Exercicis resolts 2 (2013)

Otro Español
Universidad Universidad Politécnica de Cataluña (UPC)
Grado Ingeniería de Recursos Energéticos y Mineros - 1º curso
Asignatura Mates I
Año del apunte 2013
Páginas 25
Fecha de subida 22/03/2015
Descargas 0
Subido por

Descripción

Taula de derivades i integrals immediates.

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TEMA 5 INTEGRAL DE RIEMANN PER FUNCIONS REALS DE VARIABLE REAL (Exercicis resolts) 1. Calculeu les seg¨ uents primitives, directament o fent un canvi de variable: ∫ ∫ √ ∫ 3x 2 √ (3x + 1) dx x x2 + 1 dx dx x2 + 1 ∫ 3 ∫ ∫ x −1 sin x dx sin x cos x dx dx x−1 cos2 x ∫ ∫ √ ∫ arcsin x arctg x ln x dx dx dx x 1 − x2 1 + x2 ∫ ∫ ∫ dx x dx tg 2x dx 2 x + 10 x ln x ∫ √ ∫ ∫ x+1 3x2 + 2x − 5 x+1 dx dx dx 3 2 x x + x − 5x + 1 x2 − 1 ∫ ∫ ∫ 2 sin x dx 2x + 3 arcsin x x −1 √ dx dx 2 x−1 3 − 2 cos x 1−x ∫ ∫ ∫ 2 2 x3 x −x x e dx (e − e ) dx x e−2x +5 dx ∫ ∫ ∫ 2x dx 1 + 2x e2arctg x dx 1 + x2 √ e x √ dx x ∫ ∫ (2x + 1)e ∫ ∫ ∫ ∫ e1/x dx x2 cos x esin x dx ∫ 2 x cos(x + 1) dx ∫ 1 + cos 2x dx sin2 2x dx cos2 (2x + 1) dx 4x2 + 1 e2−x dx dx ∫ sin(3x − 1) dx ∫ ∫ x2 +x sec x tg x dx ∫ 2 ex sin ex dx cos x dx ∫ ∫ ∫ 2x cos dx 3 dx 1 − cos2 4x dx 4 + x2 1 ∫ ∫ ∫ ex dx cos2 ex x3 dx 1 + x8 x2 dx 1 + 25x6 ∫ ∫ −4x √ dx 1 − 4x4 ∫ √ ∫ cos x 3 − sin x 2 dx ∫ x + 2x3 dx 1 + x4 ex √ dx 1 − e2x x+1 dx x2 + 1 ∫ dx dx − 2x + 2 x2 Soluci´ o: ∫ • 1 (3x + 1) dx = 3 ∫ 2 ∫ • ∫ • ∫ • (3x + 1)3 1 (3x + 1)3 +C = +C 3(3x + 1) dx = · 3 3 9 2 ∫ √ √ 1 1 2 x x + 1 dx = 2x x2 + 1 dx = 2 2 √ (x2 + 1)3 3 2 x3 − 1 dx = x−1 ∫ (x2 + x + 1) dx = ∫ sin x cos x dx = ∫ • = ∫ • sin x dx = cos2 x ∫ • ∫ • (x2 + 1)3 +C 3 x3 x2 + +x+C 3 2 x3 − 1 = x2 + x + 1 x−1 sin2 x +C 2 ∫ −2 sin x(cos x) ∫ dx = − −2 − sin x(cos x) (cos x)−2+1 dx = − +C = −2 + 1 1 +C cos x ln x dx = x ∫ √ • +C = √ 3x √ dx = 3 x2 + 1 + C x2 + 1 Fem la divisi´o entre els polinomis i obtenim: • √ ∫ 1 ln2 x ln x· dx = +C x 2 arcsin x dx = 1 − x2 ∫ 1 (arcsin x) 2 2√ (arcsin x) · √ dx = + C = (arcsin x)3 + C 3 3 1 − x2 2 3 1 2 arctg x arctg 2 x dx = +C 1 + x2 2 1 x dx = x2 + 10 2 ∫ 2x 1 dx = ln(x2 + 10) + C 2 x + 10 2 2 ∫ • ∫ sin 2x 1 dx = − cos 2x 2 tg 2x dx = ∫ • dx = x ln x ∫ ∫ −2 sin 2x 1 dx = − ln(cos 2x) + C cos 2x 2 1 1 · dx = ln(ln x) + C ln x x ) ) ∫ √ ∫ (√ ∫ ( √ 1 1 x+1 x 1 √ + • dx = + dx = dx = 2 x + ln x + C x x x x x ∫ • ∫ • ∫ • ∫ • ∫ • ∫ 3x2 + 2x − 5 dx = ln(x3 + x2 − 5x + 1) + C x3 + x2 − 5x + 1 x+1 dx = x2 − 1 ∫ 2x + 3 arcsin x √ dx = 1 − x2 x2 − 1 dx = x−1 ∫ • (x + 1) dx = 1 x e dx = 3 • ∫ √ arcsin x arcsin2 x √ +C dx = −2 1 − x2 +3 2 1 − x2 (x + 1)2 +C 2 ∫ 1 3 3 3x2 ex = ex + C 3 (ex − e−x ) dx = ex + e−x + C xe ∫ 2x √ dx+3 1 − x2 sin x dx 1 = ln(3 − 2 cos x) + C 3 − 2 cos x 2 ∫ • ∫ ∫ 2 x3 • 1 dx = ln(x − 1) + C x−1 −2x2 +5 1 dx = − 4 2x 1 dx = x 1+2 ln 2 ∫ ∫ −4xe−2x 2 +5 1 2 dx = − e−2x +5 + C 4 2x ln 2 1 dx = ln(1 + 2x ) + C x 1+2 ln 2 ∫ 2 +x • (2x + 1)ex ∫ • • 2 +x +C ∫ e ∫ dx = ex 2−x dx = − e2arctg x 1 dx = 2 1+x 2 −e2−x = −e2−x + C ∫ 2 1 ·e2arctg x dx = e2arctg x + C 2 1+x 2 3 ∫ • 1 e1/x dx = −e x + C 2 x ∫ • cos x esin x dx = esin x + C ∫ • ∫ • √ e x √ dx = 2 x ∫ √ 1 √ √ e x dx = 2e x + C 2 x 1 x cos(x + 1) dx = 2 ∫ 2 ∫ • ∫ 1 sin x · dx = cos x cos x sec x tg x dx = ∫ • 1 sin(3x − 1) dx = − 3 ∫ • ∫ 2x cos(x2 + 1)dx = ∫ cos x dx = sin x 1 = +C 2 cos x cos x 1 −3 sin(3x − 1) = − cos(3x − 1) + C 3 1 + cos 2x 1 dx = 2 2 2 ∫ 1 sin(x2 + 1) + C 2 ( ) sin 2x x+ +C 2 ∫ • ex sin ex dx = − cos ex + C ∫ • ∫ • ∫ • ∫ • ∫ • ∫ • ∫ • 1 + cos 2x dx = sin2 2x ∫ ∫ 2x 3 cos dx = 3 2 ∫ 1 + sin2 2x cos 2x 1 1 1 dx = − cotg 2x − +C 2 2 2 sin 2x sin 2x 2 2x 3 2x cos dx = sin +C 3 3 2 3 ex dx = tg ex + C cos2 ex dx 1 = 2 cos (2x + 1) 2 ∫ dx = 1 − cos2 4x x3 dx = 1 + x8 dx = 4x2 + 1 ∫ ∫ ∫ 2 cos2 (2x 1 dx = tg (2x + 1) + C + 1) 2 1 dx = 2 4 sin 4x ∫ 1 4 dx = − cotg 4x + C 2 4 sin 4x x3 1 dx = 1 + [x4 ]2 4 1 dx = 2 1 + (2x) 2 ∫ ∫ 4x3 1 dx = arctg x4 + C 4 2 1 + [x ] 4 1 2 dx = arctg 2x + C 2 1 + (2x) 2 4 ∫ • ∫ • ∫ • ∫ • dx = 4 + x2 ∫ dx 1 )= ( x2 4 4 1+ 4 x2 dx = 1 + 25x6 ∫ −4x √ dx = 1 − 4x4 x + 2x3 dx = 1 + x4 1 1 ( x )2 dx = ·2 4 1+ √ ∫ 2 x2 1 dx = 3 2 1 + (5x ) 15 ∫ ∫ ∫ −4x 1 − (2x2 )2 ∫ x dx + 1 + x4 ∫ 1/2 1 x ( 1 )2 dx = arctg + C 2 2 1+ 2 15x2 1 dx = arctg 5x3 + C 3 2 1 + (5x ) 15 dx = − arcsin 2x2 + C 2x3 dx = 1 + x4 ∫ x 1 dx + ln(1 + x4 ) = 2 2 1 + (x ) 2 1 1 = arctg x2 + ln(1 + x4 ) + C 2 2 ∫ • x+1 dx = x2 + 1 ∫ x dx + 2 x +1 ∫ cos x √ dx = 3 − sin2 x • ∫ = √ √1 3 1− ∫ • ∫ • ∫ ∫ x2 cos x 1 √ ( dx = √ ) 2 3 3 1 − sin3 x cos x ( )2 dx = arcsin ( sin √x 3 ex √ dx = 1 − e2x ∫ dx dx = 2 x − 2x + 2 √ ∫ 1 1 dx = ln(x2 + 1) + arctg x + C +1 2 ex 1− (ex )2 sin x √ 3 ∫ ) √ cos x )2 dx = ( √x 1 − sin 3 +C dx = arcsin ex + C dx = arctg (x − 1) + C (x − 1)2 + 1 2. Calculeu les seg¨ uents primitives pel m`etode d’integraci´o per parts: 5 ∫ ∫ ln x dx arctg x dx ∫ ∫ 2 ∫ x cos x dx ∫ ln(ln x) dx x ln(x + 1) √ dx x+1 ∫ x arcsin x √ dx 1 − x2 ∫ sin2 x dx e cos x dx 2 ln x dx x ln x dx x e sin 2x dx ∫ ∫ ∫ 3x x e dx ∫ x sin x dx ∫ 2 x ∫ ∫ ∫ arcsin x dx x3 ln x dx ∫ cos(ln x) dx ∫ (x − 1)e dx 2 2x (x + 1) cos x dx Solucions: ∫ • (∗) ∫ ∫ ↓ 1 ln x dx = x ln x − x dx = x ln x − dx = x ln x − x + C x  1   u = ln x ⇒ du = dx x (∗)   dv = dx ⇒ v = x ∫ • (∗) ∫ ↓ arctg x dx = x·arctg x −    u = arctg x ⇒ du = (∗) ∫ •   dv = dx x 1 dx = x·arctg x − ln(1 + x2 ) + C 1 + x2 2 1 dx 1 + x2 ⇒ v=x (∗) ∫ ↓ x sin x dx = −x cos x + cos x dx = −x cos x + sin x + C   u=x (∗)  ⇒ du = dx dv = sin x dx ⇒ v = − cos x 6 ∫ • (∗) ∫ ↓ x2 1 x2 x2 x ln x dx = ln x − x dx = ln x − +C 2 2 2 4      u = ln x (∗) ∫ • ⇒ du = 1 dx x  2    dv = x dx ⇒ v = x 2 (∗) (∗∗) [ ] ∫ ∫ ↓ ↓ 2 x 2 x x 2 x x x x