Riemann Sums Class Work - NJCTL

32
Pre-Calc Intro to Integrals KEY ~1~ NJCTL.org Riemann Sums – Class Work 1. Consider the region between = 9 βˆ’ 3 and the x-axis for 0≀≀3. a. Sketch the graph of the region partitioned into 6 rectangles with LRAM b. Calculate the area using LRAM 2. Using the same region as question 1, follow the same step to find RRAM. 3. Using the same region as question 1, follow the same step to find MRAM. 4. Using the same region as question 1, find LRAM, RRAM, & MRAM but with 12 partitions. 5. Make a conjecture about the area of the region in question 1. . . . LRAM: . RRAM: . MRAM: . About .

Transcript of Riemann Sums Class Work - NJCTL

Page 1: Riemann Sums Class Work - NJCTL

Pre-Calc Intro to Integrals KEY ~1~ NJCTL.org

Riemann Sums – Class Work

1. Consider the region between 𝑦 = 9π‘₯ βˆ’ π‘₯3and the x-axis for 0 ≀ π‘₯ ≀ 3.

a. Sketch the graph of the region partitioned into 6 rectangles with LRAM

b. Calculate the area using LRAM

2. Using the same region as question 1, follow the same step to find RRAM.

3. Using the same region as question 1, follow the same step to find MRAM.

4. Using the same region as question 1, find LRAM, RRAM, & MRAM but with 12 partitions.

5. Make a conjecture about the area of the region in question 1.

πŸπŸ—. πŸ”πŸ–πŸ•πŸ“

πŸπŸ—. πŸ”πŸ–πŸ•πŸ“

𝟐𝟎. πŸ“πŸ‘πŸπŸπŸ“

LRAM: 𝟐𝟎. πŸπŸŽπŸ—πŸ‘πŸ•πŸ“

RRAM: 𝟐𝟎. πŸπŸŽπŸ—πŸ‘πŸ•πŸ“

MRAM: 𝟐𝟎. πŸ‘πŸπŸŽπŸ‘πŸπŸπŸ“

About 𝟐𝟎. πŸπŸ“

Page 2: Riemann Sums Class Work - NJCTL

Pre-Calc Intro to Integrals KEY ~2~ NJCTL.org

Find LRAM, RRAM, and MRAM between f(x) and the x-axis. Given are the bounds [a,b] and the

number of partitions n.

6. 𝑓(π‘₯) = √π‘₯, [0,10], 𝑛 = 5

7. 𝑓(π‘₯) = π‘₯3, [1,3], 𝑛 = 4

8. 𝑓(π‘₯) = cos π‘₯, [0,2πœ‹], 𝑛 = 8

The table shows the rate of fuel consumption of a car at given times on a 2 hour trip.

Time 10am 10:15 10:30 10:45 11am 11:15 11:30 11:45 Noon

gal/hour 2 3 3 4 3 2 2 3 4

9. Using 4 partitions and MRAM, estimate the area.

10. What does this area represent?

11. What are the appropriate units for the area?

LRAM: πŸπŸ•. πŸ‘πŸ–πŸ’

RRAM: πŸπŸ‘. πŸ•πŸŽπŸ–πŸ–

MRAM: 𝟐𝟏. πŸπŸπŸ–

LRAM: πŸπŸ’

RRAM: πŸπŸ•

MRAM: πŸπŸ—. πŸ•πŸ“

LRAM: 𝟎

RRAM: 𝟎

MRAM: 𝟎

MRAM: πŸ”

Approximate number of gallons consumed

Gallons

Page 3: Riemann Sums Class Work - NJCTL

Pre-Calc Intro to Integrals KEY ~3~ NJCTL.org

Reimann Sums – Homework

12. Consider the region between 𝑦 = 3π‘₯ βˆ’ π‘₯2and the x-axis for 0 ≀ π‘₯ ≀ 3.

a. Sketch the graph of the region and partition into 6 rectangles with LRAM

b. Calculate the area using LRAM

13. Using the same region as question 12, follow the same step to find RRAM.

14. Using the same region as question 12, follow the same step to find MRAM.

15. Using the same region as question 12, find LRAM, RRAM, & MRAM but with 12 partitions.

16. Make a conjecture about the area of the region in question 12.

πŸ’. πŸ‘πŸ•πŸ“

πŸ’. πŸ‘πŸ•πŸ“

πŸ’. πŸ‘πŸ•πŸ“

LRAM: πŸ’. πŸ’πŸ”πŸ–πŸ•πŸ“

RRAM: πŸ’. πŸ’πŸ”πŸ–πŸ•πŸ“

MRAM: πŸ’. πŸ“πŸπŸ“πŸ”πŸπŸ“

About πŸ’. πŸ“

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Pre-Calc Intro to Integrals KEY ~4~ NJCTL.org

Find LRAM, RRAM, and MRAM between f(x) and the x-axis. Given are the bounds [a,b] and the

number of partitions n.

17. 𝑓(π‘₯) = π‘₯2, [1,9], 𝑛 = 4

18. 𝑓(π‘₯) = √π‘₯3 [0,8], 𝑛 = 4

19. 𝑓(π‘₯) = sin π‘₯, [0, πœ‹], 𝑛 = 4

The table shows the rate of downloads of a new song in the first 6 hours it was available.

Time 12am 12:45 1:30 2:15 3:00 3:45 4:30 5:15 6am

downloads/min 200 100 90 80 50 20 25 35 24

20. Using 4 partitions and MRAM, estimate the area.

21. What does this area represent?

22. What are the appropriate units for the area?

LRAM: πŸπŸ”πŸ–

RRAM: πŸ‘πŸπŸ–

MRAM: πŸπŸ’πŸŽ

LRAM: πŸ—. πŸ‘πŸπŸ—

RRAM: πŸπŸ‘. πŸ‘πŸπŸ—

MRAM: 𝟏𝟐. πŸπŸ‘

LRAM: 𝟏. πŸ–πŸ—πŸ”

RRAM: 𝟏. πŸ–πŸ—πŸ”

MRAM: 𝟐. πŸŽπŸ“πŸ

MRAM: πŸπŸπŸπŸ“πŸŽ

Approximate number of downloads

Downloads

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Trapezoid Rule – Class Work

23. Consider the region between 𝑦 = 4π‘₯ βˆ’ π‘₯3and the x-axis for 0 ≀ π‘₯ ≀ 2.

a. Sketch the graph of the region and partition into 4 trapezoids.

b. Calculate the area.

24. Using the same region as question 23, apply the trapezoid rule but with 8 partitions.

25. Make a conjecture about the area of the region in question 23.

Find the area using the trapezoid rule between f(x) and the x-axis. Given are the bounds [a,b]

and the number of partitions n.

26. 𝑓(π‘₯) =1

π‘₯, [2,4], 𝑛 = 4

27. 𝑓(π‘₯) = π‘₯ βˆ’ π‘₯3, [1,3], 𝑛 = 4

28. 𝑓(π‘₯) = sin π‘₯, [0,2πœ‹], 𝑛 = 6

The table shows the speed of a car at given times on a 2 hour trip.

