Jörg Gayler, Lubov Vassilevskaya

Transkript

1 Integralrechnung: Aufgaben Jörg Gayler, Lubov Vassilevskaya

2 ii Contents 1. Unbestimmtes Integral: Aufgaben Grund- oder Stammintegrale (Tabelle Unbestimmte Integrale (Tabelle Einige Regeln Binomische Formeln Rechenregeln mit Potenzen Rechenregeln mit Logarithmen Trigonometrische Formeln Integration von Potenzfunktionen Elementarintegrale Integration rch Substitution Partielle Integration Anwenngen der Integralrechnung Flächen zwischen Kurven: Aufgaben Eingeschlossene Flächen Flächen zwischen Kurven in einem Intervall Unbestimmtes Integral: Lösungen Integration von Potenzfunktionen Elementarintegrale Integration rch Substitution Partielle Integration Anwenngen der Integralrechnung Flächen zwischen Kurven: Lösungen

3 1 Unbestimmtes Integral: Aufgaben 1. Unbestimmtes Integral: Aufgaben 1.1. Grund- oder Stammintegrale (Tabelle 1 n = n+1 n C e = e + C, sin = cos + C, cos = tan + C, sinh = cosh + C, cosh = tanh + C, (n 1, a + = 1 a arctan + C (a a a = 1 a ln a + a + C (a, a a ± = ± 1 ln a ± + C a = arcsin + C (a > a ± a = ln + ± a + C (a > a ± = ± a ± + C (a > a = = ln + C a ln a + C cos = sin + C sin = cot + C cosh = sinh + C sinh = coth + C a = ± a = a + a arcsin + C (a > a ± a ± a ln + ± a (a > 1.. Unbestimmte Integrale (Tabelle sin(a = 1 a cos(a + C, cos(a = 1 a sin(a + C (a

4 Einige Regeln. Einige Regeln.1. Binomische Formeln Binome zweiten Grades (a + b = a + ab + b (a b = a ab + b (a + b (a b = a b Binome höheren Grades (a + b = a + a b + ab + b (a b = a a b + ab b a b = (a b (a + ab + b a + b = (a + b (a ab + b.. Rechenregeln mit Potenzen a n a m = a n + m, a n b n = (a b n, a n a m = an m, a n b n = ( a b n a = 1, a n = 1 a n, an = 1 a n (a n m = a n m.. Rechenregeln mit Logarithmen log a ( y = log a + log a y ( log a = log y a log a y log a ( r = r log a log a 1 = log a

5 Einige Regeln.4. Trigonometrische Formeln sin α + cos α = 1 (1 sin ( α = sin α, cos ( α = cos α ( ( π sin α = cos α, ( π cos α = sin α ( sin α = sin (π α, cos α = cos (π α (4 sin (α = sin α cos α, cos (α = cos α sin α (5 sin (α ± β = sin α cos β ± cos α sin β, (6 cos (α ± β = cos α cos β sin α sin β (7 sin α sin β = 1 [ cos (α β cos (α + β ], (8 cos α cos β = 1 sin α cos β = 1 [ cos (α β + cos (α + β ] [ sin (α β + sin (α + β ] (9 (1 sin α = 1 (1 cos α, cos α = 1 (1 + cos α (11 sin α = 1 4 ( sin α sin (α, cos α = 1 (cos(α + cos α (1 4

6 Einige Regeln.5. Integration von Potenzfunktionen A1 Beispiel: ( = (5 1/4 + 4/ = 4 5/4 1/ 1 + C = C (, (4, ( ( 1 ( + 6, ( c ( I 1 = + 1 (, + + 5, d ( 1 I 1 = 1, ( e ( 1 I 1 = + ( +, f ( I 1 =, (5 7 A ( + 1, ( 1, ( ( 1 ( + + 4,, + 1 ( 1 ( + 1 c I 1 =,, Elementarintegrale A a 1 I 1 = (e sin, + b ( ( 5 I 1 = cos +, sin ( ( 1 c I 1 =,

7 Einige Regeln A4 Beispiel: = 1 1 = 1 1 = 1 1 = = + 1 ln C [ ] 1 = I 1 = (1 +, 1, I + = 1 A5 Beispiel: = = 5 1 ( = a = 5 14 arctan 7 + C, + a = 1 a arctan a + C, a = 7 I 1 = + 4, +, 4 +

