Difference between revisions of "Diamond integral"

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The $\Diamond$ integral was introduced to fix the issues with the [[Diamond alpha derivative | $\Diamond_{\alpha}$ integral]].
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=Properties=
 
=Properties=
 
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<strong>Theorem (Constant Multiple):</strong> The following formula holds:
 
<strong>Theorem (Constant Multiple):</strong> The following formula holds:
 
$$\int_a^b \alpha f(t) \Diamond t= \alpha \int_a^b f(t) \Diamond t.$$
 
$$\int_a^b \alpha f(t) \Diamond t= \alpha \int_a^b f(t) \Diamond t.$$
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<strong>Proof:</strong> █
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<strong>Theorem (Mean Value Theorem):</strong> Let $f,g \colon \mathbb{T} \rightarrow \mathbb{R}$ be bounded and $\Diamond$-integrable on $[a,b] \cap \mathbb{T}$, and let $g$ be nonnegative or nonpositive on $[a,b] \cap \mathbb{T}$. Let $m$ and $M$ be the infimum and supremum respectively of $f$. Then there exists a real number $K$ satisfying $m \leq K \leq M$ such that
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$$\displaystyle\int_a^b (fg)(t) \Diamond t = K \displaystyle\int_a^b g(t) \Diamond t.$$
 
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<strong>Proof:</strong> █  
 
<strong>Proof:</strong> █  

Latest revision as of 08:05, 10 March 2015

The $\Diamond$ integral was introduced to fix the issues with the $\Diamond_{\alpha}$ integral.

Properties

Theorem: The following formula holds: $$\int_a^a f(t) \Diamond t=0.$$

Proof:

Theorem: The following formula holds: $$\int_a^b f(t) \Diamond t= \int_a^c f(t) \Diamond t + \int_c^b f(t) \Diamond t.$$

Proof:

Theorem: The following formula holds: $$\int_a^b f(t) \Diamond t= -\int_b^a f(t) \Diamond t.$$

Proof:

Theorem (Sum Rule): The following formula holds: $$\int_a^b (f+g)(t) \Diamond t= \int_a^b f(t) \Diamond t + \int_a^b g(t) \Diamond t.$$

Proof:

Theorem (Constant Multiple): The following formula holds: $$\int_a^b \alpha f(t) \Diamond t= \alpha \int_a^b f(t) \Diamond t.$$

Proof:

Theorem (Mean Value Theorem): Let $f,g \colon \mathbb{T} \rightarrow \mathbb{R}$ be bounded and $\Diamond$-integrable on $[a,b] \cap \mathbb{T}$, and let $g$ be nonnegative or nonpositive on $[a,b] \cap \mathbb{T}$. Let $m$ and $M$ be the infimum and supremum respectively of $f$. Then there exists a real number $K$ satisfying $m \leq K \leq M$ such that $$\displaystyle\int_a^b (fg)(t) \Diamond t = K \displaystyle\int_a^b g(t) \Diamond t.$$

Proof:

References

The Diamond Integral on Time Scales