Difference between revisions of "Fundamental theorems of calculus"

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(Fundamental theorems of calculus)
 
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'''First fundamental theorem of exact differential and integral calculus for line integrals:''' The function <math>F(z)=\int\limits_{\gamma }{f(\zeta )dB\zeta }</math> ist mit <math>\gamma: [d, x[ \; \cap \; C \rightarrow A \subseteq {}^{(\omega)}\mathbb{K}, C \subseteq \mathbb{R}, f: A \rightarrow {}^{(\omega)}\mathbb{K}, d \in [a, b[ \; \cap \; C</math>, and choosing <math>\curvearrowright B \gamma(x) = \gamma(\curvearrowright D x)</math> is exactly <math>B</math>-differentiable, and for all <math>x \in [a, b[ \; \cap \; C</math> and <math>z = \gamma(x)</math>
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'''First fundamental theorem of exact differential and integral calculus for line integrals:''' The function <math>F(z)={\uparrow}_{\gamma }{f(\zeta ){\downarrow}\zeta }</math> where <math>\gamma: [d, x[ \, \cap \, C \rightarrow A \subseteq {}^{(\omega)}\mathbb{K}, C \subseteq \mathbb{R}, f: A \rightarrow {}^{(\omega)}\mathbb{K}, d \in G = [a, b[ \, \cap \, C</math>, and choosing <math>\overset{\rightharpoonup}{\gamma}(x) = \gamma(\overset{\rightharpoonup}{x})</math> is exactly differentiable, and for all <math>x \in G</math> and <math>z = \gamma(x)</math>
  
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<div style="text-align:center;"><math>F' \curvearrowright B(z) = f(z).</math></div>
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<div style="text-align:center;"><math>F^{\prime}(z) = f(z).</math></div>
  
  
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<table style="width:100%"><tr><td style="vertical-align: top; padding-top: 1em;">'''Proof:'''</td><td style="text-align: center;"><math>\begin{aligned}dB(F(z))&amp;=\int\limits_{t\in [d,x] \; \cap \; C}{f(\gamma (t)){{{{\gamma }'}}_{\curvearrowright }}D(t)dDt}\;\,\;\;-\int\limits_{t\in [d,x[ \; \cap \; C}{f(\gamma (t)){{{{\gamma }'}}_{\curvearrowright }}D(t)dDt} \\ &amp;=\int\limits_{x}{f(\gamma (t))\frac{\gamma (\curvearrowright Dt)-\gamma (t)}{\curvearrowright Dt-t}dDt}=f(\gamma (x)){{{\gamma }'}_{\curvearrowright }}D(x)dDx=\,f(\gamma (x))(\curvearrowright B\gamma (x)-\gamma (x))=f(z)dBz.\square\end{aligned}</math></td></tr></table>
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<table style="width:100%"><tr><td style="vertical-align: top; padding-top: 1em;">'''Proof:'''</td><td style="text-align: center; font-size: 84%;"><math>\begin{aligned}{\downarrow}(F(z)) &amp;={\uparrow}_{s\in [d,x] \cap C}{f(\gamma (s)){{\gamma}^{\prime}}(s){\downarrow}s}-{\uparrow}_{s\in [d,x[ \, \cap \, C}{f(\gamma (s)){{\gamma }^{\prime}}(s){\downarrow}s} ={\uparrow}_{x}{f(\gamma (s))\tfrac{\gamma (\overset{\rightharpoonup}{s})-\gamma (s)}{\overset{\rightharpoonup}{s}-s}{\downarrow}s} \\ &amp;=f(\gamma (x)){{\gamma}^{\prime}}(x){\downarrow}x =\,f(\gamma (x))(\overset{\rightharpoonup}{\gamma}(x)-\gamma (x))=f(z){\downarrow}z.\square\end{aligned}</math></td></tr></table>
  
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'''Second fundamental theorem of exact differential and integral calculus for line integrals:''' According to the conditions from above, it holds with <math>\gamma: [a, b[ \; \cap \; C \rightarrow {}^{(\omega)}\mathbb{K}</math> that
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'''Second fundamental theorem of exact differential and integral calculus for line integrals:''' Conditions above imply with <math>\gamma: G \rightarrow {}^{(\omega)}\mathbb{K}</math> that
  
  
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<div style="text-align:center;"><math> F(\gamma (b))-F(\gamma (a))=\int\limits_{\gamma }{{{{{F}'}}_{\curvearrowright }}B(\zeta )dB\zeta }.</math></div>
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<div style="text-align:center;"><math> F(\gamma (b))-F(\gamma (a))={\uparrow}_{\gamma }{{F^{\prime}}(\zeta ){\downarrow}\zeta }.</math></div>
  
