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### Taller 4 Integral definida - Ejercicios resueltos - Calculo Integral

• 1. TALLER DIRIGIDO 4 TAREA 3 INTEGRAL DEFINIDA CALCULO INTEGRAL 5. ∫ 𝟔𝒙 𝒅𝒙 = 𝟐 𝟎 6𝑥2 2 ] 0 2 = 3(2)2 − 3(0)2 = 𝟏𝟐 6. ∫ 𝟓𝒅𝒗 𝟗 𝟒 =5 ∫ 𝑑𝑣 = 9 4 5𝑣]4 9 = 5(9) − 5(4 = 𝟐𝟓 7. ∫ ( 𝟐𝒙 − 𝟏) 𝒅𝒙 𝟎 −𝟏 = ( 𝟐𝒙 𝟐 − 𝑥) −1 0 = (02 − 0) − ((−1)2 − (−1)) = (0) − (1 + 1) = −𝟐
• 2. 8. ∫ (−𝟑𝒗 + 𝟒) 𝒅𝒗 𝟓 𝟐 = ( −3𝑣2 2 + 4𝑣) 2 5 = ( −3(5)2 2 + 4(5)) − ( −3(2)2 2 + 4(2)) = ( −75 2 + 20) − (−6 + 8) = −35 2 − 2 = −39 2 = −𝟏𝟗. 𝟓 9. ∫ ( 𝒕 𝟐 − 𝟐) 𝒅𝒕 𝟏 −𝟏 = ( 𝑡3 2 − 2𝑡) −1 1 = ( 1 3 − 2) − (− 1 3 + 2) = − 5 3 − 5 3 = − 10 3 = −𝟑. 𝟑𝟑
• 3. 10. ∫ ( 𝟔𝒙 𝟐 + 𝟐𝒙 − 𝟑) 𝒅𝒙 𝟕 𝟏 = ( 6𝑥3 3 + 2𝑥2 2 − 3𝑥) 1 7 = (2𝑥3 + 𝑥2 − 3𝑥 )1 7 = (2(7)3 + (7)2) − 3(7)) − (2(1)3 + (1)2 − 3(1)) = (686 + 49 − 21) − (2 + 1 − 3) = 𝟕𝟏𝟒 11. ∫ ( 𝟐𝒕 − 𝟏) 𝒅𝒕 𝟏 𝟎 = ∫ (4𝑡2 − 4𝑡 + 1) 𝑑𝑡 = ( 4𝑡3 3 − 4𝑡2 2 + 𝑡) 0 1 1 0 = ( 4 3 − 2 + 1) − (0 − 0 + 0) = 1 3 = 𝟎. 𝟑𝟑 12. ∫ ( 𝒕 𝟐 − 𝟗𝒕) 𝒅𝒕 = ( 𝑡4 4 − 9𝑡2 2 ) −1 1 = ( 1 4 − 9 2 ) − ( 1 4 − 9 2 ) 𝟏 −𝟏 = 𝟎
