Exercícios - Integrais - Cálculo 2 2021 2

laislufu.br sites.google.com/site/laislufu Laís Rodrigues Página 2 Universidade Federal de Uberlândia Integrais (6) Resolver as seguintes integrais utilizando a técnica de integração por partes: a) ∫ (1 - x) ln(x) dx b) ∫ x^e^x dx c) ∫ (x + 1) cos (2x) dx d) ∫ e^x cos (3) dx e) ∫ √x ln xdx f) ∫ x² sen (4x) dx g) ∫ x cos² xdx h) ∫ (x + 3)² e^x dx i) ∫ x⁵ ln x dx (7) Calcule as seguintes integrais: a) ∫ x² e⁻^x dx b) ∫ x² sen x dx c) ∫ (3x² + 1) e^x⁺¹ dx d) ∫ ln ( x ) dx ( x + 1 ) e) ∫ sen (2x) e^sen(2x) dx f) ∫ 3x⁴ + 1 dx jx² + x 3 + 1 h) ∫ x⁻² e⁻² dx f) ∫ x cos (4 ) dx 5 (8) Calcule as integrais utilizando o método de substituição trigonométrica: a) ∫ √9 - x² dx 2x - dx b) ∫ dx 1 √x²∙√x²-16 c) ∫ 4√x² dx (9) Calcule as integrais utilizando o método de funções racionais: a) ∫ 2x + 1 dx 2x² + 3x - 2 c) ∫ x dx x³ + 6x² + 5 dx b) ∫ x³ + x (10) Calcule as seguintes integrais a) ∫ ∞ (x - 2x) dx b) ∫ (3 + x√x) dx c) ∫ ∞⁻¹ dx √x - 1 dx d) ∫ sec x tg xdx e) ∫ √x⁺e dx -1 dx f) ∫ e^x² dx g) ∫ 3x dx 3 (3 - 5x)² h) ∫ xsenx dx -3 aislufu.br sites.google.com/site/laislufu lais@ufu.br Integras Universidade Federal de Uberlândia Página 3 (11) Encontre a área da região sombreada: 3 a) y = 5x - x² y = x (4, 4) e) y = x y = √x c) y = x² + 1 2 f) y = √x g) x = y² - 2 x = e^y y = -1 y = 1 c) ( x + 1) x = e² y = 2 y = √x + 2 x= 1 x + 1 d) x x - 3x (0) e) y = x aislufu br sites.google.com/site/laislufu Laís Rodrigues 6- Resolva as seguintes integrais utilizando a técnica de integração por partes: a) ∫ ln (1−x )dx u=1−x→du=−dx→dx=−du ∫ (ln (1−x ))dx=∫ (lnu) (−du)=∫ (−lnu)du u=−lnu→du=−1 u du dv=du→v=u ∫ (−lnu)du=−ln (u)⋅u−∫ u( −1 u )du=−u⋅lnu+ ∫ du=−u⋅lnu+u+c=−(1−x )⋅ln (1−x )+1ox+c b) ∫ x e 4 xdx u=x→du=dx dv=e 4 xdx→v=e 4 x 4 ∫ x e 4 xdx=x⋅ e 4 x 4 −∫( e 4 x 4 )dx= x e 4 x 4 −e 4 x 16 +c c) ∫ (x+1)cos (2 x )dx u=x+1→du=dx dv=cos2 x dx→v= sen (2 x ) 2 ∫ (x+1)cos2 x dx=(x+1)⋅ sen (2 x ) 2 −∫ sen (2 x ) 2 dx= (x+1) sen (2 x ) 2 −( −cos (2 x ) 4 )+c= (x+1) sen (2 x ) 2 + cos (2 x ) 4 +c d) ∫ e x⋅cos( x 2)dx u=cos( x 2)→du= −sen( x 2) 2 dx dv=e xdx→v=e x ∫ e x⋅cos( x 2)dx=cos( x 2)⋅e x−∫ e x(−sen( x 2)) 2 dx ∫ e x⋅cos( x 2)dx=cos( x 2)⋅e x+ 1 2∫e x sen( x 2)dx u=sen( x 2)→du= cos( x 2) 2 dx dv=e xdx→v=e x ∫ e x⋅cos( x 2)dx=cos( x 2)⋅e x+ 1 2[sen( x 2)⋅e x−∫e x⋅((cos( x 2)) 2 )dx] ∫ e x⋅cos( x 2)dx=e x xos( x 2)+ 1 2 ⋅[e x sen( x 2)−1 2∫e xcos( x 2)dx] ∫ e x⋅cos( x 2)dx=e xcos( x 2)+ e x sen( x 2) 2 − 1 4∫e xcos( x 2)dx 5 4∫e xcos( x 2)dx=e xcos( x 2)+ e x sen( x 2) 2 ∫e xcos( x 2)dx= 4 5 ⋅[e xcos( x 2)+ e x sen( x 2) 2 ] e) ∫ √x⋅ln ⁡( x)dx u=ln (x )→du= 1 x dx dv=√x dx→v=2 3 ⋅ x 3 2 ∫√x⋅ln (x )dx=ln (x )⋅ 2 3 x 3 2−∫ 2 3 x 3 2 ⋅ 1 x dx=2 x 3 2 ln (x ) 3 −2 3∫ x 1 2 dx=2 x 3 2 ln (x ) 3 − 4 9 x 3 2+c f) ∫ x 3⋅sen (4 x )dx u=x 3→du=3 x 2dx dv=sen (4 x )dx→v=−cos (4 x ) 4 ∫ x 3 sen (4 x )dx=x 3⋅( −cos (4 x ) 4 )−∫ −cos (4 x ) 4 ⋅3 x 2dx=−x 3cos (4 x ) 4 + 3 4∫ x 2cos (4 x )dx u=x 2→du=2 xdx dv=cos (4 x )→v= sen (4 x ) 4 ∫ x 3 sen (4 x )dx=−x 3cos (4 x ) 4 + 3 4[x 2⋅ sen (4 x ) 4 −∫ sen (4 x ) 4 ⋅2 xdx]=−x 3cos (4 x ) 4 + 3 4[ x 2sen (4 x ) 4 −1 2∫ xsen (4 x )dx]=−x 3cos (4 x ) 4 + 3 x 2sen (4 x ) 16 −3 8∫ xsen (4 x )dx u=x→du=dx dv=sen (4 x )dx→v=−cos (4 x ) 4 ∫ x 3 sen (4 x )dx=−x 3cos (4 x ) 4 + 3 x 2sen (4 x ) 16 −3 8[x⋅( −cos (4 x ) 4 )−∫ −cos (4 x ) 4 dx]=−x 3cos (4 x ) 4 + 3 x 2sen (4 x ) 16 −3 8[ −xcos (4 x ) 4 + 1 4∫cos (4 x )dx]=−x 3cos (4 x ) 4 + 3 x 2sen (4 x ) 16 + 3 xcos (4 x ) 4 − 3 32 ⋅ sen (4 x ) 4 +c ¿− x 3cos (4 x ) 4 + 3 x 2sen (4 x ) 16 + 3 xcos (4 x ) 4 −3 sen (4 x ) 128 +c g) ∫ x⋅cos 2 x dx u=x→du=dx dv=cos 2 x dx→v= x 2 + sen (2 x ) 4 ∫ xco s 2 (x )dx=x⋅( x 2 + sen (2 x ) 4 )−∫( x 2 + sen (2 x ) 4 )dx= x 2 2 + xsen (2 x ) 4 −( x 2 4 −cos (2 x ) 8 )+c= x 2 2 + xsen (2 x ) 4 − x 2 2 + cos (2 x ) 8 +c= xsen (2 x ) 4 + cos (2 x ) 8 +c h) ∫ (x+3) 2⋅e xdx u=(x+3) 2→du=2 (x+3)dx dv=e xdx→v=e x ¿ (x+3) 2⋅e x−∫e x⋅2 (x+3)dx=(x+3) 2e x−2∫ (x+3)e x dx u=x+3→du=dx dv=e x dx→v=e x ∫ (x+3) 2e x dx=(x+3) 2e x−2⋅[(x+3)⋅e x−∫e x dx]=(x+3) 2e x−2 (x+3)e x+4 e x+c=e x [(x+3) 2−2 x−6+4]+c=e x [(x+3) 2−2 x−2]+c i) ∫ 9 x 2⋅ln (x )dx u=ln (x )→du= 1 x dx dv=9 x 2→v=3 x 3 ∫9 x 2ln x dx=ln( x)⋅3 x 2−∫3 x 3⋅ 1 x dx=3 x 2ln (x )−∫3 x 2dx=3 x 2ln (x )−x 3+c j) ∫ x 5⋅ln (x )dx u=ln (x )→du= 1 x dx dv=x 5→v= x 6 6 ∫ x 5ln (x )dx=ln (x )⋅ x 6 6 −∫ x 6 6 ⋅ 1 x dx= x 6 ln (x ) 6 −∫ x 5 6 dx= x 6ln (x ) 6 − x 6 36 +c 7- Calcule as seguintes integrais: a) ∫ x⋅e 2 xdx u=x→du=dx dv=e 2 xdx→v=e 2 x 2 ∫ x e 2 xdx=x⋅ e 2 x 2 −∫ e 2 x 2 dx= x e 2 x 2 −e 2 x 4 +c b) ∫ x 2⋅sen (x )dx u=x 2→du=2 xdx dv=sen (x )dx→v=−cos ⁡( x) ∫ x 2sen (x )dx=x 2⋅(−cos (x ))−∫−cos (x )⋅2 xdx=−x 2cos (x )+2∫ xcos (x )dx u=x→du=dx dv=cos (x )dx→v=sen( x) ∫ x 2sen (x )dx=−x 2cos (x )+2[x⋅sen (x )−∫ sen (x )dx]=−x 2cos (x )+2 xsen (x )+2cos (x )+c c) ∫ (3 x 2+1)⋅e x 3+x+1dx u=x 3+x+1→du=(3 x 2+1)dx ∫ (3 x 2+1)e x 3+x+1dx=∫e udu=e u+c=e x 3+x+1+c d) ∫ ln( x x+1)dx u=ln( x x+1)→du= 1 x x+1 ⋅ 1 (x+1)−1⋅ x (x+1) 2 = x+1 x ⋅ 1 (x+1) 2= 1 x (x+1) 2 dx dv=dx→v=x ∫ ln( x x+1)dx=ln( x x+1)⋅ x−∫ x⋅ 1 x (x+1) 2 dx=xln( x x+1)−∫ 1 (x+1) 2 dx u=x+1→du=dx ∫ ln( x x+1)dx=xln( x x+1)−∫ 1 u 2 du=x⋅ln( x x+1)+ 1 u +c=x⋅ln( x x+1)+ 1 x+1 +c e) ∫ sen (2 x )ecos (2 x )dx u=cos (2 x )→du=−sen (2 x ) 2 dx→sen (2 x )dx=−2du ∫ sen (2 x )ecos (2 x )dx=∫e u(−2du)=−2e u+c=−2ecos (2 x )+c f) ∫( 3 x 2+1 x 3+x )dx u=x 3+x →du=(3 x 2+1)dx ∫ 3 x 2+1 x 3+x dx=∫ 1 u du=lnu+c=ln( x 3+x)+c g) ∫ sen (x )⋅cos (x )dx u=sen (x )→du=cos (x )dx ∫ sen (x )cos (x )dx=∫udu=u 2 2 +c= sen 2 (x ) 2 +c h) ∫( x 3−2 x 2 )dx=∫( x 3 x 2− 2 x 2)dx=¿∫(x− 2 x 2)dx= x 2 2 + 2 x +c ¿ i) ∫ x 2⋅e −( x 3)dx u=x 2→du=2 xdx dv=e −x 3 →v=−3e −x 3 ∫ x 2e −x 3 dx=x 2⋅(−3e −x 3 )−∫−3e −x 3 ⋅2 xdx=−3 x 2e −x 3 +6∫ x e −( x 3)dx u=x→du=dx dv=e −x 3 →v=−3e −x 3 ∫ x 2e −x 3 dx=−3 x 2e −x 3 +6[x⋅(−3e −x 2 )−∫−3e −x 3 dx]=−3 x 2e −x 3 −18 x e −x 3 +18 ∫ e −x 3 dx=−3 x 2e −x 3 −18 x e −x 3 −54 e −x 3 +c j) ∫ x⋅cos( x 4)dx u=x→du=dx dv=cos( x 4)dx→v=4 sen( x 4) ∫ xcos( x 4)dx=x⋅4 sen( x 4)−∫ 4 sen( x 4)dx=4 xsen( x 4)+16cos( x 4)+c 8- Calcule as integrais utilizando o método de substituição trigonométrica: a) ∫( √9−x 2 2 x 2 )dx x=3 sen (θ)→dx=3cos (θ) √9−x 2 2 x 2 =√9−(3 sen (θ)) 2 2(3 sen (θ)) 2 =√9−9 se n 2 (θ) 2⋅9 se n 2(θ) =√9(1−se n 2 (θ)) 18 se n 2 (θ) =√9cos 2 (θ) 18 se n 2 (θ) = 3cos (θ) 18 se n 2 (θ) = cos (θ) 6 sen 2 (θ) ∫ √9−x 2 2 x 2 dx=∫ cos (θ) 6 sen 2 (θ) 3cos (θ)dθ=1 2∫ cos 2 (θ) sen 2 (θ) dθ=1 2∫ 1−sen 2 (θ) sen 2 (θ) dθ=1 2 ∫( 1 sen 2 (θ) −1)dθ=1 2∫(cossec 2 (θ)−1)dθ=1 2 (−cotg (θ)−θ)+c=1 2(−cotg(sen −1( x 3))−sen −1 (θ))+c b) ∫( 1 x 3⋅√x 2−16)dx x=4 sec (θ)→dx=4 sec(θ)tg(θ) 1 x 3√x 2−16 = 1 sec 3 (θ)√16 sec 2 (θ)−16 = 1 sec 3 (θ)√16 (sec 2 (θ)+1) = 1 sec 3 (θ)√16t g 2 (θ) = 1 sec 3 (θ)⋅4 tg (θ) ∫ 1 x 3√x 2−16 dx=∫ 1 4 sec 3 (θ)tg (θ) 4 sec (θ)tg (θ)dθ=∫ 1 sec 2 (θ) dθ=∫cos 2 (θ)dθ=θ 2 + sen (2θ) 2 +c= sec −1( x 4) 2 + sen(sec −1( x 4)) 2 +c c) ∫ √4+x 2dx x=2tg (θ )→dx=2sec 2(θ)dθ √4+x 2=√4+4 t g 2 (θ)=√4 (1+t g 2 (θ))=√4 sec 2 (θ)=2sec ⁡(θ) ∫√4+x 2dx=∫2sec(θ)2sec 2(θ)dθ=∫ 4 sec (θ)(sec 2(θ))dθ u=4 sec (θ)→du=4 sec (θ)tg(θ) dv=sec 2 (θ)dθ→v=tg (θ) ∫ 4 sec 3(θ)=4 sec (θ)tg (θ)−∫tg (θ)⋅4 sec (θ)tg (θ)dθ ∫ 4 sec 3 (θ)=4 sec (θ)tg (θ)−∫ 4 sec (θ)t g 2 (θ)dθ ∫ 4 sec 3 (θ )dθ=4 sec (θ )tg (θ )−∫ 4 sec (θ )(1+sec 2 (θ ))dθ ∫ 4 sec 3 (θ)dθ=4 sec (θ)tg (θ)−∫ 4 sec (θ)dθ−∫ 4 sec 3(θ)dθ 2∫ 4 sec 3 (θ)dθ=4 sec (θ)tg (θ)−∫ 4 sec (θ)⋅[ sec (θ)+tg (θ) sec (θ)+tg (θ)]dθ 2∫ 4 sec 3 (θ)dθ=4 sec (θ)tg (θ)−ln∣sec (θ)+tg (θ)∣ ∫ 4 sec 3 (θ)dθ=2sec (θ)tg (θ)−ln∣sec (θ)+tg (θ)∣ 9- Calcule as integrais utilizando o método de funções racionais: a) ∫( 2 x+1 2 x 2+3 x−2)dx 2 x+1 2 x 2+3 x−2 = 2 x+1 (x−1 2)(x+2) = A x−1 2 + B x+2= A (x+2)+B(x−1 2) 2 x 2+3 x−2 { A+B=2 2 A−1 2 B=1 2 A−1 2 B=1→4 A−B=1→B=4 A−1→B=7 5 A+B=2→ A+4 A−1=2→5 A=3→ A=3 5 ∫ 2 x+1 2 x 2+3 x−1 dx=∫ 6 5 (2 x−1) dx+∫ 7 5 (x+2) dx=6 5 ln∣2 x−1∣+ 7 5 ln∣ x+2∣+c b) ∫( 2 x 3 x 2+x)dx 2 x 3 x 2+x = 2 x 3 x (x+1)= 2 x 2 x+1 ∫ 2 x 3 x 2+x dx=∫ 2 x 2 x+1 dx u=x+1→ x=u−1 ∫ 2 x 2 x+1 dx=∫ 2 (u−1) 2 u du=∫ 2(u 2−2u+1) u du=∫(2u−2+ 1 u)du=u 2−2u+ln∣u∣+c=(x+1) 2−2 (x+1)+ln∣ x+1∣+c 10- Calcule as seguintes integrais: a) ∫ −1 2 (x 3−2 x)dx=( x 4 4 −x 2)∨ 2 −1=(4−4 )−( 1 4 −1)=3 3 b) ∫ 0 1 (3+x √x )dx=(3 x+ 2 x 5 2 5 )∨1 0=3+ 2 5=17 5 c) ∫ 0 9 x−1 √x dx=∫ 0 9 (√x−x −1 2 )dx=( 2 x 3 2 3 −2 x 1 2)∨9 0=18−6=12 d) ∫ 0 π 4 sec (x )tg (x )dx=sec (x )∨ π 4 0 = 2 √2−1=√2−1 e) ∫ 0 1 (x e+e x)dx=( x e−1 e−1 +e x)∨1 0=( 1 e−1 +e)−(1)=1+e 2−e−e+1 e−1 =e 2−2e+2 e−1 f) ∫ −1 1 e 2 x dx=e 2 x 2 ∨ 1 −1=e 2 2 −e −2 2 =e 2−e −2 2 g) ∫ 1 2 ( 1 (3−5 x ) 2)dx u=3−5 x→du=−5dx→−du 5 =dx ∫ 1 2 1 (3−5 x ) 2 dx=∫ −2 7 1 u 2( −du 5 )=( 1 5u)∨ 7 −2= 1 35 + 1 10= 9 70 h) ∫ −π 3 π 3 x⋅sen (x )dx=xcos (x )| π 3 −π 3 −∫ −π 3 π 3 −cos (x )dx= π 3 ⋅ 1 2−( −π 3 )( 1 2)+sen (x )| π 3 −π 3 = π 6 + π 6 +( √3 2 + √3 2 )= π 3 +√3 11- Encontre a área da região sombreada: A=∫ 0 4 (4 x−x 2)dx=(2 x 2− x 3 3 )∨4 0=32 3 A=∫ 0 3 (x 2−3 x)dx+∫ 3 4 (x 2−3 x)dx=( x 3 3 −3 x 2 2 )| 3 0+( x 3 3 −3 x 2 2 )| 4 3=13,5−8 3−13,5=8 3 A=∫ 0 2 x 2dx= x 3 3 ∨2 0=8 3 A=∫ 0 2 (√x+2−( 1 x+1))dx=∫ 0 2 √x+2dx−∫ 0 2 1 x+1 dx=2 (x+2) 3 2 3 ∨2 0−ln ⁡( x+1)∨2 0=16 3 −2 5 2 3 −ln ⁡(3) A=∫ 0 1 (√x−x )dx=( 2 3 x 3 2− x 2 2 )∨1 0=2 3−1 2=1 3