  
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<table style="width:100%"><tr><td style="vertical-align: top; padding-top: 0.4em;">'''Proof:'''</td><td style="text-align: center;"><math>\begin{aligned}F(\gamma (b))-F(\gamma (a))&amp;=\sum\limits_{t\in [a,b[ \; \cap \; C}{F(\curvearrowright B\,\gamma (t))}-F(\gamma (t))\;\,=\sum\limits_{t\in [a,b[ \; \cap \; C}{{{{{F}'}}_{\curvearrowright }}B(\gamma (t))(\curvearrowright B\,\gamma (t)-\gamma (t))} \\ &amp;=\int\limits_{t\in [a,b[ \; \cap \; C}{{{{{F}'}}_{\curvearrowright }}B(\gamma (t)){{{{\gamma }'}}_{\curvearrowright }}D(t)dDt}=\int\limits_{\gamma }{{{{{F}'}}_{\curvearrowright }}B(\zeta )dB\zeta }.\square\end{aligned}</math></td></tr></table>
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<table style="width:100%"><tr><td style="vertical-align: top; padding-top: 0.4em;">'''Proof:'''</td><td style="text-align: center; font-size: 84%;"><math>F(\gamma (b))-F(\gamma (a))</math> <math>={\Large{+}}_{s\in G}{F(\overset{\rightharpoonup}{\gamma}(s))}-F(\gamma (s))</math> <math>={\Large{+}}_{s\in G}{{{F}^{\prime}}(\gamma (s))(\overset{\rightharpoonup}{\gamma}(s)-\gamma(s))}</math> <math>={\uparrow}_{s\in G}{{{F}^{\prime}}(\gamma (s)){{\gamma }^{\prime}}(s){\downarrow}s}</math> <math>={\uparrow}_{\gamma }{{{F}^{\prime}}(\zeta ){\downarrow}\zeta }.\square</math></td></tr></table>
  
 
== See also ==
 
== See also ==

Latest revision as of 18:04, 1 May 2024

First fundamental theorem of exact differential and integral calculus for line integrals: The function [math]\displaystyle{ F(z)={\uparrow}_{\gamma }{f(\zeta ){\downarrow}\zeta } }[/math] where [math]\displaystyle{ \gamma: [d, x[ \, \cap \, C \rightarrow A \subseteq {}^{(\omega)}\mathbb{K}, C \subseteq \mathbb{R}, f: A \rightarrow {}^{(\omega)}\mathbb{K}, d \in G = [a, b[ \, \cap \, C }[/math], and choosing [math]\displaystyle{ \overset{\rightharpoonup}{\gamma}(x) = \gamma(\overset{\rightharpoonup}{x}) }[/math] is exactly differentiable, and for all [math]\displaystyle{ x \in G }[/math] and [math]\displaystyle{ z = \gamma(x) }[/math]

[math]\displaystyle{ F^{\prime}(z) = f(z). }[/math]


Proof:[math]\displaystyle{ \begin{aligned}{\downarrow}(F(z)) &={\uparrow}_{s\in [d,x] \cap C}{f(\gamma (s)){{\gamma}^{\prime}}(s){\downarrow}s}-{\uparrow}_{s\in [d,x[ \, \cap \, C}{f(\gamma (s)){{\gamma }^{\prime}}(s){\downarrow}s} ={\uparrow}_{x}{f(\gamma (s))\tfrac{\gamma (\overset{\rightharpoonup}{s})-\gamma (s)}{\overset{\rightharpoonup}{s}-s}{\downarrow}s} \\ &=f(\gamma (x)){{\gamma}^{\prime}}(x){\downarrow}x =\,f(\gamma (x))(\overset{\rightharpoonup}{\gamma}(x)-\gamma (x))=f(z){\downarrow}z.\square\end{aligned} }[/math]

Second fundamental theorem of exact differential and integral calculus for line integrals: Conditions above imply with [math]\displaystyle{ \gamma: G \rightarrow {}^{(\omega)}\mathbb{K} }[/math] that


[math]\displaystyle{ F(\gamma (b))-F(\gamma (a))={\uparrow}_{\gamma }{{F^{\prime}}(\zeta ){\downarrow}\zeta }. }[/math]


Proof:[math]\displaystyle{ F(\gamma (b))-F(\gamma (a)) }[/math] [math]\displaystyle{ ={\Large{+}}_{s\in G}{F(\overset{\rightharpoonup}{\gamma}(s))}-F(\gamma (s)) }[/math] [math]\displaystyle{ ={\Large{+}}_{s\in G}{{{F}^{\prime}}(\gamma (s))(\overset{\rightharpoonup}{\gamma}(s)-\gamma(s))} }[/math] [math]\displaystyle{ ={\uparrow}_{s\in G}{{{F}^{\prime}}(\gamma (s)){{\gamma }^{\prime}}(s){\downarrow}s} }[/math] [math]\displaystyle{ ={\uparrow}_{\gamma }{{{F}^{\prime}}(\zeta ){\downarrow}\zeta }.\square }[/math]

See also

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