• 4. 13. ∫ ( 𝟑 𝒙 𝟐 − 𝟏) 𝒅𝒙 = 𝟐 𝟏 ∫ (3𝑥−2 − 1) 2 1 𝒅𝒙 = (− 3 𝑥 − 𝑥) 1 2 = (− 3 2 − 2) — 3 − 1 = − 7 2 + 4 = 1 2 = 𝟎. 𝟓 14. ∫ (𝒖 − 𝟏 𝒖 𝟐) 𝒅𝒖 = ( 𝒖 𝟐 𝟐 + 𝟏 𝒖 ) −𝟐 −𝟏 = ( 𝟏 𝟐 − 𝟏) − (𝟐 − 𝟏 𝟐 ) −𝟏 −𝟐 = − 1 2 − 3 2 = − 4 2 = −𝟐 15. ∫ 𝒖−𝟐 √𝒖 𝒅𝒖 = ∫ ( 𝒖 − 𝟐). (𝒖− 𝟏 𝟐 ) 𝟒 𝟏 𝟒 𝟏 𝒅𝒖 = ∫ (𝒖 𝟏 𝟐 − 𝟐𝒖− 𝟏 𝟐)𝒅𝒖 𝟒 𝟏 = ( 2 3 𝑢 3 2 − 4𝑢 1 2) 1 4 = ( 2 3 (4) 3 2 − 4 (4) 1 2) − ( 2 3 (1) 3 2 − 4 (1) 1 2) = ( 16 3 − 8) − ( 2 3 − 4) = − 8 3 + 10 3 = 2 3 = 𝟎. 𝟔𝟕
• 5. 16. ∫ 𝒗 𝟏 𝟑 𝟑 −𝟑 𝒅𝒗 = 𝟑 𝟒 𝒗 𝟒 𝟑 ] −𝟑 𝟑 = ( 𝟑 𝟒 . ( 𝟑) 𝟒 𝟑 − 𝟑 𝟒 . (−𝟑) 𝟒 𝟑) = 3.24 − 3.24 = 𝟎 17. ∫ (√ 𝒕 𝟑 − 𝟐)𝒅𝒕 = ∫ (𝒕 𝟏 𝟑 − 𝟐) 𝒅𝒕 = ( 𝟑 𝟒 𝒕 𝟒 𝟑 − 𝟐𝒕) −𝟏 𝟏 𝟏 −𝟏 𝟏 −𝟏 = ( 𝟑 𝟒 . ( 𝟏) 𝟒 𝟑 − 𝟐 ( 𝟏)) − ( 𝟑 𝟒 . (−𝟏) 𝟒 𝟑 − 𝟐 (−𝟏)) = ( 𝟑 𝟒 − 𝟐) − ( 𝟑 𝟒 + 𝟐) = − 𝟓 𝟒 − 𝟏𝟏 𝟒 = − 𝟏𝟔 𝟒 = −𝟒 18. ∫ √ 𝟐 𝒙 𝒅𝒙 = ∫ √𝟐 √𝒙 𝒅𝒙 = √𝟐 𝟖 𝟏 ∫ 𝒙− 𝟏 𝟐 𝒅𝒙 𝟖 𝟏 𝟖 𝟏 = 𝟐√𝟐 𝒙 𝟏 𝟐] 𝟏 𝟖 = 𝟐√𝟐 . (𝟖) 𝟏 𝟐 − 𝟐√𝟐 . ( 𝟏) 𝟏 𝟐 = 𝟖 − 𝟐√𝟐 = 𝟓. 𝟏𝟕
• 6. 19. ∫ 𝒙−√𝒙 𝟑 𝟏 𝟎 𝒅𝒙 = 𝟏 𝟑 ∫ (𝒙 − √ 𝒙)𝒅𝒙 = 𝟏 𝟑 ( 𝒙 𝟐 𝟐 − 𝟐 𝟑 𝒙 𝟑 𝟐) 𝟎 𝟏 𝟏 𝟎 = 1 3 ( 1 2 − 2 3 (1) 3 2) = − 1 18 = −𝟎. 𝟎𝟔 20. ∫ ( 𝟐 − 𝒕)√𝒕. 𝒅𝒕 = ∫ (𝟐√ 𝒕 − 𝒕√ 𝒕)𝒅𝒕 𝟐 𝟎 𝟐 𝟎 = ∫ (2√ 𝑡 − 𝑡 3 2) 𝑑𝑡 = 4 3 𝑡 3 2 − 2 5 𝑡 5 2] 0 2 2 0 = 4 3 (2) 3 2 − 2 5 (2) 5 2 = 3.77 − 2.26 = 𝟏. 𝟓𝟏 21. ∫ (𝒕 𝟏 𝟑 − 𝒕 𝟐 𝟑) 𝒅𝒕 = 𝟑 𝟒 𝒕 𝟒 𝟑 − 𝟑 𝟓 𝒕 𝟓 𝟑] −𝟏 𝟎 𝟎 −𝟏 = 0 − ( 3 4 (−1) 4 3 − 3 5 (−1) 5 3) = 0 − ( 3 4 + 3 5 ) = −𝟏. 𝟑𝟓
