restart;Grundbegriffe

Maple f\344llt NICHT auf die falsche Integralauswertung herein!Int(1/x^2,x=-1..2)=int(1/x^2,x=-1..2);

NiMvLUkkSW50RzYkSSpwcm90ZWN0ZWRHRidJKF9zeXNsaWJHNiI2JCokSSJ4R0YpISIjL0YsOyEiIiIiI0kpaW5maW5pdHlHRic=

Integrationstechniken

Partielle Integration wird korrekt ausgef\374hrt:(Bemerkung: Die Integrationskonstanten fehlen jeweils.)Int(x^2*sin(x),x)=int(x^2*sin(x),x);

NiMvLUkkSW50RzYkSSpwcm90ZWN0ZWRHRidJKF9zeXNsaWJHNiI2JComSSJ4R0YpIiIjLUkkc2luR0YmNiNGLCIiIkYsLCgqJkYsRi0tSSRjb3NHRiZGMEYxISIiRjRGLSomRixGMUYuRjFGLQ==

Variante: Hier wird mit Hilfe eines Student-Packages die partielle Integration schrittweise ausgef\374hrt:with(Student:-Calculus1): infolevel[Student[Calculus1]] := 1:Rule[parts, x^2,-cos(x)](Int(x^2*sin(x), x));

Creating problem #1 NiMvLUkkSW50RzYkSSpwcm90ZWN0ZWRHRidJKF9zeXNsaWJHNiI2JComSSJ4R0YpIiIjLUkkc2luR0YmNiNGLCIiIkYsLCYqJkYsRi0tSSRjb3NHRiZGMEYxISIiLUYlNiQsJComRixGMUY0RjEhIiNGLEY2

Das Integral \374ber -2xcos(x) wird (mittels partieller Integration) ausgewertet:Rule[parts, -2*x,sin(x)](%);

NiMvLUkkSW50RzYkSSpwcm90ZWN0ZWRHRidJKF9zeXNsaWJHNiI2JComSSJ4R0YpIiIjLUkkc2luR0YmNiNGLCIiIkYsLCgqJkYsRi0tSSRjb3NHRiZGMEYxISIiKiZGLEYxRi5GMUYtLUYlNiQsJEYuISIjRixGMQ==

Konstante werden vor das Integral gezogen:Rule[constantmultiple](%);

NiMvLUkkSW50RzYkSSpwcm90ZWN0ZWRHRidJKF9zeXNsaWJHNiI2JComSSJ4R0YpIiIjLUkkc2luR0YmNiNGLCIiIkYsLCgqJkYsRi0tSSRjb3NHRiZGMEYxISIiKiZGLEYxRi5GMUYtLUYlNiRGLkYsISIj

Das Integral \374ber sin(x) wird ausgewertet:Rule[sin](%);

NiMvLUkkSW50RzYkSSpwcm90ZWN0ZWRHRidJKF9zeXNsaWJHNiI2JComSSJ4R0YpIiIjLUkkc2luR0YmNiNGLCIiIkYsLCgqJkYsRi0tSSRjb3NHRiZGMEYxISIiKiZGLEYxRi5GMUYtRjRGLQ==

Int((sin(x))^n, x)=int((sin(x))^n, x);

NiMvLUkkSW50RzYkSSpwcm90ZWN0ZWRHRidJKF9zeXNsaWJHNiI2JCktSSRzaW5HRiY2I0kieEdGKUkibkdGKUYvLUkkaW50R0YmRio=

Das ist offenbar auch f\374r Maple nicht so einfach!with(Student:-Calculus1): infolevel[Student[Calculus1]] := 1:Zun\344chst partielle Integration:Rule[parts, (sin(x))^(n-1), -cos(x)](Int((sin(x))^n, x));

Creating problem #2 NiMvLUkkSW50RzYkSSpwcm90ZWN0ZWRHRidJKF9zeXNsaWJHNiI2JCktSSRzaW5HRiY2I0kieEdGKUkibkdGKUYvLCYqJilGLCwmISIiIiIiRjBGNkY2LUkkY29zR0YmRi5GNkY1LUYlNiQqKiwmRjZGNkYwRjVGNkY3IiIjRiwhIiNGK0Y2Ri9GNQ==

Konstante werden vor das Integral gezogen:Rule[constantmultiple](%);