Time 10am 10:15 10:30 10:45 11am 11:15 11:30 11:45 Noon

miles/hour 65 50 60 45 70 60 55 60 0

29. Using 4 partitions and trapezoid rule to estimate the area.

30. What does this area represent?

31. What are the appropriate units for the area?

πŸ‘. πŸ•πŸ“

πŸ‘. πŸ—πŸ‘πŸ•πŸ“

About πŸ’

𝟎. πŸ”πŸ—πŸ•

βˆ’πŸπŸ”. πŸ“

𝟎

πŸπŸŽπŸ–. πŸ•πŸ“

Approximate number of miles driven

Miles

Page 6: Riemann Sums Class Work - NJCTL

Pre-Calc Intro to Integrals KEY ~6~ NJCTL.org

Trapezoid Rule – Homework

32. Consider the region between 𝑦 = 8 βˆ’ 4π‘₯ and the x-axis for 0 ≀ π‘₯ ≀ 2.

a. Sketch the graph of the region and partition into 4 trapezoids.

b. Calculate the area using the trapezoid rule

33. Using the same region as question 32, apply the trapezoid rule but with 10 partitions.

34. Make a conjecture about the area of the region in question 32.

Find the area using the trapezoid rule between f(x) and the x-axis. Given are the bounds [a,b]

and the number of partitions n.

35. 𝑓(π‘₯) = π‘₯2 βˆ’ 4, [1,3], 𝑛 = 6

36. 𝑓(π‘₯) = √4 βˆ’ π‘₯[0,4], 𝑛 = 4

37. 𝑓(π‘₯) = cos π‘₯, [0, πœ‹], 𝑛 = 6

The table shows the rate of typists typing a manuscript over a 6 hour period.

Time Noon 12:45 1:30 2:15 3 3:45 4:30 5:15 6pm

words/min 200 100 90 80 50 20 25 35 24

38. Using 4 partitions and trapezoid rule to estimate the area.

39. What does this area represent?

40. What are the appropriate units for the area?

πŸ–

πŸ–

Exactly πŸ–

πŸπŸ—

πŸπŸ•

πŸ“. πŸπŸ’πŸ”

𝟎

πŸπŸ’πŸ—πŸ‘πŸŽ

Approximate number of words typed

Words

Page 7: Riemann Sums Class Work - NJCTL

Pre-Calc Intro to Integrals KEY ~7~ NJCTL.org

Accumulation Functions – Class Work

Use the graph of 𝑓′(π‘₯) to answer the following questions. 𝑓(0) = 2

41. ∫ 𝑓′(π‘₯)𝑑π‘₯3

0

42. ∫ 𝑓′(π‘₯)𝑑π‘₯5

0

43. ∫ 𝑓′(π‘₯)𝑑π‘₯0

βˆ’4

44. ∫ 𝑓′(π‘₯)𝑑π‘₯0

3

45. ∫ 𝑓′(π‘₯)𝑑π‘₯βˆ’4

5

46. 𝑓′(2)

47. 𝑓(2)

48. 𝑓"(2)

49. When is 𝑓"(π‘₯) > 0?

πŸ‘

𝟐. πŸ“

𝟎

βˆ’πŸ‘

βˆ’πŸ. πŸ“

𝟏

πŸ’

𝟎

[βˆ’πŸ, βˆ’πŸ]𝒂𝒏𝒅 [πŸ’, πŸ“]

Page 8: Riemann Sums Class Work - NJCTL

Pre-Calc Intro to Integrals KEY ~8~ NJCTL.org

Homework

Use the graph of 𝑓′(π‘₯), a semi-circle and two lines to answer the following questions. 𝑓(1) = 0

50. ∫ 𝑓′(π‘₯)𝑑π‘₯3

0

51. ∫ 𝑓′(π‘₯)𝑑π‘₯5

0

52. ∫ 𝑓′(π‘₯)𝑑π‘₯0

βˆ’4

53. ∫ 𝑓′(π‘₯)𝑑π‘₯βˆ’4

0

54. ∫ 𝑓′(π‘₯)𝑑π‘₯βˆ’4

5

55. 𝑓′(3)

56. 𝑓(4)

57. 𝑓"(βˆ’2)

58. When is 𝑓(π‘₯) increasing?

𝟐. πŸ“

πŸ’. πŸ“

πŸ”. πŸπŸ–

βˆ’πŸ”. πŸπŸ–

βˆ’πŸπŸŽ. πŸ•πŸ–

βˆ’πŸ

πŸ‘

𝟎

[𝟎, πŸ“]

Page 9: Riemann Sums Class Work - NJCTL

Pre-Calc Intro to Integrals KEY ~9~ NJCTL.org

Anti-Derivatives – Class Work

∫ 𝑓(π‘₯)𝑑π‘₯ = 42

βˆ’2

, ∫ 𝑔(π‘₯)𝑑π‘₯ = βˆ’32

βˆ’2

, ∫ 𝑓(π‘₯)𝑑π‘₯ = 8,5

2

∫ 𝑓(π‘₯)𝑑π‘₯ = 32

0

, ∫ 𝑓(π‘₯)𝑑π‘₯ = 25

8

59. ∫ (𝑓(π‘₯) + 𝑔(π‘₯))𝑑π‘₯2

βˆ’2 60. ∫ (𝑓(π‘₯) βˆ’ 𝑔(π‘₯))𝑑π‘₯

2

βˆ’2 61. ∫ (3𝑓(π‘₯) + |𝑔(π‘₯)|)𝑑π‘₯

2

βˆ’2

62. ∫ 𝑓(π‘₯)𝑑π‘₯5

βˆ’2 63. ∫ 4𝑓(π‘₯)𝑑π‘₯

8

2 64. ∫ 𝑓(π‘₯)𝑑π‘₯

8

βˆ’2 65. ∫ 𝑓(π‘₯)𝑑π‘₯

0

βˆ’2

Find the value of following definite integrals.

66. ∫ 3𝑑π‘₯4

1 67. ∫ π‘₯𝑑π‘₯

5

2 68. ∫ 4π‘₯3𝑑π‘₯

3

βˆ’2

69. ∫1

π‘₯

5

1𝑑π‘₯ 70. ∫ 𝑒π‘₯𝑑π‘₯

6

0 71. ∫ (3π‘₯2 + 6π‘₯ βˆ’ 5)𝑑π‘₯

1

βˆ’2

72. ∫1

π‘₯2 𝑑π‘₯2

1 73. ∫ 𝑠𝑒𝑐2π‘₯

πœ‹

40

𝑑π‘₯ 74. ∫ sin π‘₯ 𝑑π‘₯2πœ‹

0

75. ∫1

1+π‘₯2 𝑑π‘₯1

0

𝟏 πŸ• πŸπŸ“

𝟏𝟐 πŸπŸ’ 𝟏𝟎 𝟏

πŸ— 𝟏𝟎. πŸ“ πŸ”πŸ“

π₯𝐧 πŸ“ β‰ˆ 𝟏. πŸ”πŸŽπŸ— π’†πŸ” βˆ’ 𝟏 β‰ˆ πŸ’πŸŽπŸ. πŸ’πŸ‘ βˆ’πŸπŸ“

𝟏

𝟐 𝟏 𝟎

𝝅

πŸ’