e dx = x e − 2 xe dx = x e − 2 xe − e = = x2 ex − 2xex + 2ex = (x2 − 2x + 2)ex + C   u = x2 (∗)  dv = ex dx ⇒ v = ex   u=x (∗∗) ⇒ du = 2x dx  ⇒ du = dx dv = ex dx ⇒ v = ex (∗) (∗∗) ∫ ↓ ↓ 1 1 3 e3x sin 2x dx = = − e3x cos 2x + e3x cos 2x dx = − e3x cos 2x + • 2 ( )2 ∫ 2 3 3 1 3x e sin 2x − e3x sin 2x dx + 2 2 2 ∫  3x   u=e (∗) ⇒ du = 3e3x dx   dv = sin 2x dx ⇒ v = − cos 2x 2  3x   u=e ⇒ du = 3e3x dx (∗∗)   dv = cos 2x dx ⇒ v = sin 2x 2 ∫ Si anomenem I = e3x sin 2x tenim: 1 3 9 I = − e3x cos 2x + e3x sin 2x − I 2 4 4 Aillant I: 7 2e3x I= 13 ∫ • ( ) 3 − cos 2x + sin 2x + C 2 (∗) (∗∗) ∫ ∫ ↓ ↓ x x x x x e cos x dx = e cos x + e sin x dx = e cos x + e sin x − ex cos x dx   u = cos x (∗)  ⇒ du = − sin x dx dv = ex dx ⇒ v = ex   u = sin x (∗∗)  ⇒ du = cos x dx dv = ex dx ⇒ v = ex ∫ Si anomenem I = ex cos x dx tenim: I = ex (cos x + sin x) − I I aillant I: ex (cos x + sin x) I= +C 2 ∫ • (∗) (∗∗) [ ] ∫ ∫ ↓ ↓ 1 2 2 2 ln x dx = x ln x − 2 ln x dx = x ln x − 2 x ln x − x dx = x ln2 x− x −2x ln x + 2x + C  1   u = ln2 x ⇒ du = 2 ln x· dx x (∗)   dv = dx ⇒ v = x  1   u = ln x ⇒ du = , dx x (∗∗)   dv = dx ⇒ v = x ∫ • (∗) (∗∗) [ ] ∫ ∫ ↓ ↓ 2 2 2 x cos x dx = x sin x−2 x sin x dx = x sin x−2 −x cos x + cos x dx = 8 = x2 sin x + 2x cos x − 2 sin x + C  ⇒ du = 2x dx  u = x2 (∗)  dv = cos x dx ⇒ v = sin x   u=x (∗∗) ∫ •  ⇒ du = dx dv = sin xdx ⇒ v = cos x (∗) ∫ √ ↓ x arcsin x dx = x· arcsin x − √ dx = x arcsin x + 1 − x2 +C 1 − x2  1   dx  u = arcsin x ⇒ du = √ 2 1 − x (∗)    dv = dx ⇒ v=x ∫ • (∗) ∫ ↓ x4 1 x4 x4 3 x ln x dx = ln x − x3 dx = ln x − +C 4 4 4 16      u = ln x 1 dx x (∗)  4    dv = x3 dx ⇒ v = x 4 ∫ (∗) ∫ ↓ ln(ln x) 1 dx = ln x· ln(ln x) − dx = ln x· ln(ln x) − ln x + C x x • (∗) ∫ • ⇒ du =  1 1   · dx u = ln(ln x) ⇒ du =   ln x x   1   dv = dx x ⇒ v = ln x (∗) ∫ √ √ ↓ x arcsin x 2 √ dx = − 1 − x · arcsin x + dx = − 1 − x2 · arcsin x + x + C 1 − x2 9    u = arcsin x   (∗) ∫ • 1 ⇒ du = √ dx 1 − x2  √ x    dv = √ dx ⇒ v = − 1 − x2 1 − x2 (∗) (∗∗) ∫ ∫ ↓ ↓ cos(ln x) dx = x· cos(ln x)+ sin(ln x) dx = x cos(ln x)+x sin(ln x)− cos(ln x) dx  1   u = cos(ln x) ⇒ du = − sin(ln x)· dx x (∗)   dv = dx ⇒ v=x  1   u = sin(ln x) ⇒ du = cos(ln x)· dx x (∗∗)   dv = dx ⇒ v=x ∫ Si I = cos(ln x) dx tenim: I = x[cos(ln x) + sin(ln x)] − I I, aillant I, I= x[cos(ln x) + sin(ln x)] +C 2 (∗) ∫ √ √ √ ↓ ln(x + 1) 1 √ • x + 1· dx = 2 x + 1 ln(x + 1)− dx = 2 x + 1· ln(x + 1) − 2 x+1 x+1 ∫ √ √ 1 −2 √ dx = 2 x + 1 ln(x + 1) − 4 x + 1 + C x+1 ∫    u = ln(x + 1)   (∗) ∫ • ⇒ du = 1 dx x+1  √ 1    dv = √ dx ⇒ v = 2 x + 1 x+1 (∗) (∗∗) [ ] ∫ ∫ ↓ 1 2 ↓ 1 2 x 2x 1 2 2x 2x 2x 2x 2x (x −1)e dx = (x −1)e − xe dx = (x −1)e − e − e dx = 2 2 2 2 =( x2 x 1 2x − − )e + C 2 2 4 10  2   u = x − 1 ⇒ du = 2x dx (∗)   dv = e2x dx ⇒ v = 1 e2x 2    u=x (∗∗) ∫ • ⇒ du = dx   dv = e2x dx ⇒ v = 1 e2x 2 (∗) ∫ ↓ (x + 