Monotonie des Integrals von Treppenfunktionen
About points...
We associate a certain number of points with each exercise.
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.
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:
Beweisen Sie folge Aussage: Sind fg in mathcalTFab zwei Treppenfunktionen mit fleq g. Dann gilt _a^b f ddxleq _a^b g ddx Insbesondere impliziert f in mathcalTFab und f geq dass _a^b f ddx geq .
Solution:
Beweis. Wie schon bei Lemma . kann man für fg in mathcalTFab eine gemeinsame Zerlegung zeta ax_ x_ ... x_n- x_n b in Konstanzervalle finden. Man schreibt wieder c_...c_nd_...d_n für die Konstanzwerte von f resp. g bezüglich zeta. Falls nun f leq g ist dann ist c_k leq d_k für alle k in ...n und man erhält _a^b f ddx Ifzeta_k^n c_kDelta x_k leq _k^n d_kDelta x_k Igzeta _a^b g ddx Die zweite Aussage folgt aus der ersten angewet auf und f.
Beweisen Sie folge Aussage: Sind fg in mathcalTFab zwei Treppenfunktionen mit fleq g. Dann gilt _a^b f ddxleq _a^b g ddx Insbesondere impliziert f in mathcalTFab und f geq dass _a^b f ddx geq .
Solution:
Beweis. Wie schon bei Lemma . kann man für fg in mathcalTFab eine gemeinsame Zerlegung zeta ax_ x_ ... x_n- x_n b in Konstanzervalle finden. Man schreibt wieder c_...c_nd_...d_n für die Konstanzwerte von f resp. g bezüglich zeta. Falls nun f leq g ist dann ist c_k leq d_k für alle k in ...n und man erhält _a^b f ddx Ifzeta_k^n c_kDelta x_k leq _k^n d_kDelta x_k Igzeta _a^b g ddx Die zweite Aussage folgt aus der ersten angewet auf und f.
Meta Information
Exercise:
Beweisen Sie folge Aussage: Sind fg in mathcalTFab zwei Treppenfunktionen mit fleq g. Dann gilt _a^b f ddxleq _a^b g ddx Insbesondere impliziert f in mathcalTFab und f geq dass _a^b f ddx geq .
Solution:
Beweis. Wie schon bei Lemma . kann man für fg in mathcalTFab eine gemeinsame Zerlegung zeta ax_ x_ ... x_n- x_n b in Konstanzervalle finden. Man schreibt wieder c_...c_nd_...d_n für die Konstanzwerte von f resp. g bezüglich zeta. Falls nun f leq g ist dann ist c_k leq d_k für alle k in ...n und man erhält _a^b f ddx Ifzeta_k^n c_kDelta x_k leq _k^n d_kDelta x_k Igzeta _a^b g ddx Die zweite Aussage folgt aus der ersten angewet auf und f.
Beweisen Sie folge Aussage: Sind fg in mathcalTFab zwei Treppenfunktionen mit fleq g. Dann gilt _a^b f ddxleq _a^b g ddx Insbesondere impliziert f in mathcalTFab und f geq dass _a^b f ddx geq .
Solution:
Beweis. Wie schon bei Lemma . kann man für fg in mathcalTFab eine gemeinsame Zerlegung zeta ax_ x_ ... x_n- x_n b in Konstanzervalle finden. Man schreibt wieder c_...c_nd_...d_n für die Konstanzwerte von f resp. g bezüglich zeta. Falls nun f leq g ist dann ist c_k leq d_k für alle k in ...n und man erhält _a^b f ddx Ifzeta_k^n c_kDelta x_k leq _k^n d_kDelta x_k Igzeta _a^b g ddx Die zweite Aussage folgt aus der ersten angewet auf und f.
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