Page 10: Riemann Sums Class Work - NJCTL

Pre-Calc Intro to Integrals KEY ~10~ NJCTL.org

Anti-Derivatives – Homework

∫ 𝑓(π‘₯)𝑑π‘₯ = 52

βˆ’2

, ∫ 𝑔(π‘₯)𝑑π‘₯ = 92

βˆ’2

, ∫ 𝑓(π‘₯)𝑑π‘₯ = βˆ’6,5

2

∫ 𝑓(π‘₯)𝑑π‘₯ = βˆ’12

0

, ∫ 𝑓(π‘₯)𝑑π‘₯ = βˆ’75

8

76. ∫ (𝑓(π‘₯) + 2𝑔(π‘₯))𝑑π‘₯2

βˆ’2 77. ∫ (2𝑓(π‘₯) βˆ’ 2𝑔(π‘₯))𝑑π‘₯

2

βˆ’2 78. ∫ (3𝑓(π‘₯) + |𝑔(π‘₯)|)𝑑π‘₯

2

βˆ’2

79. ∫ (𝑓(π‘₯) + 1)𝑑π‘₯5

2 80. ∫ 3𝑓(π‘₯)𝑑π‘₯

8

2 81. ∫ 𝑓(π‘₯)𝑑π‘₯

8

βˆ’2 82. ∫ 𝑓(π‘₯)𝑑π‘₯

0

βˆ’2

Find the value of following definite integrals.

83. ∫ 4𝑑π‘₯4

2 84. ∫ 2π‘₯𝑑π‘₯

5

2 85. ∫ π‘₯3𝑑π‘₯

6

βˆ’1

86. ∫2

π‘₯

6

1𝑑π‘₯ 87. ∫ (4𝑒π‘₯ + 1)𝑑π‘₯

5

0 88. ∫ (3π‘₯2 + 6π‘₯ βˆ’ 5)𝑑π‘₯

1

βˆ’2

89. ∫6

π‘₯3 𝑑π‘₯2

1 90. ∫ 𝑠𝑒𝑐π‘₯π‘‘π‘Žπ‘›π‘₯

πœ‹

40

𝑑π‘₯ 91. ∫ cos π‘₯ 𝑑π‘₯2πœ‹

0

92. ∫1

√1βˆ’π‘₯2𝑑π‘₯

1

20

πŸπŸ‘ βˆ’πŸ– πŸπŸ’

βˆ’πŸ‘ πŸ‘ πŸ” πŸ”

πŸ– 𝟐𝟏 πŸ‘πŸπŸ‘. πŸ•πŸ“

𝟐 π₯𝐧 πŸ” β‰ˆ πŸ‘. πŸ“πŸ– πŸ’π’†πŸ“ + 𝟏 β‰ˆ πŸ“πŸ—πŸ’. πŸ”πŸ“ πŸ’

𝟐. πŸπŸ“ √𝟐 βˆ’ 𝟏 β‰ˆ 𝟎. πŸ’πŸπŸ’ 𝟎

𝝅

πŸ”

Page 11: Riemann Sums Class Work - NJCTL

Pre-Calc Intro to Integrals KEY ~11~ NJCTL.org

Fundamental Theorem of Calculus – Class Work

Find 𝑑𝑦

𝑑π‘₯

93. 𝑦 = ∫ (4𝑑 βˆ’ 2)𝑑𝑑π‘₯

1 94. 𝑦 = ∫ (3𝑒2 βˆ’ 4𝑒)𝑑𝑒

2π‘₯

2

95. 𝑦 = ∫ 𝑙𝑛(𝑣)𝑑𝑣4

π‘₯ 96. 𝑦 = ∫ (4𝑑3 βˆ’ 2𝑑)𝑑𝑑

0

π‘₯2

97. 𝑦 = ∫ (7𝑒)𝑑𝑒2π‘₯

3π‘₯2 98. 𝑦 = ∫ 𝑒𝑣𝑑𝑣π‘₯ 𝑙𝑛 π‘₯

𝑙𝑛 π‘₯

99. Let 𝐹(π‘₯) = ∫ 𝑓(𝑑)𝑑𝑑,π‘₯

0 where 𝑓(𝑑) is defined by the graph.

a. 𝑓(2)

b. 𝐹(2)

c. 𝐹′(2)

d. 𝑓′(2)

100. ∫ (5𝑒 βˆ’ 6)𝑑𝑒 + 𝐾π‘₯

βˆ’2= ∫ (5𝑒 βˆ’ 6)𝑑𝑒

π‘₯

4, find K

π’…π’š

𝒅𝒙= πŸ’π’™ βˆ’ 𝟐

π’…π’š

𝒅𝒙= πŸπŸ’π’™πŸ βˆ’ πŸπŸ”π’™

π’…π’š

𝒅𝒙= βˆ’ π₯𝐧 𝒙

π’…π’š

𝒅𝒙= βˆ’πŸ–π’™πŸ• + πŸ’π’™πŸ‘

π’…π’š

𝒅𝒙= πŸπŸ–π’™ βˆ’ πŸπŸπŸ”π’™πŸ‘ π’…π’š

𝒅𝒙= π₯𝐧 𝒙 𝒆𝒙 π₯𝐧 𝒙 + 𝒆𝒙 π₯𝐧 𝒙 βˆ’

𝒆π₯𝐧 𝒙

𝒙

𝒇(𝟐) = πŸ’

𝑭(𝟐) = πŸ—

𝑭′(𝟐) = πŸ’

𝒇′(𝟐) = βˆ’πŸ

𝟐

𝑲 = πŸ”

Page 12: Riemann Sums Class Work - NJCTL

Pre-Calc Intro to Integrals KEY ~12~ NJCTL.org

Fundamental Theorem of Calculus – Homework

Find 𝑑𝑦

𝑑π‘₯

101. 𝑦 = ∫ 𝑒𝑒𝑑𝑒π‘₯

2 102. 𝑦 = ∫ 𝑑2𝑑𝑑

√π‘₯

3

103. 𝑦 = ∫ 𝑠𝑖𝑛 𝑣 π‘‘π‘£πœ‹

π‘₯ 104. 𝑦 = ∫ √5 βˆ’ 𝑑 𝑑𝑑

5

4βˆ’π‘₯

105. 𝑦 = ∫ (𝑒2 βˆ’ 4𝑒 + 2)𝑑𝑒7π‘₯

2π‘₯ 106. 𝑦 = ∫ βˆšπ‘£

π‘₯2

1

π‘₯

𝑑𝑣

107. Let 𝐹(π‘₯) = ∫ 𝑓(𝑑)𝑑𝑑,π‘₯

0 where 𝑓(𝑑) is defined by the graph.

a. 𝑓(2)

b. 𝐹(2)

c. 𝐹′(2)

d. 𝑓′(2)

108. ∫ (3𝑒2 + 2𝑒 + 1)𝑑𝑒 + 𝐾π‘₯

1= ∫ (3𝑒2 + 2𝑒 + 1)𝑑𝑒

π‘₯

3, find K

π’…π’š

𝒅𝒙= 𝒆𝒙 π’…π’š

𝒅𝒙=

βˆšπ’™

𝟐

π’…π’š

𝒅𝒙= βˆ’ 𝐬𝐒𝐧 𝒙

π’…π’š

𝒅𝒙= √𝟏 + 𝒙

π’…π’š

𝒅𝒙= πŸ‘πŸ‘πŸ“π’™πŸ βˆ’ πŸπŸ–πŸŽπ’™ + 𝟏𝟎 π’…π’š

𝒅𝒙= πŸπ’™πŸ +

βˆšπ’™

π’™πŸ‘

𝒇(𝟐) = πŸ’