NiMvLUkkSW50RzYkSSpwcm90ZWN0ZWRHRidJKF9zeXNsaWJHNiI2JCktSSRzaW5HRiY2I0kieEdGKUkibkdGKUYvLCYqJilGLCwmISIiIiIiRjBGNkY2LUkkY29zR0YmRi5GNkY1KiYsJkY2RjZGMEY1RjYtRiU2JCooRjciIiNGLCEiI0YrRjZGL0Y2RjU=

Es wird ausgenutzt: cos^2=1-sin^2:Rule[rewrite,(cos(x))^2=1-(sin(x))^2](%);

NiMvLUkkSW50RzYkSSpwcm90ZWN0ZWRHRidJKF9zeXNsaWJHNiI2JCktSSRzaW5HRiY2I0kieEdGKUkibkdGKUYvLCYqJilGLCwmISIiIiIiRjBGNkY2LUkkY29zR0YmRi5GNkY1KiYsJkY2RjZGMEY1RjYtRiU2JCwkKihGLCEiI0YrRjYsJkY1RjYqJEYsIiIjRjZGNkY1Ri9GNkY1

Es wird ausgenutzt: sin^n/sin^2=sin^(n-2):Rule[rewrite,(sin(x))^n/(sin(x))^2=(sin(x))^(n-2)](%);

NiMvLUkkSW50RzYkSSpwcm90ZWN0ZWRHRidJKF9zeXNsaWJHNiI2JCktSSRzaW5HRiY2I0kieEdGKUkibkdGKUYvLCYqJilGLCwmISIiIiIiRjBGNkY2LUkkY29zR0YmRi5GNkY1KiYsJkY2RjZGMEY1RjYtRiU2JCwkKiYpRiwsJiEiI0Y2RjBGNkY2LCZGNUY2KiRGLCIiI0Y2RjZGNUYvRjZGNQ==

Die Sinus-Terme im Integranden rechts werden ausmultipliziert:Rule[rewrite,(sin(x))^(n-2)*(-1+(sin(x))^2)=-(sin(x))^(n-2)+(sin(x))^n](%);

NiMvLUkkSW50RzYkSSpwcm90ZWN0ZWRHRidJKF9zeXNsaWJHNiI2JCktSSRzaW5HRiY2I0kieEdGKUkibkdGKUYvLCYqJilGLCwmISIiIiIiRjBGNkY2LUkkY29zR0YmRi5GNkY1KiYsJkY2RjZGMEY1RjYtRiU2JCwmKUYsLCYhIiNGNkYwRjZGNkYrRjVGL0Y2RjU=

Die Additivit\344t des Integrals wird ausgenutzt:Rule[sum](%);

NiMvLUkkSW50RzYkSSpwcm90ZWN0ZWRHRidJKF9zeXNsaWJHNiI2JCktSSRzaW5HRiY2I0kieEdGKUkibkdGKUYvLCYqJilGLCwmISIiIiIiRjBGNkY2LUkkY29zR0YmRi5GNkY1KiYsJkY2RjZGMEY1RjYsJi1GJTYkKUYsLCYhIiNGNkYwRjZGL0Y2LUYlNiQsJEYrRjVGL0Y2RjZGNQ==

a:=simplify(%);

NiM+SSJhRzYiLy1JJEludEc2JEkqcHJvdGVjdGVkR0YqSShfc3lzbGliR0YlNiQpLUkkc2luR0YpNiNJInhHRiVJIm5HRiVGMSwsKiYpRi4sJiEiIiIiIkYyRjhGOC1JJGNvc0dGKUYwRjhGNy1GKDYkKUYuLCYhIiNGOEYyRjhGMUY3RidGOComRjJGOEY7RjhGOComRjJGOEYnRjhGNw==

Jetzt wird das Integral \374ber sin^n isoliert:isolate(Int((sin(x))^n,x)=rhs(a),Int((sin(x))^n,x));

NiMvLUkkSW50RzYkSSpwcm90ZWN0ZWRHRidJKF9zeXNsaWJHNiI2JCktSSRzaW5HRiY2I0kieEdGKUkibkdGKUYvKiYsKComKUYsLCYhIiIiIiJGMEY3RjctSSRjb3NHRiZGLkY3RjYtRiU2JClGLCwmISIjRjdGMEY3Ri9GNiomRjBGN0Y6RjdGN0Y3RjBGNg==

UFF!!