1) cos x dx = (x + 1) sin x − sin x dx = (x + 1) sin x + cos x + C   u=x+1 (∗)  ⇒ du = dx dv = cos x dx ⇒ v = sin x uents integrals de funcions racionals: 3. Calculeu les seg¨ ∫ ∫ ∫ ∫ ∫ ∫ x5 dx x−2 x2 + 1 dx x3 − 4x2 + 5x − 2 x2 − x + 14 dx (x − 2)(x − 4)2 x+1 dx 2 x − 2x + 5 ∫ ∫ ∫ ∫ x5 + 2 dx x2 − 1 x+2 dx x2 − 4x + 3 3x3 + 5x2 + 2x dx (x − 1)3 2x + 3 dx 2 x + 2x + 2 2x − 1 dx (x + 1)2 (x − 2)2 x4 dx x3 − 1 ∫ ∫ ∫ ∫ ∫ x3 + x2 − x dx x2 − 1 x2 + 1 dx x3 − x 2x + 4 dx x2 + 16 x2 dx x4 − 1 1 dx x4 − 16 Solucions: ∫ • x5 dx x−2 Fem pr`eviament la divisi´o: 32 x5 = x4 + 2x3 + 4x2 + 8x + 16 + x−2 x−2 11 ∫ Llavors, ∫ • x5 x4 4 3 x5 dx = + + x + 4x2 + 16x + 32 ln(x − 2) + C x−2 5 2 3 x+2 dx x2 − 4x + 3 Descomposem Q(x) = x2 − 4x + 3 en factors: x2 − 4x + 3 = (x − 1)(x − 3).
Fem la descomposici´o en fraccions simples: x2 A B x+2 = + − 4x + 3 x−1 x−3 3 5 i B = . Llavors, 2 2 ∫ ∫ ∫ x+2 3 1 5 1 3 5 dx = − dx + dx = − ln(x − 1) + ln(x − 3) + C 2 x − 4x + 3 2 x−1 2 x−3 2 2 Tenim A = − ∫ • x3 + x2 − x dx x2 − 1 Fem pr`eviament la divisi´o: x3 + x2 − x 1 =x+1+ 2 2 x −1 x −1 ∫ 3 ∫ ∫ ∫ 2 x +x −x 1 (x + 1)2 1 Llavors: dx = (x + 1) dx + dx = + dx x2 − 1 x2 − 1 2 x2 − 1 ∫ 1 1 dx descomposem en fraccions simples tenint en compte que x2 − 1 x2 − 1 x2 − 1 = (x − 1)(x + 1).
Per fer Tenim x2 ∫ Llavors x2 ∫ ∫ • 1 1/2 1/2 = − −1 x−1 x−1 1 1 1 dx = ln(x − 1) − ln(x + 1) + C i, −1 2 2 x3 + x2 − x (x + 1)2 1 1 dx = + ln(x − 1) − ln(x + 1) + C 2 x −1 2 2 2 x2 + 1 dx x3 − 4x2 + 5x − 2 Descomposem Q(x) = x3 − 4x2 + 5x − 2 en factors. Tenim Q(x) = (x − 2)(x − 1)2 12 La descomposici´o en fraccions simples ´es: x2 + 1 A B C = + + x3 − 4x2 + 5x − 2 x − 2 x − 1 (x − 1)2 , on A = 5 , B = −4 i C = −2 Llavors, ∫ ∫ • x2 + 1 2 dx = 5 ln(x − 2) − 4 ln(x − 1) + +C 3 2 x − 4x + 5x − 2 x−1 3x3 + 5x2 + 2x dx (x − 1)3 Fem pr`eviament la divisi´o i obtenim 3x3 + 5x2 + 2x 14x2 − 7x + 3 = 3 + (x − 1)3 (x − 1)3 Llavors, ∫ 3x3 + 5x2 + 2x dx = (x − 1)3 ∫ ∫ 3 dx + 14x2 − 7x + 3 dx = 3x + (x − 1)3 ∫ 14x2 − 7x + 3 dx (x − 1)3 Per fer aquesta nova integral farem descomposici´o en fraccions simples: A B C 14x2 − 7x + 3 = + + (x − 1)3 x − 1 (x − 1)2 (x − 1)3 on A = 14, B = 21, C = 10 Llavors ∫ 14x2 − 7x + 3 dx = 14 (x − 1)3 = 14 ln(x − 1) − ∫ 1 dx + 21 x−1 ∫ 1 + 10 (x − 1)2 ∫ 1 dx = (x − 1)3 5 21 − +C x − 1 (x − 1)2 i, finalment: ∫ 21 5 3x3 + 5x2 + 2x dx = 3x + 14 ln(x − 1) − − +C 3 (x − 1) x − 1 (x − 1)2 13 ∫ • x2 + 1 dx x3 − x Q(x) = x3 − x = x(x − 1)(x + 1). La descomposici´o en fraccions simples ´es: x2 + 1 A B C = + + on A = −1, B = 1, C = 1 3 x −x x x−1 x+1 Llavors, ∫ ∫ • x2 + 1 dx = − x3 − x ∫ 1 dx+ x ∫ 1 dx+ x−1 ∫ 1 dx = − ln x+ln(x−1)+ln(x+1)+C x+1 x2 − x + 14 dx (x − 2)(x − 4)2 La descomposici´o en fraccions simples ´es x2 − x + 14 A B C = + + 2 (x − 2)(x − 4) x − 2 x − 4 (x − 4)2 on A = 4, B = −3 i C = 13.