𝑭(𝟐) = πŸ”

𝑭′(𝟐) = πŸ’

𝒇′(𝟐) = π’–π’π’…π’†π’‡π’Šπ’π’†π’…

𝑲 = βˆ’πŸ‘πŸ”

Page 13: Riemann Sums Class Work - NJCTL

Pre-Calc Intro to Integrals KEY ~13~ NJCTL.org

Substitution Method – Class Work

Evaluate the indefinite integral using the Substitution Method

109. ∫ 2π‘₯√π‘₯2 + 1𝑑π‘₯ 110. ∫π‘₯3

(π‘₯4+1)4 𝑑π‘₯

111. ∫ sin5 π‘₯ cos π‘₯ 𝑑π‘₯ 112. ∫ π‘₯ cos(π‘₯2) 𝑑π‘₯

113. ∫ π‘₯π‘’βˆ’π‘₯2𝑑π‘₯ 114. ∫

𝑑π‘₯

π‘₯ ln π‘₯

Evaluate the definite integral

115. ∫π‘₯

(π‘₯2+1)3 𝑑π‘₯1

0 116. ∫ (π‘₯ + 1)(π‘₯2 + 2π‘₯)3𝑑π‘₯

2

1

117. ∫ π‘₯ tan(π‘₯2) 𝑑π‘₯1

0 118. ∫ √5π‘₯ + 6

2

βˆ’1𝑑π‘₯

𝟐

πŸ‘(π’™πŸ + 𝟏)πŸ‘/𝟐 + π‘ͺ βˆ’

𝟏

𝟏𝟐(π’™πŸ’+𝟏)πŸ‘ + π‘ͺ

𝟏

πŸ”π¬π’π§πŸ” 𝒙 + π‘ͺ

𝟏

𝟐𝐬𝐒𝐧(π’™πŸ) + π‘ͺ

βˆ’πŸ

πŸπ’†βˆ’π’™πŸ

+ π‘ͺ π₯𝐧(π₯𝐧 𝒙) + π‘ͺ

πŸ‘

πŸπŸ”

πŸ’πŸŽπŸπŸ“

πŸ–= πŸ“πŸŽπŸ. πŸ–πŸ•πŸ“

𝟏

πŸ‘(𝟐√𝟐 βˆ’ 𝟏) β‰ˆ 𝟎. πŸ”πŸŽπŸ— πŸ–. πŸ’

Page 14: Riemann Sums Class Work - NJCTL

Pre-Calc Intro to Integrals KEY ~14~ NJCTL.org

Substitution Method – Homework

Evaluate the indefinite integral using the Substitution Method

119. ∫ π‘₯2(π‘₯3 + 1)4𝑑π‘₯ 120. ∫1

(π‘₯+2)2 𝑑π‘₯

121. ∫2π‘₯2+π‘₯

(4π‘₯3+3π‘₯2)5 𝑑π‘₯ 122. ∫ sin(2π‘₯ βˆ’ 4) 𝑑π‘₯

123. ∫π‘₯

√π‘₯2+9𝑑π‘₯ 124. ∫ sec2 π‘₯ (4 tan3 π‘₯ βˆ’ 3 tan2 π‘₯)𝑑π‘₯

Evaluate the definite integral

125. ∫ π‘₯√π‘₯2 + 94

0𝑑π‘₯ 126. ∫

π‘₯+3

(π‘₯2+6π‘₯+1)3 𝑑π‘₯2

0

127. ∫ (π‘₯ βˆ’ 9)βˆ’2/3𝑑π‘₯17

10 128. ∫ tan2 π‘₯ sec2 π‘₯ 𝑑π‘₯

πœ‹/4

0

𝟏

πŸπŸ“(π’™πŸ‘ + 𝟏)πŸ“ + π‘ͺ βˆ’

𝟏

(𝒙+𝟐)+ π‘ͺ

βˆ’πŸ

πŸπŸ’(πŸ’π’™πŸ‘+πŸ‘π’™πŸ)πŸ’ + π‘ͺ βˆ’

𝟏

𝟐𝐜𝐨𝐬(πŸπ’™ βˆ’ πŸ’) + π‘ͺ

βˆšπ’™πŸ + πŸ— + π‘ͺ π­πšπ§πŸ’ 𝒙 βˆ’ π­πšπ§πŸ‘ 𝒙 + π‘ͺ

πŸ—πŸ–

πŸ‘β‰ˆ πŸ‘πŸ. πŸ”πŸ•

πŸ•πŸ

πŸπŸ–πŸ—β‰ˆ 𝟎. πŸπŸ’πŸ—

πŸ‘ 𝟏

πŸ‘

Page 15: Riemann Sums Class Work - NJCTL

Pre-Calc Intro to Integrals KEY ~15~ NJCTL.org

Area Between Curves – Class Work

Find the total area between the functions.

129. 𝑦 = π‘₯4 βˆ’ 6π‘₯2 π‘Žπ‘›π‘‘ 𝑦 = 6 βˆ’ π‘₯2 130. 𝑦 = cos π‘₯ π‘Žπ‘›π‘‘ 𝑦 = 2cos π‘₯ π‘“π‘œπ‘Ÿ 0 ≀ π‘₯ β‰€πœ‹

2

131. 𝑦 = π‘₯3 βˆ’ 3π‘₯ π‘Žπ‘›π‘‘ 𝑦 = π‘₯ π‘“π‘œπ‘Ÿ π‘₯ β‰₯ 0 132. π‘₯ = 𝑦2 βˆ’ 6𝑦 π‘Žπ‘›π‘‘ π‘₯ + 𝑦 = 6

133. π‘₯ = 𝑦2 π‘Žπ‘›π‘‘ π‘₯ = 𝑦 134. 𝑦 = cos π‘₯ π‘Žπ‘›π‘‘ 𝑦 = π‘₯2 βˆ’ πœ‹2

Area Between Curves – Homework

Find the total area between the functions.

135. 𝑦 = 2 + 3π‘₯2 π‘Žπ‘›π‘‘ 𝑦 = 6 βˆ’ 2π‘₯2 136. 𝑦 = sin π‘₯ π‘Žπ‘›π‘‘ 𝑦 = cos π‘₯ π‘“π‘œπ‘Ÿ 0 ≀ π‘₯ ≀ πœ‹

137. 𝑦 = βˆ’2π‘₯3 + 10π‘₯ π‘Žπ‘›π‘‘ 𝑦 = βˆ’8π‘₯ π‘“π‘œπ‘Ÿ π‘₯ ≀ 0 138. π‘₯ = βˆ’π‘¦2 + 9 π‘Žπ‘›π‘‘ π‘₯ + 𝑦 = 3

139. π‘₯ = 𝑦2 π‘Žπ‘›π‘‘ π‘₯ = 𝑦2

3⁄ 140. 𝑦 = sin π‘₯ π‘Žπ‘›π‘‘ 𝑦 = π‘₯2 βˆ’ πœ‹π‘₯

πŸ–πŸ–

πŸ“βˆšπŸ” β‰ˆ πŸ’πŸ‘. 𝟏𝟏𝟏 𝟏

πŸ’ πŸ‘πŸ’πŸ‘

πŸ”β‰ˆ πŸ“πŸ•. πŸπŸ•

𝟏

πŸ” πŸ’πŸ. πŸ“

πŸπŸ”

πŸ‘β‰ˆ πŸ“. πŸ‘πŸ‘ 𝟐√𝟐 β‰ˆ 𝟐. πŸ–πŸπŸ–

πŸ’πŸŽ. πŸ“ πŸπŸπŸ“

πŸ”β‰ˆ 𝟐𝟎. πŸ–πŸ‘

πŸ’

πŸπŸ“β‰ˆ 𝟎. πŸπŸ”πŸ• 𝟐 +

𝟏

πŸ”π…πŸ‘ β‰ˆ πŸ•. πŸπŸ•

Page 16: Riemann Sums Class Work - NJCTL

Pre-Calc Intro to Integrals KEY ~16~ NJCTL.org

Volume: Disk Method – Class Work

Find the volume of the solid.

