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 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:
- $\ds \int_{a_0}^{a_n} \map f t \rd t = \sum_{i \mathop = 0}^{n - 1} \int_{a_i}^{a_{i + 1} } \map f t \rd t$
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:
- $\ds \int_{a_0}^{a_k} \map f t \rd x = \sum_{i \mathop = 0}^{k - 1} \int_{a_i}^{a_{i + 1} } \map f t \rd t$
Now we need to show true for $n = k + 1$:
- $\ds \int_{a_0}^{a_{k + 1} } \map f t \rd t = \sum_{i \mathop = 0}^k \int_{a_i}^{a_{i + 1} } \map f t \rd t$
Induction Step
This is our induction step:
\(\ds \int_{a_0}^{a_{k + 1} } \map f t \rd t\) | \(=\) | \(\ds \int_{a_0}^{a_k} \map f t \rd t + \int_{a_k}^{a_{k + 1} } \map f t \rd t\) | Sum of Integrals on Adjacent Intervals for Integrable Functions | |||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{i \mathop = 0}^{k - 1} \int_{a_i}^{a_{i + 1} } \map f t \rd t + \int_{a_k}^{a_{k + 1} } \map f t \rd t\) | Induction Hypothesis | |||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{i \mathop = 0}^k \int_{a_i}^{a_{i + 1} } \map f t \rd t\) |
The result follows by induction.
$\blacksquare$
Sources
- 1961: I.N. Sneddon: Fourier Series ... (previous) ... (next): Chapter Two: $\S 1$. Piecewise-Continuous Functions: $(5)$
- 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 15$: General Formulas involving Definite Integrals: $15.7$