8 Einige Regeln.7. Integration rch Substitution f (g ( g ( = f (u, u = g (, = g ( (1 Das Integral eines Proktes lässt sich immer dann berechnen, wenn ein Faktor f (g( eine Funktion einer inneren Funktion und der andere Faktor die Ableitung g ( der inneren Funktion ist, sofern für die Funktion f unter Beachtung der Substitution u = g( das Integral f (u gelöst werden kann. A6 Beispiel 1: ( + 5, u = +, ( + 5 = 1 u 5 = 1 =, = 1 u C = 1 1 ( C Beispiel : (6 4, u = 6, (6 4 = 1 u 4 = 1 =, = u C = 1 15 (6 5 + C ( +, (1 + 4, (4 5 ( + 1, (, ( 6 c I 1 = (1 +, 5 (7 4 4, 5 (5 6 A7 Beispiel: (6 9 ( + 11, u = + 11, (6 9 ( + 11 = =, u = u + C = ( C ( =

9 Einige Regeln I 1 = I 4 = ( 1 ( + 1 ( + ( ( + 1 ( ( 1 ( A8 Beispiel: 6 = 1 = 1 + = 1 + = = + ln + C I 1 = +,, + 4 A9 Beispiel: (9 4 ( = u = + 7, ( + 1 ( +, u 5 = 1 4u 4 + C = 1 4 ( C = 9 4 = (9 4, = (9 4 ( ( 4 11, ( ( A1 Beispiel: (4 1 = 1 u 1/ = 1 4 (1 1 u = 1, = 1 = (4, = (4, 6, + 7 (4, +, ( ,

10 Einige Regeln A11 Beispiel 1: e 7 = 1 e u = 1 eu + C u = 7, Beispiel : 4e = 1 e 1 = 1 = 1 ( u = 1 ( u + 1 u =, = 1 u (u + 1 = 1 ( 1 u 1 = u + 1 = 1 (ln u ln v + C 1 = 1 (ln u ln(u C 1 dv v = 1 ( ln( e 1 ln( e + C 1 = 1 ln( e 1 1 (ln + ln(e + C 1 = = 1 ln( e C, C = 1 ln + C 1 u = e 1, v = u + 1, = e = u + 1, = u + 1 dv = e, e, e m+n (m, n e, e, ( + 1 e + A1 c I 1 = e 1, π π/4 e, sin ( cos, sin ( sin, π/ π/ e e + 5 sin ( cos sin (4 sin ( c I 1 = π/4 cos ( cos, π/6 cos (4 cos ( A1 I a = sin (a, I b = cos (a

11 Einige Regeln A14 Beispiel: sin( cos(4 = 1 sin(4 4 + cos(4 = 1 sin(4 4 u = 1 u = 1 ln u + C = 1 ln + cos(4 + C, u = + cos(4, = sin(4 4 = 8 sin(4, = 1 8 ( 1 8 sin(4 sin(4 = cos 1 + sin, sin 7 cos, sin 1 + cos, sin( 1 + cos(, cos + sin cos( 4 + sin( A15 Beispiel 1: u = sin, sin cos = u cos = cos, = cos cos = u = u4 4 + C = 1 4 sin4 + C Beispiel : cos 6 ( sin(4 = = u = cos(, ( u 7 sin( sin( cos 6 ( sin( cos( = = = sin(, = sin( u 7 = u8 8 + C = 1 8 cos8 ( + C cos 7 ( sin( = sin 5 cos, cos 5 sin, sin 7 cos sin sin(, cos 5 sin(, sin 8 ( sin(4 Hinweis: sin( = sin cos A16

12 Einige Regeln Beispiel: cos 1 + sin = = arctan (sin + C, (sin 1 + sin = u = sin, = cos, = cos Grundintegral: 1 + = arctan + C cos 1 + u cos = 1 + u = arctan u + C = A17 sin 1 + cos, sin( 1 + cos (, cos 4 + sin, cos( 4 + sin (, sin 6 + cos sin( + 5 cos ( Beispiel: 1 + = ( = dv v = ln C u =, v = 1 + u, dv = u (u u = 1 + u = ( 1 1 = 1 + u = (u ln v + C = (u ln 1 + u + C = = 1 1/ = 1, = = u A18 A19 I 1 = + + 1, + 4, 4 +, + a, ,, , + +, + 1, + 5, 5.