141. 𝑦 = 6π‘₯ βˆ’ π‘₯2 revolved about the x-axis

142. Area between 𝑦 = 6π‘₯ βˆ’ π‘₯2 and the x-axis revolved about 𝑦 = βˆ’5

143. 𝑦 = π‘₯, 𝑦 = βˆ’2π‘₯ + 4, π‘Žπ‘›π‘‘ 𝑦 = 0 revolved about the x-axis

144. 𝑦 = π‘₯, 𝑦 = βˆ’2π‘₯ + 4, π‘Žπ‘›π‘‘ π‘₯ = 0 revolved about the y-axis

𝝅 ∫ (πŸ”π’™ βˆ’ π’™πŸ)πŸπ’…π’™πŸ”

𝟎

πŸπŸπŸ—πŸ”

πŸ“π… β‰ˆ πŸ–πŸπŸ’. πŸ‘

𝝅 ∫ (πŸ“ + πŸ”π’™ βˆ’ π’™πŸ)πŸπ’…π’™πŸ”

𝟎

πŸ‘πŸ–πŸ’πŸ”

πŸ“π… β‰ˆ πŸπŸ’πŸπŸ”. πŸ“πŸ

𝝅 ∫ (𝒙)πŸπ’…π’™πŸ’/πŸ‘

𝟎+ 𝝅 ∫ (πŸ’ βˆ’ πŸπ’™)πŸπ’…π’™

𝟐

πŸ’/πŸ‘

πŸ‘πŸ

πŸπŸ•π… β‰ˆ πŸ‘. πŸ•πŸπŸ‘

𝝅 ∫ (π’š)πŸπ’…π’šπŸ’/πŸ‘

𝟎+ 𝝅 ∫ (𝟐 βˆ’

𝟏

πŸπ’š)

𝟐

π’…π’šπŸ’

πŸ’/πŸ‘

πŸ”πŸ’

πŸπŸ•π… β‰ˆ πŸ•. πŸ’πŸ“

Page 17: Riemann Sums Class Work - NJCTL

Pre-Calc Intro to Integrals KEY ~17~ NJCTL.org

145. 𝑦 = 2π‘₯, π‘₯ = 3, π‘Žπ‘›π‘‘ π‘₯ βˆ’ π‘Žπ‘₯𝑖𝑠 revolved about the x-axis

146. 𝑦 = 2π‘₯, π‘₯ = 3, π‘Žπ‘›π‘‘ π‘₯ βˆ’ π‘Žπ‘₯𝑖𝑠 revolved about the π‘₯ = 3

147. 𝑦 = 6 βˆ’ π‘₯, 𝑦 = π‘₯2, π‘Žπ‘›π‘‘ 𝑦 = 0 revolved about the x-axis

148. 𝑦 = 6 βˆ’ π‘₯, 𝑦 = π‘₯2, π‘Žπ‘›π‘‘ π‘₯ = 0 revolved about the y-axis

𝝅 ∫ (πŸπ’™)πŸπ’…π’™πŸ‘

𝟎

πŸ‘πŸ”π… β‰ˆ πŸπŸπŸ‘. πŸŽπŸ—πŸ•

𝝅 ∫ (πŸ‘ βˆ’πŸ

πŸπ’š)

𝟐

π’…π’šπŸ”

𝟎

πŸπŸ–π… β‰ˆ πŸ“πŸ”. πŸ“πŸ’πŸ—

𝝅 ∫ (π’™πŸ)πŸπ’…π’™πŸ

𝟎+ 𝝅 ∫ (πŸ” βˆ’ 𝒙)πŸπ’…π’™

πŸ”

𝟐

πŸ’πŸπŸ”

πŸπŸ“π… β‰ˆ πŸ–πŸ•. πŸπŸπŸ•

𝝅 ∫ (βˆšπ’š)𝟐

π’…π’šπŸ’

𝟎+ 𝝅 ∫ (πŸ” βˆ’ π’š)πŸπ’…π’š

πŸ”

πŸ’

πŸ’πŸ’πŸ–

πŸ‘π… β‰ˆ πŸ’πŸ”πŸ—. πŸπŸ’πŸ“

Page 18: Riemann Sums Class Work - NJCTL

Pre-Calc Intro to Integrals KEY ~18~ NJCTL.org

Volume: Disk Method – Homework

Find the volume of the solid.

149. 𝑦 = 8π‘₯ βˆ’ 2π‘₯2 revolved about the x-axis

150. 𝑦 = 8π‘₯ βˆ’ 2π‘₯2 and the x-axis revolved about 𝑦 = βˆ’3

151. 𝑦 = 3π‘₯, 𝑦 = βˆ’π‘₯2 + 4, π‘Žπ‘›π‘‘ 𝑦 = 0 revolved about the x-axis

152. 𝑦 = 3π‘₯, 𝑦 = βˆ’π‘₯2 + 4, π‘Žπ‘›π‘‘ π‘₯ = 0 revolved about the y-axis

𝝅 ∫ (πŸ–π’™ βˆ’ πŸπ’™πŸ)πŸπ’…π’™πŸ’

𝟎

πŸπŸŽπŸ’πŸ–

πŸπŸ“π… β‰ˆ πŸ’πŸπŸ–. πŸ—πŸ‘πŸ

𝝅 ∫ (πŸ‘ + πŸ–π’™ βˆ’ πŸπ’™πŸ)πŸπ’…π’™πŸ’

𝟎

πŸ’πŸ“πŸŽπŸ–

πŸπŸ“π… β‰ˆ πŸ—πŸ’πŸ’. πŸπŸ“πŸ‘

𝝅 ∫ (πŸ‘π’™)πŸπ’…π’™πŸ

𝟎+ 𝝅 ∫ (πŸ’ βˆ’ π’™πŸ)πŸπ’…π’™

𝟐

𝟏

πŸ—πŸ–

πŸπŸ“π… β‰ˆ 𝟐𝟎. πŸ“πŸπŸ“

𝝅 ∫ (𝟏

πŸ‘π’š)

𝟐

π’…π’šπŸ‘

𝟎+ 𝝅 ∫ (βˆšπŸ’ βˆ’ π’š)

πŸπ’…π’š

πŸ’