Llavors, ∫ x2 − x + 14 dx = 4 (x − 2)(x − 4)2 ∫ = 4 ln(x − 2) − 3 ln(x − 4) − ∫ • x2 1 dx − 3 x−2 ∫ 1 dx + 13 x−4 ∫ 1 dx = (x − 4)2 13 +C x−4 2x + 3 dx + 2x + 2 Observem que Q(x) = x2 + 2x + 2 no t´e arrels reals. En aquest cas el podem escriure de la forma Q(x) = (x + 1)2 + 1.
∫ 2x + 3 dx(∗) = (x + 1)2 + 1 ∫ 2t + 1 dt = 1 + t2 ∫ 2t dt + 1 + t2 ∫ 1 dt = ln(1 + t2 )+ 1 + t2 +arctg t + C = ln(1 + (x + 1)2 ) + arctg (x + 1) + C = ln(x2 + 2x + 2) + arctg (x + 1) + C (∗) Fem el canvi de variable: { x + 1 = t (⇔ x = t − 1) dx = dt 14 ∫ • 2x + 4 dx = x2 + 16 ∫ • x2 ∫ 2x dx + 2 x + 16 ∫ x2 4 x = ln(x2 + 16) + arctg ( ) + C + 16 4 x+1 dx − 2x + 5 Q(x) = x2 − 2x + 5 no t´e arrels reals. El podem escriure com Q(x) = (x − 1)2 + 4.
∫ = x+1 dx = x2 − 2x + 5 ∫ x+1 dx(∗) = (x − 1)2 + 4 ∫ t+2 dt = t2 + 4 ∫ t dt+2 t2 + 4 ∫ 1 dt = t2 + 4 1 t 1 x−1 ln(t2 + 4) + arctg + C = ln(x2 − 2x + 5) + arctg ( )+C 2 2 2 2 (∗) Fem el canvi de variable { x − 1 = t (⇔ x = t + 1) dx = dt ∫ • 2x − 1 dx (x + 1)2 (x − 2)2 La descomposci´o en fraccions simples ´es: 2x − 1 A B C D = + + + 2 2 2 (x + 1) (x − 2) x + 1 (x + 1) x − 2 (x − 2)2 1 1 on A = 0, B = − , C = 0 i D = 3 3 LLavors, ∫ ∫ • 2x − 1 −1 dx = 2 2 (x + 1) (x − 2) 3 ∫ 1 1 dx+ 2 (x + 1) 3 ∫ 1 1 1 1 1 dx = − +C 2 (x − 2) 3x+1 3x−2 x2 dx x4 − 1 Q(x) = (x4 − 1) = (x2 + 1)(x + 1)(x − 1). La descomposici´o en fraccions simples ´es: Ax + B C D x2 = + + x4 − 1 x2 + 1 x−1 x+1 1 1 −1 on A = 0, B = , C = i D = 2 4 4 Llavors, 15 ∫ x2 1 dx = 4 x −1 2 ∫ 1 1 dx + 2 1+x 4 ∫ 1 1 dx − x−1 4 ∫ 1 1 dx = arctg x+ x+1 2 1 1 + ln(x − 1) − ln(x + 1) + C 4 4 ∫ • x5 + 2 dx x2 − 1 Fem pr`eviament la divisi´o i obtenim: x5 + 2 x+2 = x3 + x + 2 2 x −1 x −1 Llavors, ∫ x5 + 2 dx = x2 − 1 ∫ ∫ 3 (x + x) dx + x+2 x4 x2 dx = + + x2 − 1 4 2 ∫ x+2 dx x2 − 1 Per fer la nova integral fem descomposici´o en fraccions simples, tenint en compte que x2 − 1 = (x − 1)(x + 1).
