Riemann-Integral mit additiver Intervallfunktion
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That being said... How many "default points" should you associate with an exercise upon creation?
As with difficulty, there is no straight forward and generally accepted way.
But as a guideline, we tend to give as many points by default as there are mathematical steps to do in the exercise.
Again, very vague... But the number should kind of represent the "work" required.
When you click an exercise into a collection, this number will be taken as points for the exercise, kind of "by default".
But once the exercise is on the collection, you can edit the number of points for the exercise in the collection independently, without any effect on "points by default" as represented by the number here.
That being said... How many "default points" should you associate with an exercise upon creation?
As with difficulty, there is no straight forward and generally accepted way.
But as a guideline, we tend to give as many points by default as there are mathematical steps to do in the exercise.
Again, very vague... But the number should kind of represent the "work" required.
About difficulty...
We associate a certain difficulty with each exercise.
When you click an exercise into a collection, this number will be taken as difficulty for the exercise, kind of "by default".
But once the exercise is on the collection, you can edit its difficulty in the collection independently, without any effect on the "difficulty by default" here.
Why we use chess pieces? Well... we like chess, we like playing around with \(\LaTeX\)-fonts, we wanted symbols that need less space than six stars in a table-column... But in your layouts, you are of course free to indicate the difficulty of the exercise the way you want.
That being said... How "difficult" is an exercise? It depends on many factors, like what was being taught etc.
In physics exercises, we try to follow this pattern:
Level 1 - One formula (one you would find in a reference book) is enough to solve the exercise. Example exercise
Level 2 - Two formulas are needed, it's possible to compute an "in-between" solution, i.e. no algebraic equation needed. Example exercise
Level 3 - "Chain-computations" like on level 2, but 3+ calculations. Still, no equations, i.e. you are not forced to solve it in an algebraic manner. Example exercise
Level 4 - Exercise needs to be solved by algebraic equations, not possible to calculate numerical "in-between" results. Example exercise
Level 5 -
Level 6 -
When you click an exercise into a collection, this number will be taken as difficulty for the exercise, kind of "by default".
But once the exercise is on the collection, you can edit its difficulty in the collection independently, without any effect on the "difficulty by default" here.
Why we use chess pieces? Well... we like chess, we like playing around with \(\LaTeX\)-fonts, we wanted symbols that need less space than six stars in a table-column... But in your layouts, you are of course free to indicate the difficulty of the exercise the way you want.
That being said... How "difficult" is an exercise? It depends on many factors, like what was being taught etc.
In physics exercises, we try to follow this pattern:
Level 1 - One formula (one you would find in a reference book) is enough to solve the exercise. Example exercise
Level 2 - Two formulas are needed, it's possible to compute an "in-between" solution, i.e. no algebraic equation needed. Example exercise
Level 3 - "Chain-computations" like on level 2, but 3+ calculations. Still, no equations, i.e. you are not forced to solve it in an algebraic manner. Example exercise
Level 4 - Exercise needs to be solved by algebraic equations, not possible to calculate numerical "in-between" results. Example exercise
Level 5 -
Level 6 -
Question
Solution
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Exercise:
Seien a b in mathbbR f:ab rightarrow mathbbR eine R-bare Funktion und mathcalI eine additive Intervallfunktion auf ab. Angenommen es gilt beta -alpha textinf_x in alphabeta fx leq mathcalIalphabeta leq beta-alpha textsup_x in alphabeta fx für alle alpha beta in ab. Dann ist mathcalIalphabeta _alpha^beta fx ddx für alle alphabeta in ab.
Solution:
Beweis. Sei uleq f eine Treppenfunktion auf ab mit Zerlegung zeta ax_ ... x_nbeta in Konstanzervalle von u. Seien c_...c_n die Konstanzwerte von u bezüglich zeta. Auf Grund der Annahme uleq f folgt c_k leq textinf_x in x_k-x_kfx für alle k in ...n. Unter Verwung der Additivität von mathcalI erhält man damit für die Unterme _alpha^beta ux ddx _k^n c_kx_k-x_k- &leq _k^n x_k-x_k- textinf_x in x_k-x_kfx &leq _k^n mathcalIx_k--x_k mathcalIalpha beta Ebenso ergibt sich mathcalIalpha beta leq _alpha^beta ox ddx für jede Treppenfunktion o mit f leq o. Daher gelten für das untere Integral underlineI und das obere Integral overlineI von f über alpha beta die Ungleichungen underlineIleq mathcalIalphabetaleq overlineI. Da f aber R-bar ist gilt underlineI overlineI und somit mathcalIalphabeta _alpha^beta fx ddx.