πŸ‘

πŸ‘

πŸπ… β‰ˆ πŸ’. πŸ•πŸπŸ

Page 19: Riemann Sums Class Work - NJCTL

Pre-Calc Intro to Integrals KEY ~19~ NJCTL.org

153. 𝑦 = 3π‘₯, π‘₯ = 4, π‘Žπ‘›π‘‘ π‘₯ βˆ’ π‘Žπ‘₯𝑖𝑠 revolved about the x-axis

154. 𝑦 = 3π‘₯, π‘₯ = 4, π‘Žπ‘›π‘‘ π‘₯ βˆ’ π‘Žπ‘₯𝑖𝑠 revolved about the π‘₯ = 4

155. 𝑦 = π‘₯2 + 2, 𝑦 = 10 βˆ’ π‘₯2, 𝑦 = 0, π‘₯ = 0, π‘Žπ‘›π‘‘ π‘₯ = 3 revolved about the x-axis

156. 𝑦 = π‘₯2 + 2, 𝑦 = 10 βˆ’ π‘₯2, π‘₯ = 0, π‘Žπ‘›π‘‘ π‘₯ = 3 revolved about the x-axis

𝝅 ∫ (πŸ‘π’™)πŸπ’…π’™πŸ’

𝟎

πŸπŸ—πŸπ… β‰ˆ πŸ”πŸŽπŸ‘. πŸπŸ–πŸ”

𝝅 ∫ (πŸ’ βˆ’πŸ

πŸ‘π’š)

𝟐

π’…π’šπŸπŸ

𝟎

πŸ”πŸ’π… β‰ˆ 𝟐𝟎𝟏. πŸŽπŸ”

𝝅 ∫ (π’™πŸ + 𝟐)πŸπ’…π’™πŸ

𝟎+ 𝝅 ∫ (𝟏𝟎 βˆ’ π’™πŸ)πŸπ’…π’™

πŸ‘

𝟐

πŸπŸŽπŸ‘

πŸ“π… β‰ˆ πŸπŸπŸ•. πŸ“πŸ“

𝝅 ∫ ((𝟏𝟎 βˆ’ π’™πŸ) βˆ’ (π’™πŸ + 𝟐))𝟐

π’…π’™πŸ

𝟎+ 𝝅 ∫ ((π’™πŸ + 𝟐) βˆ’ (𝟏𝟎 βˆ’ π’™πŸ))

𝟐

π’…π’™πŸ‘

𝟐

πŸ’πŸ—πŸ

πŸ“π… β‰ˆ πŸ‘πŸŽπŸ—. πŸπŸ‘πŸ‘

Page 20: Riemann Sums Class Work - NJCTL

Pre-Calc Intro to Integrals KEY ~20~ NJCTL.org

Volume: Washer Method – Class Work

Find the solid created by rotating the region 𝑦 = π‘₯2, 𝑦 = π‘₯, π‘₯ = 1, π‘Žπ‘›π‘‘ π‘₯ = 4 about

157. 𝑦 = 0

158. 𝑦 = 1

159. 𝑦 = 100

160. 𝑦 = βˆ’2

161. π‘₯ = 0

𝝅 ∫ (π’™πŸ)𝟐 βˆ’ (𝒙)πŸπ’…π’™πŸ’

𝟏

πŸ—πŸπŸ–

πŸ“π… β‰ˆ πŸ“πŸ•πŸ”. πŸ•πŸ—πŸ”

𝝅 ∫ (π’™πŸ βˆ’ 𝟏)𝟐 βˆ’ (𝒙 βˆ’ 𝟏)πŸπ’…π’™πŸ’

𝟏

πŸ•πŸ–πŸ‘

πŸ“π… β‰ˆ πŸ’πŸ—πŸ. πŸ—πŸ•πŸ‘

𝝅 ∫ (𝟏𝟎𝟎 βˆ’ 𝒙)𝟐 βˆ’ (𝟏𝟎𝟎 βˆ’ π’™πŸ)πŸπ’…π’™πŸ’

𝟏

πŸπŸπŸ“πŸ–πŸ

πŸ“π… β‰ˆ πŸ•πŸ—πŸŽπŸ“. πŸ“πŸŽπŸ’

𝝅 ∫ (𝟐 + π’™πŸ)𝟐 βˆ’ (𝟐 + 𝒙)πŸπ’…π’™πŸ’

𝟏

πŸπŸπŸ–πŸ–

πŸ“π… β‰ˆ πŸ•πŸ’πŸ”. πŸ’πŸ’πŸ

𝝅 ∫ (π’š)𝟐 βˆ’ (βˆšπ’š)𝟐

π’…π’šπŸ’

𝟏+ 𝝅 ∫ (πŸ’)𝟐 βˆ’ (βˆšπ’š)

πŸπ’…π’š

πŸπŸ”

πŸ’

πŸπŸ•πŸ

πŸπ… β‰ˆ πŸπŸ”πŸ–. πŸ”πŸŽπŸ”

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Pre-Calc Intro to Integrals KEY ~21~ NJCTL.org

162. π‘₯ = βˆ’2

163. π‘₯ = 1

164. π‘₯ = 4

165. π‘₯ = 10

𝝅 ∫ (𝟐 + π’š)𝟐 βˆ’ (𝟐 + βˆšπ’š)𝟐

π’…π’šπŸ’

𝟏+ 𝝅 ∫ (πŸ”)𝟐 βˆ’ (𝟐 + βˆšπ’š)

πŸπ’…π’š

πŸπŸ”

πŸ’

πŸπŸ•πŸ—

πŸπ… β‰ˆ πŸ’πŸ‘πŸ–. πŸπŸ“πŸ

𝝅 ∫ (π’š βˆ’ 𝟏)𝟐 βˆ’ (βˆšπ’š βˆ’ 𝟏)𝟐

π’…π’šπŸ’

𝟏+ 𝝅 ∫ (πŸ“)𝟐 βˆ’ (βˆšπ’š βˆ’ 𝟏)

πŸπ’…π’š

πŸπŸ”

πŸ’

πŸ“πŸŽπŸ

πŸπ… β‰ˆ πŸ•πŸ–πŸ”. πŸ—πŸ”πŸ—

𝝅 ∫ (πŸ’ βˆ’ βˆšπ’š)𝟐

βˆ’ (πŸ’ βˆ’ π’š)πŸπ’…π’šπŸ’

𝟏+ 𝝅 ∫ (πŸ’ βˆ’ βˆšπ’š)

πŸπ’…π’š

πŸπŸ”

πŸ’

πŸ’πŸ“

πŸπ… β‰ˆ πŸ•πŸŽ. πŸ”πŸ–πŸ”

𝝅 ∫ (𝟏𝟎 βˆ’ βˆšπ’š)𝟐

βˆ’ (𝟏𝟎 βˆ’ π’š)πŸπ’…π’šπŸ’

𝟏+ 𝝅 ∫ (𝟏𝟎 βˆ’ βˆšπ’š)

πŸβˆ’ (πŸ”)πŸπ’…π’š

πŸπŸ”

πŸ’

πŸ‘πŸ”πŸ—

πŸπ… β‰ˆ πŸ“πŸ•πŸ—. πŸ”πŸπŸ’

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Pre-Calc Intro to Integrals KEY ~22~ NJCTL.org