Tenim doncs, x+2 A B = + 2 x −1 x−1 x+1 on A = ∫ 3 −1 iB= . Llavors, 2 2 x+2 3 dx = 2 x −1 2 ∫ 1 1 dx − x−1 2 ∫ 1 3 1 dx = ln(x − 1) − ln(x + 1) + C x+1 2 2 i, i, finalment, ∫ ∫ • x4 x2 3 1 x5 + 2 dx = + + ln(x − 1) − ln(x + 1) + C 2 x −1 4 2 2 2 x4 dx x3 − 1 Fem pr`eviament la divisi´o i tenim: 16 x4 x =x+ 3 3 x −1 x −1 A partir d’aqu´ı: ∫ x4 dx = x3 − 1 ∫ ∫ x dx + x x2 dx = + x3 − 1 2 ∫ x3 x dx −1 Tenint en compte que x3 − 1 = (x − 1)(x2 + x + 1), fem descomposici´o en fraccions simples a la segona integral.
x3 x A Bx + C = + 2 −1 x−1 x +x+1 1 −1 1 on, A = , B = iC= .
3 3 3 ∫ ∫ ∫ ∫ 1 x−1 x−1 x 1 1 1 1 dx = dx− dx = ln(x−1)− ( )2 3 2 x −1 3 x−1 3 x +x+1 3 3 x + 12 + 1 1 (∗) = ln(x−1)− 3 3 ∫ 3 4 ( ) √ ( ) t − 32 3 2t 1 1 3 2 dt = ln(x−1)− ln t + + arctg √ +C = 3 6 4 3 t2 + 43 3 √ ( ) 3 2x + 1 1 1 2 √ = ln(x − 1) − ln(x + x + 1) + arctg +C 3 6 3 3 { x + 12 = t (∗) Hem fet el canvi de variable dx = dt Finalment, ∫ ∫ • dx = √ ( ) x4 x2 1 1 3 2x + 1 2 √ dx = + ln(x − 1) − ln(x + x + 1) + arctg +C x3 − 1 2 3 6 3 3 1 dx x4 − 16 Q(x) = x4 − 16 = (x2 + 4)(x − 2)(x + 2) La descomposici´o en fracions simples ´es: Ax + B C D 1 = 2 + + − 16 x +4 x−2 x+2 1 −1 −1 ,C= iD= on, A = 0, B = 8 32 32 x4 17 Llavors, ∫ ∫ ∫ ∫ (x) 1 1 1 1 1 1 1 1 dx = − dx+ dx− dx = − arctg + x4 − 16 8 x2 + 4 32 x−2 32 x+2 16 2 + 1 1 ln(x − 2) − ln(x + 2) + C 32 32 uents integrals trigonom`etriques: 4. Calculeu les seg¨ ∫ ∫ ∫ 3 3 sin x dx cos 2x dx sin2 3x dx ∫ ∫ 4 ∫ 2 cos x dx cos4 2x dx sin 4x dx Soluci´ o: ∫ ∫ ∫ ∫ 3 2 2 • sin x dx = sin x· sin x dx = (1 − cos x) sin x dx = sin x dx− ∫ − cos2 x· sin x dx = − cos x + ∫ • ∫ ∫ 3 2 cos 2x dx = cos 2x· cos 2x dx = ∫ ∫ cos 2x dx − = ∫ • cos3 x +C 3 ∫ 2 sin 3x dx = sin2 2x· cos 2x dx = 1 − cos 6x 1 dx = 2 2 ∫ (1 − sin2 2x) cos 2x dx = 1 1 sin 2x − sin3 2x + C 2 6 1 (1 − cos 6x) dx == 2 ( ) sin 6x x− +C 6 )2 ∫ ) 1 + cos 2x 1 ( • cos x dx = (cos x) dx = dx = 1 + 2 cos 2x + cos2 2x dx = 2 4 ) ( [ ( )] ∫ 1 1 1 + cos 4x 1 sin 4x = dx = x + sin 2x + x + sin 2x + x+ = 4 2 4 2 4 ∫ 1 = 4 ∫ 2 sin 4x dx = ∫ • 2 2 ) ) ( ( 1 1 3 1 1 x + sin 2x + sin 4x + C x + sin 2x + x + sin 4x = 2 8 4 2 8 ∫ • ∫ ( ∫ 4 1 1 − cos 8x dx = 2 2 ∫ ( ∫ 4 cos 2x dx = ( ) sin 8x x− +C 8 2 2 (cos x) dx = 1 + cos 4x 2 18 )2 1 dx = 4 ∫ (1+2 cos 4x+cos2 4x) dx = [ ] ( ( )) ∫ 1 1 1 + cos 8x 1 1 1 sin 8x = x + sin 4x + dx = x + sin 4x + x+ +C = 4 2 2 4 2 2 8 ( ) 1 1 3x 1 + sin 4x + sin 8x + C = 4 2 2 16 5. Calculeu les seg¨ uents integrals impr`opies.