13 Einige Regeln.8. Partielle Integration u v = u v u v (14 A Beispiel 1: ln = 1 ln = u = ln, u = 1, v =, v = = ( ln 1 + C, Beispiel : ln = ln ln = ln ln + + C ln = (ln 1 + C u = ln, u = ln, v = 1, v = Beispiel : ln = / ln 1/ = ( / ln + C = = ( ln + C u = ln, u = 1, v =, v = = 1/ = / ln, ln 4, ln 5 ln, ln, ln ( c I 1 = ( + 1 ln, ( + 1 ln, ln ( + 1 d I 1 = ln(, ln, 5 ln A1 I 1 = ln, ln, ln

14 Einige Regeln A Beispiel 1: e = e u =, u = 1, v = e, v = e = ( 1 e + C, e = e I 1 = e, e, 4 e A sin, cos, sin (, I 4 = cos ( sin, cos, cos(, I 4 = sin (

15 Anwenngen der Integralrechnung. Anwenngen der Integralrechnung.1. Flächen zwischen Kurven: Aufgaben.1.1. Eingeschlossene Flächen Berechnen Sie die von den Kurven f ( und g( eingeschlossene Fläche A4 f ( =, g( = A5 f ( =, A6 f ( =, g( = g( =.1.. Flächen zwischen Kurven in einem Intervall Welche Fläche schließen die Kurven im Intervall I A7 f ( =, g( =, [, ] A8 f ( =, g( =, [ 1, ] A9 f ( =, g( = + 1, [ 1, ]

16 Unbestimmtes Integral: Lösungen 4. Unbestimmtes Integral: Lösungen 4.1. Integration von Potenzfunktionen L1 ( = + C, (4 = 4 + C ( ( 1 = C ( + 6 = C, ( = ( C ( 1 ( c I 1 = + = 1/ + 1/ = ( + + C, ( = / + 4 4/ /5 + C = C, ( 5 7 = 5/ 7/ + C = ( + C ( 1 d I 1 = 1 = C, ( = 9 / + 5 4/5 + C ( 1 e I 1 = + + = / + 5 5/ + + C = ( = /4 7/ + C = /4 + C ( f I 1 = = = /6 = C, ( /4 1/ = 4 7 /4 4 =

17 Unbestimmtes Integral: Lösungen L ( + 1 = C, ( 1 = C, ( = = 1 + ln + C, ( 1 ( + = ( 1 ( + = C, + 4 = ( C, 5 c I 1 = ( 1 = ( C, 5 ( + = / C = C, = ( 1 = / + C 4.. Elementarintegrale L (e sin = e + cos + C, 1 + = ( cos + ( sin = = 9 9 ( 1 c I 1 = = ln + + C ( cos + ( sin = sin + arcsin + C, 1 5 = cos 5 1 arcsin + C 1 arcsin + ln + C, ( = 1 ln(1 + + arctan + C

18 Unbestimmtes Integral: Lösungen L4 I 1 = (1 + = = 1 arctan + C 1 = = 1 1 ln C + ( 1 = = 1 ln C L5 I 1 = + 4 = 1 ( arctan + C, + = 1 ( arctan + C, ( arctan + C 4 + =

19 Unbestimmtes Integral: Lösungen 4.. Integration rch Substitution L6 c I 1 = ( + ( + 4 = + C, 4 u = +, (1 + 4 (1 + 5 = + C, 15 u = 1 +, (4 5 (4 6 = + C, 1 u = 4 ( + 1 = 1 (, u =, ( 6, u = (1 +, u = 1 +, 5 (7 4 4, u = 7 4, 5 (5 6, u = 5 6 u = ( C, u = + 1, 6 L7 I 1 = ( 1 ( + 1 = 1 4 ( C, u = + 1, = ( + ( = 1 4 ( C, ( 1 u = + 4 7, = ( + ( ( = 1 6 u = 4 + 1, = I 4 = ( + 1 ( 1 ( = 1 1 ( , u = 6 + 9, = ( 1 ( 4 + 1,

20 Unbestimmtes Integral: Lösungen L8 I 1 = = = ln + + C, = ln + C, = = 4 ln C L9 ( + 1 ( + = 1 ( + + C, u = +, ( ( 4 11 = 1 4 ( C, u = 4 11, ( ( = 1 6 ( C, u = L1 = ( / + C = ( + C, u =, 6 = ( 6 + C, u = 6, + 7 = 8 ( + 74/ + C = 8 ( C, u = + 7 (4 = 1 5 ( 4 (4 + C, u = 4, + = 1 ( + / + C = 1 ( C, u = +, ( = 8 5 ( /4 + C = = 8 5 ( C, u = + 11

