f t \rd t$ $=$ $\displaystyle \int_{a_0}^{a_k} \map f t \rd t + \int_{a_k}^{a_{k + 1} } \map f t \rd t$ Sum of Integrals on Adjacent Intervals for Integrable Functions $\displaystyle$ $=$ $\displaystyle \sum_{i \mathop = 0}^{k - 1} \int_{a_i}^{a_{i + 1} } \map f t \rd t + \int_{a_k}^{a_{k + 1} } \map f t \rd t$ Induction Hypothesis $\displaystyle$ $=$ $\displaystyle \sum_{i \mathop = 0}^k \int_{a_i}^{a_{i + 1} } \map f t \rd t$

The result follows by induction.

$\blacksquare$