Volume: Washer Method – Homework

Find the solid created by rotating the region 𝑦 = βˆ’π‘₯3, 𝑦 = βˆ’π‘₯, π‘₯ = 1, π‘Žπ‘›π‘‘ π‘₯ = 3 about

166. 𝑦 = 0

167. 𝑦 = 10

168. 𝑦 = βˆ’1

169. 𝑦 = βˆ’50

170. π‘₯ = 0

𝝅 ∫ (βˆ’π’™πŸ‘)𝟐 βˆ’ (βˆ’π’™)πŸπ’…π’™πŸ‘

𝟏

πŸ”πŸ‘πŸ•πŸ”

πŸπŸπ… β‰ˆ πŸ—πŸ“πŸ‘. πŸ–πŸ’πŸ•

𝝅 ∫ (𝟏𝟎 + π’™πŸ‘)𝟐 βˆ’ (𝟏𝟎 + 𝒙)πŸπ’…π’™πŸ‘

𝟏

πŸπŸ‘πŸŽπŸ—πŸ”

πŸπŸπ… β‰ˆ πŸπŸ—πŸ“πŸ—. πŸπŸ“πŸ•

𝝅 ∫ (𝟏 βˆ’ π’™πŸ‘)𝟐 βˆ’ (𝟏 βˆ’ 𝒙)πŸπ’…π’™πŸ‘

𝟏

πŸ“πŸ•πŸŽπŸ’

πŸπŸπ… β‰ˆ πŸ–πŸ“πŸ‘. πŸ‘πŸπŸ”

𝝅 ∫ (πŸ“πŸŽ βˆ’ 𝒙)𝟐 βˆ’ (πŸ“πŸŽ βˆ’ π’™πŸ‘)πŸπ’…π’™πŸ‘

𝟏

πŸπŸ•πŸπŸπŸ’

πŸπŸπ… β‰ˆ πŸ’πŸŽπŸ•πŸ. πŸ•πŸŽπŸ

𝝅 ∫ (π’š)𝟐 βˆ’ (π’šπŸ/πŸ‘)𝟐

π’…π’šπŸ‘

𝟏+ 𝝅 ∫ (πŸ‘)𝟐 βˆ’ (π’šπŸ/πŸ‘)

πŸπ’…π’š

πŸπŸ•

πŸ‘

πŸπŸπŸ—πŸ

πŸπŸ“π… β‰ˆ πŸπŸ’πŸ—. πŸ”πŸ“πŸ

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171. π‘₯ = βˆ’2

172. π‘₯ = 1

173. π‘₯ = 3

174. π‘₯ = 10

𝝅 ∫ (𝟐 + π’š)𝟐 βˆ’ (𝟐 + π’šπŸ/πŸ‘)𝟐

π’…π’šπŸ‘

𝟏+ 𝝅 ∫ (πŸ“)𝟐 βˆ’ (𝟐 + π’šπŸ/πŸ‘)

πŸπ’…π’š

πŸπŸ•

πŸ‘

πŸπŸπŸ“πŸ

πŸπŸ“π… β‰ˆ πŸ’πŸ“πŸŽ. πŸ•πŸπŸ’

𝝅 ∫ (π’š βˆ’ 𝟏)𝟐 βˆ’ (π’šπŸ/πŸ‘ βˆ’ 𝟏)𝟐

π’…π’šπŸ‘

𝟏+ 𝝅 ∫ (𝟐)𝟐 βˆ’ (π’šπŸ/πŸ‘ βˆ’ 𝟏)

πŸπ’…π’š

πŸπŸ•

πŸ‘

πŸ•πŸπŸ

πŸπŸ“π… β‰ˆ πŸπŸ’πŸ—. 𝟏𝟐𝟏

𝝅 ∫ (πŸ‘ βˆ’ π’šπŸ/πŸ‘)𝟐

βˆ’ (πŸ‘ βˆ’ π’š)πŸπ’…π’šπŸ‘

𝟏+ 𝝅 ∫ (πŸ‘ βˆ’ π’šπŸ/πŸ‘)

πŸπ’…π’š

πŸπŸ•

πŸ‘

πŸπŸ’πŸ–

πŸπŸ“π… β‰ˆ πŸ“πŸ. πŸ—πŸ’πŸ

𝝅 ∫ (𝟏𝟎 βˆ’ π’šπŸ/πŸ‘)𝟐

βˆ’ (𝟏𝟎 βˆ’ π’š)πŸπ’…π’šπŸ‘

𝟏+ 𝝅 ∫ (𝟏𝟎 βˆ’ π’šπŸ/πŸ‘)

πŸβˆ’ (πŸ•)πŸπ’…π’š

πŸπŸ•

πŸ‘

πŸ‘πŸ”πŸŽπŸ–

πŸπŸ“π… β‰ˆ πŸ•πŸ“πŸ“. πŸ”πŸ“πŸ–

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Volume: Shell Method – Class Work

Use the Shell Method to calculate the volume of the object created by rotating the described

region about the given axis.

175. 𝑦 = 1 + π‘₯2, 𝑦 = 0, π‘₯ = 1, π‘₯ = 3 revolved about the y-axis

176. 𝑦 = 8 βˆ’ π‘₯3, 𝑦 = 8 βˆ’ 4π‘₯ revolved about the y-axis

177. 𝑦 = π‘₯βˆ’4, 𝑦 = 0, π‘₯ = βˆ’3, π‘₯ = βˆ’1 revolved about π‘₯ = 4

178. 𝑦 = π‘₯2, 𝑦 = 8 βˆ’ π‘₯2, π‘₯ = 0 revolved about π‘₯ = βˆ’3

πŸπ… ∫ 𝒙(𝟏 + π’™πŸ)π’…π’™πŸ‘

𝟏

πŸ’πŸ–π… β‰ˆ πŸπŸ“πŸŽ. πŸ•πŸ—πŸ”

πŸπ… ∫ 𝒙 ((πŸ– βˆ’ π’™πŸ‘) βˆ’ (πŸ– βˆ’ πŸ’π’™)) π’…π’™πŸ

𝟎

πŸπŸπŸ–

πŸπŸ“π… β‰ˆ πŸπŸ”. πŸ–πŸŽπŸ–

πŸπ… ∫ (βˆ’π’™ + πŸ’)(π’™βˆ’πŸ’)π’…π’™βˆ’πŸ

βˆ’πŸ‘

πŸπŸ–πŸŽ

πŸ–πŸπ… β‰ˆ 𝟏𝟎. πŸ–πŸ”

πŸπ… ∫ (𝒙 + πŸ‘) ((πŸ– βˆ’ π’™πŸ) βˆ’ (π’™πŸ)) π’…π’™πŸ

𝟎

πŸ–πŸŽπ… β‰ˆ πŸπŸ“πŸ. πŸ‘πŸπŸ•

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179. 𝑦 = π‘₯, π‘₯ = 0, π‘₯ = 1 revolved about the x-axis