21 Unbestimmtes Integral: Lösungen L11 e = 1 e + C, u =, e = 1 e + C, u =, e m+n = 1 m em+n + C, u = m + n (m, n e = 1 e + C, u =, e = 1 6 e + C, u =, ( + 1 e + = 1 + e + C, u = + c I 1 = e 1 = ln e 1 ln (e + C = = ln e 1 + C, ln (e =, u = e 1, e = u (u + = 1 ( 1 u 1 = u + = 1 ( u u + = 1 (ln u ln u + + C 1, = 1 ln e 1 ln (e + C 1 = 1 ln e 1 ( ln + ln (e + C 1 = = 1 ln e 1 1 ln + C 1 = 1 ln e 1 + C, u = e, u = e + 5, = e, e e + 5 = 1 = e, = e = u +, ln(e =, C = C 1 1 u = 1 ln(u + C = 1 ln(e C, e = ln,

22 Unbestimmtes Integral: Lösungen L1 π sin ( cos sin α cos β = 1 [ ] sin (α β + sin (α + β (Eq. (1, Seite sin ( cos = 1 (sin + sin ( I 1 = 1 π [ 1 = 1 4 π/ sin ( cos = 1 ] π cos ( + cos = 4 [ 1 sin ( cos = 1 cos (4 + cos ( π π/ ] π/ (sin + sin ( = (sin ( + sin (4 = = 1 Variante : sin α = 1 4 ( sin α sin (α (Eq. (1, Seite sin( = sin 4 sin, π/ π/ sin ( cos = ( sin 4 sin cos = π/ = = π/ sin cos 4 [ cos sin 4 π/4 1 = = 1 6 π/ ] π/ sin ( sin = [ sin ] π/4 = 4 = 1, sin cos = [ sin sin (] π/4 = 1, sin (4 sin ( = 1 4 [sin ( 1 ] π/ sin (6 = 8

23 Unbestimmtes Integral: Lösungen L1 I a = c I 1 = π/4 π/6 cos ( cos = 1 [sin + 1 ] π/4 sin ( = cos (4 cos ( = 1 [sin ( + 1 ] π/6 4 sin (6 = 8 sin (a = 1 4a sin(a + C, I b = cos (a = + 1 4a sin(a + C L14 L15 cos = ln 1 + sin + C, u = 1 + sin, 1 + sin sin = ln 1 + cos + C, u = 1 + cos, 1 + cos cos + sin = 1 sin 7 cos = 1 sin( 1 + cos( = 1 ln + sin + C, u = + sin ln 7 cos + C, u = 7 cos, ln 1 + cos( + C, u = 1 + cos(, cos( 4 + sin( = 1 6 ln + sin( + C 1 = 1 6 ln 4 + sin( + C, C 1 = C + ln 6, u = + sin( sin 5 cos = 1 6 sin6 + C, u = sin, cos 5 sin = 1 6 cos6 + C, u = cos, sin 7 cos = 1 8 sin8 + C, u = sin sin sin( = cos 5 sin( = sin 8 ( sin(4 = sin 4 cos = 5 sin5 + C, u = sin, cos 6 sin = 7 cos7 + C, u = cos, sin 9 ( cos( = 1 1 sin1 ( + C, u = sin(

24 Unbestimmtes Integral: Lösungen L16 sin 1 + cos = arctan (cos + C, u = cos, cos 4 + sin = 1 arctan sin 6 + cos = sin( 1 + cos ( = 1 cos( 4 + sin ( = 1 4 arctan ( 1 sin + C, u = sin, ( 1 arctan cos + C, u = cos, arctan (cos (, u = cos (, ( 1 sin( + 5 cos ( = 1 1 arctan sin ( + C, u = sin (, 5 cos ( + C, u = cos ( L17 ( = = ln C, u = + 1, 4 + = 8 ln C, u =, 1 + = ln C, u = + 4 = ( 4 arctan + C, u =, + a, u =, =, u + a = (u a + u = + a a a + u = u a a a + u = u a a + u + a = ( a arctan + C, u =, a ( = = = 18 ( + 54 arctan =

25 Unbestimmtes Integral: Lösungen L = 5 6 5/6 + + C = C, = 6 7 ( + 17/6 + C = 6 7 ( C, u = + 1, = 5 ( + 55/4 + C = 5 ( C, u = = 4 7 ( + 7/6 9 ( + / + C = = 4 7 ( ( + + C, u = +, + 1 = ln C u = = arctan + C, u = L19 I 1 = + = 15 ( + / ( 4 + C, + 5 = 15 ( + 5/ ( 1 + C, 5 = 1 1 ( 5/ ( C.