Int((sin(t))^2*cos(t),t)=int((sin(t))^2*cos(t),t);

NiMvLUkkSW50RzYkSSpwcm90ZWN0ZWRHRidJKF9zeXNsaWJHNiI2JComLUkkc2luR0YmNiNJInRHRikiIiMtSSRjb3NHRiZGLiIiIkYvLCQqJEYsIiIkI0YzRjY=

Int(tan(x),x)=int(tan(x),x);

NiMvLUkkSW50RzYkSSpwcm90ZWN0ZWRHRidJKF9zeXNsaWJHNiI2JC1JJHRhbkdGJjYjSSJ4R0YpRi4sJC1JI2xuR0YmNiMtSSRjb3NHRiZGLSEiIg==

Int(x^2/sqrt(1-x^2),x)=int(x^2/sqrt(1-x^2),x);

NiMvLUkkSW50RzYkSSpwcm90ZWN0ZWRHRidJKF9zeXNsaWJHNiI2JComSSJ4R0YpIiIjLCYiIiJGLyokRixGLSEiIiNGMUYtRiwsJiomRixGL0YuI0YvRi1GMi1JJ2FyY3NpbkdGJjYjRixGNQ==

Uneigentliche Integrale

Int(exp(-x),x=0..infinity)=int(exp(-x),x=0..infinity);

NiMvLUkkSW50RzYkSSpwcm90ZWN0ZWRHRidJKF9zeXNsaWJHNiI2JC1JJGV4cEdGJjYjLCRJInhHRikhIiIvRi87IiIhSSlpbmZpbml0eUdGJyIiIg==

Int(1/sqrt(x),x=0..1)=int(1/sqrt(x),x=0..1);

NiMvLUkkSW50RzYkSSpwcm90ZWN0ZWRHRidJKF9zeXNsaWJHNiI2JCokSSJ4R0YpIyEiIiIiIy9GLDsiIiEiIiJGLw==

Int(1/x^2,x=-1..0)=int(1/x^2,x=-1..0);

NiMvLUkkSW50RzYkSSpwcm90ZWN0ZWRHRidJKF9zeXNsaWJHNiI2JCokSSJ4R0YpISIjL0YsOyEiIiIiIUkpaW5maW5pdHlHRic=

Int(1/x^2,x=0..2)=int(1/x^2,x=0..2);

NiMvLUkkSW50RzYkSSpwcm90ZWN0ZWRHRidJKF9zeXNsaWJHNiI2JCokSSJ4R0YpISIjL0YsOyIiISIiI0kpaW5maW5pdHlHRic=

Mehrfachintegrale

Int( Int(1,y=0..sqrt(1-x^2)), x=0..1)=Int( int(1,y=0..sqrt(1-x^2)), x=0..1);

NiMvLUkkSW50RzYkSSpwcm90ZWN0ZWRHRidJKF9zeXNsaWJHNiI2JC1GJTYkIiIiL0kieUdGKTsiIiEqJCwmRi1GLSokSSJ4R0YpIiIjISIiI0YtRjYvRjU7RjFGLS1GJTYkRjJGOQ==

Int( int(1,y=0..sqrt(1-x^2)), x=0..1)=int( int(1,y=0..sqrt(1-x^2)), x=0..1);

NiMvLUkkSW50RzYkSSpwcm90ZWN0ZWRHRidJKF9zeXNsaWJHNiI2JCokLCYiIiJGLSokSSJ4R0YpIiIjISIiI0YtRjAvRi87IiIhRi0sJEkjUGlHRicjRi0iIiU=

Fl\344che:=int( int(1,y=0..sqrt(1-x^2)), x=0..1);

NiM+SSdGbHxfeWNoZUc2IiwkSSNQaUdJKnByb3RlY3RlZEdGKCMiIiIiIiU=

Zun\344chst wird die Fl\344che A berechnet (vgl. auch Aufgabe 1):Int( Int(1,y=0..sqrt(1-x^2)), x=0..1)=int( int(1,y=0..sqrt(1-x^2)), x=0..1);

NiMvLUkkSW50RzYkSSpwcm90ZWN0ZWRHRidJKF9zeXNsaWJHNiI2JC1GJTYkIiIiL0kieUdGKTsiIiEqJCwmRi1GLSokSSJ4R0YpIiIjISIiI0YtRjYvRjU7RjFGLSwkSSNQaUdGJyNGLSIiJQ==

A:=int( int(1,y=0..sqrt(1-x^2)), x=0..1);