∫ 1 ∫ +∞ ∫ +∞ dx dx dx x x2 −1 x 0 0 ∫ +∞ −1 ∫ 1 0 ∫ 1 −1 ∫ 1 ∫ dx x5 −3x e ∫ ∫ dx √ 1 − x2 0 ∫ dx x ln x 1 x dx √ x +∞ dx √ x sin x dx ∫ √ x e− √ 0 ∫ ∫ −∞ +∞ +∞ 1 +∞ dx −∞ ln x √ dx x +∞ +∞ ∫ 2 +∞ dx −∞ ∫ xdx 1 + x4 1 ln x dx 0 ( ) ∫ +∞ ∫ +∞ 1 dx −x x ln dx e cos x dx 2 x x +x−2 0 2 +∞ xe −x ∫ +∞ dx 0 0 ∫ x dx 1+x e−t dt √ 1 − e−t +∞ 0 Soluci´ o: ∫ 1 ∫ t ∫ 1 dx 1 1 • = lim− dx + lim+ dx = lim− [ln |x|]t−1 + lim+ [ln |x|]1z = t→0 z→0 t→0 z→0 x x x −1 −1 z lim ln |t| − lim+ |z| t→0− ∫ +∞ • 0 divergent z→0 ∫ dx = lim x t → 0+ m t m → +∞ 1 dx = lim x t → 0+ [ln |x|]tm = m → +∞ lim t → 0+ m → +∞ ln |m| − ln |t| = = +∞ ∫ +∞ • 0 ∫ dx = lim x2 t → 0+ t m → +∞ ∫ +∞ • 1 ∫ • +∞ −1 dx √ = lim x m→+∞ dx = lim− t→0 x5 ∫ ∫ t −1 1 dx = lim x2 t → 0+ [ m → +∞ +∞ 1 m 1 − x ]t = m lim t→0 m → +∞ + 1 1 − = +∞ t m [ √ ]m √ 1 √ dx = lim 2 x 1 = lim 2 m − 2 = +∞ m→+∞ m→+∞ x ∫ 1 dx+ lim x5 z → 0+ z m → +∞ 19 m ]t [ 1 1 + lim dx = lim− − 4 t→0 x5 4x −1 z → 0+ m → +∞ [ 1 − 4 4x ]m = z  [ ] [ 1 1 =−  lim −1 + lim 4 t→0− t4 z → 0+  ] 1 1   − m4 z 4  divergent m → +∞ ∫ • +∞ e −3x ∫ dx = −∞ =− m lim −3x e m → +∞ n → −∞ n ]m [ 1 −3x dx = lim − e = 3 m → +∞ n n → −∞ [ −3m ] 1 lim e − e−3n = +∞ 3 m → +∞ n → −∞ ∫ • +∞ sin x dx = −∞ ∫ +∞ 2 1 • 0 m → +∞ n → −∞ dx √ = lim x m→+∞ • ∫ lim ln x √ dx = lim t→0+ x ∫ +∞ 2 ∫ t 1 [cos x]m n = lim m → +∞ n → −∞ [cos m − cos n] divergent √ [ √ ]m √ 1 √ dx = lim 2 x 2 = lim 2 m − 2 2 = +∞ m→+∞ m→+∞ x ln x √ dx x Calculem la primitiva aplicant integraci´o per parts:  1   u = ln x ⇒ du = dx   x  √ 1    dv = √ dx ⇒ v = 2 x x Llavors: ∫ lim+ t→0 t 1 ∫ √ ln x √ dx = 2 x ln x − 2 x ∫ √ √ 1 √ dx = 2 x ln x − 4 x + C x ( √ √) [ √ √ ]1 ln x √ dx = lim 2 x ln x − 4 x t = lim −2 t ln t − 4 + 4 t = −4 t→0+ t→0+ x Calculem el 1r l´ımit aplicant la regla de l’Hˆopital: 0 · ∞Ind √ ↓ ln t lim+ t ln t = lim+ −1/2 t→0 t→0 t ∞ Ind ∞ ↓ = = lim+ t→0 20 1 t −1 −3/2 t 2 = lim+ −2t1/2 = 0 t→0 ∫ +∞ • 0 ∫ √ e− x √ dx = lim x t → 0+ √ t m → +∞ [ = ∫ • lim t → 0+ m → +∞ +∞ −∞ = −2e− √ ∫ xdx = lim 1 + x4 m → +∞ √ −2e− x ]m = t m → +∞ + 2e− m [ e− x √ dx = lim x t → 0+ m √ t m n n → −∞ ] =2 ] xdx 1[ 2 m = lim arctg x = n 1 + x4 m → +∞ 2 n → −∞ [ ] 1 lim arctg m2 − arctg n2 = 0 2 m → +∞ n → −∞ ∫ ∫ 1 • ln x dx = lim+ ln x dx t→0 0 1 t Calculem la primitiva aplicant integraci´o per parts:  1   u = ln x ⇒ du = dx x   dv = dx ⇒ v = x ∫ ∫ ln x dx = x ln x − dx + C = x ln x − x + C Llavors: ∫ lim+ t→0 1 t ln x dx = lim+ [x ln x − x]1t = lim+ −1 − t ln t + t = −1 t→0 t→0 ja que: 0 · ∞Ind ↓ ln t lim+ t ln t = = lim+ 1 t→0 ∫ • 1 −1 t→0 dx √ = lim 1 − x2 t → −1+ z → 1− ∞ Ind ∞ ↓ = t ∫ z −t √ lim t→0+ 1 t −1 t2 = lim+ −t = 0 t→0 1 dx = lim [arcsin x]zt = 2 1−x t → −1+ z → 1− 21 arcsin z − arcsin t = π lim t → −1+ z → 1− = ∫ 1 • 0 ( ) ( ) ∫ 1 1 1 x ln dx = lim+ x ln dx t→0 x x t Fem la primitiva per parts: ( )  −1 1    ⇒ du = dx u = ln   x x   2    dv = x dx ⇒ v = x 2 ( ) ( ) ( ) ∫ ∫ 1 x2 1 1 x2 1 x2 x ln dx = ln + x dx = ln + +C x 2 x 2 2 x 4 Llavors, ∫ 1 lim t→0+ t ( ) [ ( ) ]1 ( ( ) ) 1 t2 1 1 x2 1 x2 t2 1 x ln dx = lim+ ln + = lim+ − ln − = t→0 t→0 x 2 x 4 4 2 t 4 4 t Ja que, ( ) 1 2 lim+ t ln t→0 t ∫ +∞ • e 0 −x 0 · ∞Ind ↓ ln t = − lim+ −2 t→0 t ∫ cos x dx = lim m→+∞ m ∞ Ind ∞ ↓ = − lim+ t→0 1 t −2t−3 ∞ Ind ∞ ↓ = = 1 lim t2 = 0 2 t→0+ e−x cos x dx 0 Calculem la primitiva aplicant integraci´o per parts: ∫ (∗) (∗∗) ∫ ↓ ↓ e−x cos x dx = −e−x cos x − e−x sin x dx = −e−x cos x− [ ] ∫ ∫ −x −x −x −x − −e sin x + e cos x dx = −e cos x + e sin x − e−x cos x dx.