Seien a b in mathbbR f:ab rightarrow mathbbR eine R-bare Funktion und mathcalI eine additive Intervallfunktion auf ab. Angenommen es gilt beta -alpha textinf_x in alphabeta fx leq mathcalIalphabeta leq beta-alpha textsup_x in alphabeta fx für alle alpha beta in ab. Dann ist mathcalIalphabeta _alpha^beta fx ddx für alle alphabeta in ab.
Solution:
Beweis. Sei uleq f eine Treppenfunktion auf ab mit Zerlegung zeta ax_ ... x_nbeta in Konstanzervalle von u. Seien c_...c_n die Konstanzwerte von u bezüglich zeta. Auf Grund der Annahme uleq f folgt c_k leq textinf_x in x_k-x_kfx für alle k in ...n. Unter Verwung der Additivität von mathcalI erhält man damit für die Unterme _alpha^beta ux ddx _k^n c_kx_k-x_k- &leq _k^n x_k-x_k- textinf_x in x_k-x_kfx &leq _k^n mathcalIx_k--x_k mathcalIalpha beta Ebenso ergibt sich mathcalIalpha beta leq _alpha^beta ox ddx für jede Treppenfunktion o mit f leq o. Daher gelten für das untere Integral underlineI und das obere Integral overlineI von f über alpha beta die Ungleichungen underlineIleq mathcalIalphabetaleq overlineI. Da f aber R-bar ist gilt underlineI overlineI und somit mathcalIalphabeta _alpha^beta fx ddx.
Meta Information
Exercise:
Seien a b in mathbbR f:ab rightarrow mathbbR eine R-bare Funktion und mathcalI eine additive Intervallfunktion auf ab. Angenommen es gilt beta -alpha textinf_x in alphabeta fx leq mathcalIalphabeta leq beta-alpha textsup_x in alphabeta fx für alle alpha beta in ab. Dann ist mathcalIalphabeta _alpha^beta fx ddx für alle alphabeta in ab.
Solution:
Beweis. Sei uleq f eine Treppenfunktion auf ab mit Zerlegung zeta ax_ ... x_nbeta in Konstanzervalle von u. Seien c_...c_n die Konstanzwerte von u bezüglich zeta. Auf Grund der Annahme uleq f folgt c_k leq textinf_x in x_k-x_kfx für alle k in ...n. Unter Verwung der Additivität von mathcalI erhält man damit für die Unterme _alpha^beta ux ddx _k^n c_kx_k-x_k- &leq _k^n x_k-x_k- textinf_x in x_k-x_kfx &leq _k^n mathcalIx_k--x_k mathcalIalpha beta Ebenso ergibt sich mathcalIalpha beta leq _alpha^beta ox ddx für jede Treppenfunktion o mit f leq o. Daher gelten für das untere Integral underlineI und das obere Integral overlineI von f über alpha beta die Ungleichungen underlineIleq mathcalIalphabetaleq overlineI. Da f aber R-bar ist gilt underlineI overlineI und somit mathcalIalphabeta _alpha^beta fx ddx.
Seien a b in mathbbR f:ab rightarrow mathbbR eine R-bare Funktion und mathcalI eine additive Intervallfunktion auf ab. Angenommen es gilt beta -alpha textinf_x in alphabeta fx leq mathcalIalphabeta leq beta-alpha textsup_x in alphabeta fx für alle alpha beta in ab. Dann ist mathcalIalphabeta _alpha^beta fx ddx für alle alphabeta in ab.
Solution:
Beweis. Sei uleq f eine Treppenfunktion auf ab mit Zerlegung zeta ax_ ... x_nbeta in Konstanzervalle von u. Seien c_...c_n die Konstanzwerte von u bezüglich zeta. Auf Grund der Annahme uleq f folgt c_k leq textinf_x in x_k-x_kfx für alle k in ...n. Unter Verwung der Additivität von mathcalI erhält man damit für die Unterme _alpha^beta ux ddx _k^n c_kx_k-x_k- &leq _k^n x_k-x_k- textinf_x in x_k-x_kfx &leq _k^n mathcalIx_k--x_k mathcalIalpha beta Ebenso ergibt sich mathcalIalpha beta leq _alpha^beta ox ddx für jede Treppenfunktion o mit f leq o. Daher gelten für das untere Integral underlineI und das obere Integral overlineI von f über alpha beta die Ungleichungen underlineIleq mathcalIalphabetaleq overlineI. Da f aber R-bar ist gilt underlineI overlineI und somit mathcalIalphabeta _alpha^beta fx ddx.
Contained in these collections
| Title | Creator | Matched on |
|---|---|---|
| Charakterisierung der Riemann-Integrierbarkeit | rk | tags |
| Riemann-Integrierbarkeit von Polynomen | rk | tags |
| Riemann-Integral über Riemann-Summen | rk | tags |
| Linearität des Riemann-Integrals | rk | tags |
| Monotonie des Riemann-Integrals | rk | tags |