26 Unbestimmtes Integral: Lösungen 4.4. Partielle Integration L ln = ln ln + 6 ln 6 + C = = (ln ln + 6 ln 6 + C u = ln, u = v = 1, v = ln = (ln ln + + C, aus dem Beispiel zur Aufgabe ln 4 = (ln 4 4 ln + 1 ln 4 ln C, u = ln 4, v = 1 ln 5 = (ln 5 5 ln 4 + ln 6 ln + 1 ln 1 + C, u = ln 5, v = 1 ln = 9 ( ln 1 + C, u = ln, v =, ln = 4 16 (4 ln 1 + C, u = ln, v =, ln ( = c I 1 = = ( + 1 ln = ( + 1 ln = ln + ln + ln = ln ( ln = ln + ln = ( C, ( = ln C, ( 1 ln ( + 1 = ( + 1 ln ( + 1 ( C d I 1 = ln( = 1 ln = 4 4/ / ( ln + C = 1 ( ln + C ( ln 4 5 ln = 5 7 7/5 ( ln C = 4 ( ln + C 4 + C = ( ln 5 + C 7

27 Unbestimmtes Integral: Lösungen L1 I 1 = ln = 1 ln + C, ln = 1 (ln C, ln = 1 ( ln C 4 L I 1 = e = e ( + + C, e = e ( C, 4 e = e ( C L I 4 = I 4 = sin = sin cos + C, cos = cos + sin + C, sin ( = 1 4 sin ( cos ( = 1 4 cos ( + cos ( + C, sin ( + C sin = cos + cos + sin + C, cos = sin sin + cos + C, cos( = 1 ( sin ( 1 + cos ( + C, sin ( = 1 ( cos ( + + sin ( + C 9 9

28 Anwenngen der Integralrechnung 5. Anwenngen der Integralrechnung 5.1. Flächen zwischen Kurven: Lösungen Berechnen Sie die von den Kurven f ( und g( eingeschlossene Fläche L4 f ( =, g( = f ( = g( : = = 4 1 =, =, A = ( ( = (4 = 1.67 Fig. 1: Die Fläche zwischen der Parabel f ( = und der Geraden g( =

29 Anwenngen der Integralrechnung L5 f ( =, g( = f ( = g( : = = 1 = 1, =, A = ( + = Fig. : Die Fläche zwischen der Parabel f ( = und der Geraden g( =

30 Anwenngen der Integralrechnung L6 f ( =, g( = f ( = g( : = = ( = 1 =, =, A = (g( f ( = ( = 9 = 4.5 Fig. : Die Fläche zwischen der Parabel f ( = und der Geraden g( =

31 Anwenngen der Integralrechnung L7 f ( =, g( =, [, ] f ( = g( : = = 4 1 =, =, 1 I, I, I 1 + I, I 1 = [, ], [, ] A = A 1 + A = ( ( + + ( 4 = + 7 = 1 ( = (4 + Fig. 4: Die Fläche zwischen der Parabel f ( = und der Geraden g( = im Intervall [, ]

32 Anwenngen der Integralrechnung L8 f ( =, g( =, [ 1, ] f ( = g( : = = ( = 1 =, =, 1 I, I, I 1 + I, I 1 = [ 1, ], [, ] A = 1 ( f ( g( + = = (g( f ( = 1 ( + ( = Fig. 5: Die Fläche zwischen der Parabel f ( = und der Geraden g( = im Intervall [ 1, ]

33 Anwenngen der Integralrechnung L9 f ( =, g( = + 1, [ 1, ] f ( = g( : = = 1 =., =., 1 I, I, I 1 + I, I 1 = [ 1,.], [., ]. A = ( f ( g( +. (g( f ( = ( = (?? (1 + = Fig. 6: Die Fläche zwischen der Parabel f ( = und der Geraden g( = + 1 im Intervall [ 1, ]

34 Bibliography [1] B.P. Demidovich, Aufgaben zur mathematischen Analysis (auf russisch. [] Regina Gellrich, Carsten Gellrich, Mathematik Ein Lehr- und Übungsbuch, Band. [] Vasili P. Minorski, Aufgabensammlung der höheren Mathematik [4] George B. Thomas, Ross L. Finney, Calculus and Analytic Geometry (auf englisch [5] W.W. Wawilow, I.I. Mel nikow, S.N. Olechnik, P.I. Pasitschenko, Aufgaben zur Mathematik (auf russisch.