22   u = cos x ⇒ du = − sin x dx (∗)  dv = e−x dx ⇒ v = −e−x   u = sin x ⇒ du = cos x dx (∗∗)  dv = e−x dx ⇒ v = −e−x ∫ Si I= e−x cos x dx, tenim: I = e−x (sin x − cos x) − I. Llavors 2I = e−x (sin x − cos x) i I= e−x (sin x − cos x) +C 2 Per tant, ∫ m lim m→+∞ = ∫ [ −x e cos x dx = lim m→+∞ 0 e−x (sin x − cos x) 2 ]m = 0 [ ] 1 lim e−m (sin m − cos m) + 1 = 2 m→+∞ 1 2 +∞ • x2 2 dx +x−2 Q(x) = x2 + x − 2 = 0 ⇒ x = −2 ∈ (2 + ∞) i x = 1 ∈ / (2 + ∞) Per tant, ∫ +∞ 2 Descomposem Llavors x2 m 2 ∫ 2 m x2 1 dx +x−2 1 en fraccions simples per tal de fer la primitiva: +x−2 A B 1 −1 1 = + on A = i B = x2 + x − 2 x−1 x+2 3 3 ∫ x2 ∫ dx = lim 2 x + x − 2 m→+∞ 1 1 dx = [ln |x − 1| − ln |x + 2|] + C +x−2 3 1 1 dx = lim [ln(x − 1) − ln(x + 2)]m 2 = x2 + x − 2 3 m→+∞ 23 [ ] 1 1 m−1 1 = lim [ln(m − 1) − ln(m + 2) + ln 4] = lim ln + ln 4 = ln 4 3 m→+∞ 3 m→+∞ m+2 3 ∫ +∞ • 1 ∫ dx = lim x ln x z → 1+ m z m → +∞ = ∫ • xe −x [ln(ln x)]m z = m → +∞ (ln(ln m) − ln(ln z)) = +∞ lim z → 1+ m → +∞ +∞ 1 dx = lim x ln x z → 1+ ∫ dx = 0 lim z→0 m → +∞ + m xe−x dx z Calculem la primitiva per parts:   u = x ⇒ du = dx (∗)  dv = e−x dx ⇒ v = −e−x ∫ xe −x dx = −xe −x ∫ + e−x dx = −xe−x − e−x + C Llavors, ∫ m lim z → 0+ m → +∞ = xe−x dx = z lim z → 0+ m → +∞ ∫ +∞ • 0 z → 0+ m → +∞ [ ]m −xe−x − e−x z = [ ] −me−m − e−m + ze−z + e−z = 1 Ja que lim me−m m→+∞ lim 0 · ∞Ind ↓ = = lim x dx = lim m→+∞ 1+x ∞ Ind ∞ m m→+∞ em ∫ 0 m x dx = lim m→+∞ 1+x ↓ = ∫ 0 1 =0 m→+∞ em lim m ( 1− 1 x+1 ) dx = = lim [x − ln |x + 1|]m 0 = lim [m − ln(m + 1)] = +∞ m→+∞ m→+∞ ( ) ln(m + 1) Ja que lim [m − ln(m + 1)] = lim m 1 − = +∞ m→+∞ m→+∞ m 24 ∫ +∞ • 0 e−t dt √ = lim 1 − e−t z → 0+ ∫ z m → +∞ =2 lim z → 0+ m → +∞ [√ 1 − e−m − m e−t dt √ = lim 1 − e−t z → 0+ [ √ ]m 2 1 − e−t = z m → +∞ √ ] 1 − e−z = 2 6. Exercicis per fer amb Maple.
Apliqueu les f´ormules del trapezi, de Simpson, de trapezis composta i Simpson composta amb 6 subintervals per aproximar les seg¨ uents integrals: ∫ 1 xex dx (a) ∫ (b) ∫ −1 1 4 −1 4 dx 1+x 3 ex sin xdx (c) 1 Compareu els resultats obtinguts amb el valor exacte de la integral.
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