NiM+SSJBRzYiLCRJI1BpR0kqcHJvdGVjdGVkR0YoIyIiIiIiJQ==

Nun wird schrittweise das Doppelintegral berechnet (inneres Integral in y, \344u\337eres in x):Int( Int(x,y=0..sqrt(1-x^2)), x=0..1)=Int( int(x,y=0..sqrt(1-x^2)), x=0..1);

NiMvLUkkSW50RzYkSSpwcm90ZWN0ZWRHRidJKF9zeXNsaWJHNiI2JC1GJTYkSSJ4R0YpL0kieUdGKTsiIiEqJCwmIiIiRjQqJEYtIiIjISIiI0Y0RjYvRi07RjFGNC1GJTYkKiZGLUY0RjNGOEY5

Int( int(x,y=0..sqrt(1-x^2)), x=0..1)=int( int(x,y=0..sqrt(1-x^2)), x=0..1);

NiMvLUkkSW50RzYkSSpwcm90ZWN0ZWRHRidJKF9zeXNsaWJHNiI2JComSSJ4R0YpIiIiLCZGLUYtKiRGLCIiIyEiIiNGLUYwL0YsOyIiIUYtI0YtIiIk

Jetzt wird das Doppelintegral durch die Fl\344che dividiert:1/'A'*Int( Int(x,y=0..sqrt(1-x^2)), x=0..1)= int( int(x,y=0..sqrt(1-x^2)), x=0..1) /A;

NiMvKiZJIkFHNiIhIiItSSRJbnRHNiRJKnByb3RlY3RlZEdGK0koX3N5c2xpYkdGJjYkLUYpNiRJInhHRiYvSSJ5R0YmOyIiISokLCYiIiJGNyokRjAiIiNGJyNGN0Y5L0YwO0Y0RjdGNywkKiRJI1BpR0YrRicjIiIlIiIk

Zun\344chst wird die Fl\344che A berechnet (vgl. auch Aufgabe 1):Int( Int(1,y=0..sqrt(1-x^2)), x=0..1)=int( int(1,y=0..sqrt(1-x^2)), x=0..1);

NiMvLUkkSW50RzYkSSpwcm90ZWN0ZWRHRidJKF9zeXNsaWJHNiI2JC1GJTYkIiIiL0kieUdGKTsiIiEqJCwmRi1GLSokSSJ4R0YpIiIjISIiI0YtRjYvRjU7RjFGLSwkSSNQaUdGJyNGLSIiJQ==

A:=int( int(1,y=0..sqrt(1-x^2)), x=0..1);

NiM+SSJBRzYiLCRJI1BpR0kqcHJvdGVjdGVkR0YoIyIiIiIiJQ==

Nun wird schrittweise das Doppelintegral berechnet (inneres Integral in x, \344u\337eres in y):Int( Int(x,x=0..sqrt(1-y^2)), y=0..1)=Int( int(x,x=0..sqrt(1-y^2)), y=0..1);

NiMvLUkkSW50RzYkSSpwcm90ZWN0ZWRHRidJKF9zeXNsaWJHNiI2JC1GJTYkSSJ4R0YpL0YtOyIiISokLCYiIiJGMyokSSJ5R0YpIiIjISIiI0YzRjYvRjU7RjBGMy1GJTYkLCZGOEYzRjQjRjdGNkY5

Int( int(x,x=0..sqrt(1-y^2)), y=0..1)=int( int(x,x=0..sqrt(1-y^2)), y=0..1);

NiMvLUkkSW50RzYkSSpwcm90ZWN0ZWRHRidJKF9zeXNsaWJHNiI2JCwmIyIiIiIiI0YtKiRJInlHRilGLiMhIiJGLi9GMDsiIiFGLSNGLSIiJA==

Jetzt wird das Doppelintegral durch die Fl\344che dividiert:1/'A'*Int( Int(x,x=0..sqrt(1-y^2)), y=0..1)= int( int(x,x=0..sqrt(1-y^2)), y=0..1) /A;

NiMvKiZJIkFHNiIhIiItSSRJbnRHNiRJKnByb3RlY3RlZEdGK0koX3N5c2xpYkdGJjYkLUYpNiRJInhHRiYvRjA7IiIhKiQsJiIiIkY2KiRJInlHRiYiIiNGJyNGNkY5L0Y4O0YzRjZGNiwkKiRJI1BpR0YrRicjIiIlIiIk