• 7. 22. ∫ 𝒙−𝒙 𝟐 𝟐 √𝒙 𝟑 𝒅𝒙 = 𝟏 𝟐 ∫ ( 𝒙 − 𝒙 𝟐) (𝒙− 𝟏 𝟑 ) 𝒅𝒙 −𝟏 −𝟖 −𝟏 −𝟖 = 𝟏 𝟐 ∫ (𝒙 𝟐 𝟑 − 𝒙 𝟓 𝟑 ) −𝟏 −𝟖 𝒅𝒙 = 𝟏 𝟐 ( 𝟑 𝟓 𝒙 𝟓 𝟑 − 𝟑 𝟖 𝒙 𝟖 𝟑 ) −𝟖 −𝟏 = 𝟏 𝟐 [( 𝟑 𝟓 (−𝟏) 𝟓 𝟑 − 𝟑 𝟖 (−𝟏) 𝟖 𝟑) − 𝟑 𝟓 ((−𝟖) 𝟓 𝟑 − 𝟑 𝟖 (−𝟖) 𝟖 𝟑)] = 𝟏 𝟐 [(− 𝟑 𝟓 − 𝟑 𝟖 ) − (− 𝟗𝟔 𝟓 − 𝟗𝟔)] = 𝟏 𝟐 [− 𝟑𝟗 𝟒𝟎 + 𝟓𝟕𝟔 𝟓 ] = 𝟓𝟕. 𝟏𝟏 27. ∫ ( 𝟏 + 𝒔𝒆𝒏𝒙) 𝒅𝒙 = ( 𝒙 − 𝒄𝒐𝒔𝒙) 𝟎 𝝅 = ( 𝝅 − 𝒄𝒐𝒔𝝅) − (𝟎 − 𝒄𝒐𝒔(𝟎)) 𝝅 𝟎 1 1 = 𝟓. 𝟏𝟒 28. ∫ ( 𝟐 + 𝒄𝒐𝒔𝒙) 𝒅𝒙 = (𝟐𝒙 + 𝒔𝒆𝒏𝒙) 𝟎 𝝅𝝅 𝟎 = ( 𝟐𝝅 + 𝒔𝒆𝒏𝝅) − (𝟐(𝟎) + 𝒔𝒆𝒏( 𝟎)) = 𝟐. 𝝅 = 𝟔. 𝟐𝟖
• 8. 29. ∫ 𝟏−𝒔𝒆𝒏 𝟐 𝜽 𝒄𝒐𝒔 𝟐 𝜽 𝝅 𝟒 𝟎 𝒅𝜽 = ∫ 𝒄𝒐𝒔 𝟐 𝜽 𝒄𝒐𝒔 𝟐 𝜽 𝝅 𝟒 𝟎 𝒅𝜽 = ∫ 𝒅𝜽 𝝅 𝟒 𝟎 = 𝜽] 𝟎 𝝅 𝟒 = 𝝅 𝟒 − 𝟎 = 𝝅 𝟒 = 𝟎. 𝟕𝟗 30. ∫ 𝒔𝒆𝒄 𝟐 𝜽 𝒕𝒂𝒏 𝟐 𝜽+𝟏 𝒅𝜽 = ∫ 𝒔𝒆𝒄 𝟐 𝜽 𝒔𝒆𝒄 𝟐 𝜽 𝒅𝜽 = ∫ 𝒅𝜽 𝝅 𝟎 𝝅 𝟒 𝟎 𝝅 𝟒 𝟎 = 𝜽] 𝟎 𝝅 𝟒 = 𝝅 𝟒 = 𝟎. 𝟕𝟗 31. ∫ 𝒔𝒆𝒄 𝟐 𝒙 𝒅𝒙 = 𝒕𝒂𝒏 𝒙] − 𝝅 𝟔 𝝅 𝟔 = 𝒕𝒂𝒏 𝝅 𝟔 − 𝒕𝒂𝒏 (− 𝝅 𝟔 ) 𝝅 𝟔 − 𝝅 𝟔 = 𝟎. 𝟓𝟕 + 𝟎. 𝟓𝟕 = 𝟏. 𝟏𝟓
• 9. 32. ∫ (𝟐 − 𝒄𝒔𝒄 𝟐 𝒙)𝒅𝒙 𝝅 𝟐 𝝅 𝟒 = 𝟐𝒙 + 𝒄𝒐𝒕 𝒙] 𝝅 𝟒 𝝅 𝟐 = (𝟐( 𝝅 𝟐 ) + 𝒄𝒐𝒕 ( 𝝅 𝟐 )) − (𝟐 ( 𝝅 𝟒 ) + 𝒄𝒐𝒕 ( 𝝅 𝟒 )) 0 1 = 𝝅 − ( 𝝅 𝟐 + 𝟏) = 𝟎. 𝟓𝟕 33. ∫ 𝟒𝒔𝒆𝒄 𝝅 𝟑 − 𝝅 𝟑 𝜽𝐭𝐚 𝐧 𝜽 𝒅𝝑 = 𝟒 𝒔𝒆𝒄 𝜽] − 𝝅 𝟑 𝝅 𝟑 = 𝟒 𝒔𝒆𝒄 ( 𝝅 𝟑 ) − 𝟒 𝒔𝒆𝒄 (− 𝝅 𝟑 ) = 𝟖 − 𝟖 = 𝟎 2 2 34. ∫ ( 𝟐𝒕 + 𝒄𝒐𝒔 𝒕) 𝒅𝒕 = 𝟐𝒕 𝟐 𝟐 + 𝒔𝒆𝒏 𝒕]−𝝅/𝟐 𝝅/𝟐𝝅/𝟐 −𝝅/𝟐 = (( 𝝅 𝟐 ) 𝟐 + 𝒔𝒆𝒏 ( 𝝅 𝟐 )) − ((− 𝝅 𝟐 ) 𝟐 + 𝒔𝒆𝒏 (− 𝝅 𝟐 )) 1 1 = ( 𝟐. 𝟒𝟔 + 𝟏) − ( 𝟐. 𝟒𝟔 − 𝟏) = 𝟐
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