Sum of Integrals on Adjacent Intervals for Integrable Functions/Corollary

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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:

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

The result follows by induction.

$\blacksquare$


Sources