180. 𝑦 = π‘₯1/3 βˆ’ 2, 𝑦 = 0, π‘₯ = 8, π‘₯ = 27 revolved about 𝑦 = 4

181. 𝑦 = βˆšβˆ’π‘₯ βˆ’ 2, 𝑦 = 0, π‘₯ = βˆ’6, π‘₯ = βˆ’2 revolved about 𝑦 = βˆ’1

182. 𝑦 = √π‘₯ βˆ’ 2, 𝑦 = √10 βˆ’ π‘₯, 𝑦 = 0 revolved about 𝑦 = βˆ’6

πŸπ… ∫ π’š(π’š)π’…π’šπŸ

𝟎

𝟐

πŸ‘π… β‰ˆ 𝟐. πŸŽπŸ—πŸ’

πŸπ… ∫ (πŸ’ βˆ’ π’š)(π’š + 𝟐)πŸ‘π’…π’šπŸ

𝟎

πŸ“πŸ“πŸ‘

πŸ“π… β‰ˆ πŸ‘πŸ’πŸ•. πŸ’πŸ”

πŸπ… ∫ (π’š + 𝟏)(βˆ’π’šπŸ βˆ’ 𝟐)π’…π’šπŸŽ

𝟐

πŸ–πŸ–

πŸ‘π… β‰ˆ πŸ—πŸ. πŸπŸ“πŸ‘

πŸπ… ∫ (π’š + πŸ”) ((𝟏𝟎 βˆ’ π’šπŸ) βˆ’ (π’šπŸ + 𝟐)) π’…π’šπŸ

𝟎

πŸπŸ’πŸ’π… β‰ˆ πŸ’πŸ“πŸ. πŸ‘πŸ–πŸ—

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Volume: Shell Method – Homework

Use the Shell Method to calculate the volume of the object created by rotating the described

region about the given axis.

183. 𝑦 = 1 βˆ’ 2π‘₯ + 3π‘₯2 βˆ’ 2π‘₯3, 𝑦 = 0, π‘₯ = 0, π‘₯ = 1 revolved about the y-axis

184. 𝑦 = 9 βˆ’ π‘₯2, 𝑦 = 9 βˆ’ 3π‘₯ revolved about the y-axis

185. 𝑦 = π‘₯βˆ’1/2, 𝑦 = 0, π‘₯ = 1, π‘₯ = 4 revolved about π‘₯ = βˆ’3

186. 𝑦 = sin(π‘₯2) , 𝑦 = 0, π‘₯ = 0, π‘₯ = βˆšπœ‹ revolved about y-axis

πŸπ… ∫ 𝒙(𝟏 βˆ’ πŸπ’™ + πŸ‘π’™πŸ βˆ’ πŸπ’™πŸ‘)π’…π’™πŸ

𝟎

𝟏𝟏

πŸ‘πŸŽπ… β‰ˆ 𝟏. πŸπŸ“πŸ

πŸπ… ∫ 𝒙 ((πŸ— βˆ’ π’™πŸ) βˆ’ (πŸ— βˆ’ πŸ‘π’™)) π’…π’™πŸ‘

𝟎

πŸπŸ•

πŸπ… β‰ˆ πŸ’πŸ. πŸ’πŸπŸ

πŸπ… ∫ (𝒙 + πŸ‘)(π’™βˆ’πŸ/𝟐)π’…π’™πŸ’

𝟏

πŸ”πŸ’

πŸ‘π… β‰ˆ πŸ”πŸ•. 𝟎𝟐𝟏

πŸπ… ∫ 𝒙 𝐬𝐒𝐧(π’™πŸ ) π’…π’™βˆšπ…

𝟎

πŸπ… β‰ˆ πŸ”. πŸπŸ–πŸ‘

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187. 𝑦 = 3π‘₯ βˆ’ 1, 𝑦 = 2, π‘₯ = 1, π‘₯ = 3 revolved about the x-axis

188. 𝑦 = π‘₯3, 𝑦 = π‘₯ revolved about the x-axis

189. 𝑦 = √π‘₯ βˆ’ 2, 𝑦 = 0, π‘₯ = 1, π‘₯ = 4 revolved about 𝑦 = 5

190. 𝑦 = √π‘₯ βˆ’ 1, 𝑦 = 7 βˆ’ π‘₯, 𝑦 = 0 revolved about 𝑦 = βˆ’10

πŸπ… ∫ π’š (𝟏

πŸ‘π’š +

𝟏

πŸ‘) π’…π’š

πŸ–

𝟐

πŸπŸ‘πŸπ… β‰ˆ πŸ’πŸπŸ’. πŸ”πŸ—

πŸπ… ∫ π’š(π’šπŸ/πŸ‘ βˆ’ π’š)π’…π’šπŸ

𝟎

πŸ’

πŸπŸπ… β‰ˆ 𝟎. πŸ“πŸ—πŸ–

πŸπ… ∫ (βˆ’π’š + πŸ‘)(π’š + 𝟐)πŸπ’…π’šπŸŽ

βˆ’πŸ

πŸ—πŸ“

πŸ”π… β‰ˆ πŸ’πŸ—. πŸ•πŸ’πŸ

πŸπ… ∫ (π’š + 𝟏𝟎) ((πŸ• βˆ’ π’š) βˆ’ (π’šπŸ + 𝟏)) π’…π’šπŸ

𝟎

πŸ’πŸ•πŸ

πŸ‘π… β‰ˆ πŸ’πŸ—πŸ’. πŸπŸ•πŸ•

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Unit Review Multiple Choice

1. ∫4π‘₯+2

4π‘₯

4

1𝑑π‘₯ =

a. ln 4

b. 1

2ln 4

c. 3 +1

2ln 4

d. 5 +1

2ln 4

e. 2

2. ∫ (π‘₯2 βˆ’ 4π‘₯ + 7)𝑑π‘₯2

βˆ’2

a. 0

b. 16

3

c. 8

3

d. 100

3

e. βˆ’100

3

3. The area under 𝑦 =1

π‘₯ from π‘₯ = 1 to π‘₯ = 𝑒4 is split into two equal area regions by π‘₯ = π‘˜.

Find π‘˜.

a. 2

b. 2.5

c. 𝑒2

d. ln 4

e. 𝑒

4. 𝐹(π‘₯) = ∫ (𝑑2 βˆ’ 1)2π‘₯

π‘₯𝑑𝑑, 𝐹′(π‘₯) =

a. 2π‘₯2 βˆ’ π‘₯

b. 4π‘₯2 βˆ’ π‘₯

c. 8π‘₯2 βˆ’ π‘₯

d. 3π‘₯2 βˆ’ 1

e. 7π‘₯2 βˆ’ 1

5. ∫ π‘₯3𝑑π‘₯3

1 is approximated using right rectangular approximation method (RRAM), with 4

equal partitions. Find the approximate area and state whether it is under or over estimate.

a. 17.641 u2; under estimate

b. 17.641 u2; over estimate

c. 20 u2; over estimate

d. 20 u2; under estimate

e. 27 u2; over estimate

C

D

C

E

E

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6. Using the trapezoid rule and 𝑛 = 8, approximate ∫ (π‘₯2 βˆ’ 6)𝑑π‘₯4

2

a. 6.688

b. 3.839

c. 7.647

d. 6.667

e. 13.366

7. ∫ 𝑓(π‘₯)𝑑π‘₯6

3= 4, ∫ 𝑓(π‘₯) = βˆ’8𝑑π‘₯

10

6, π‘Žπ‘›π‘‘ ∫ 𝑓(π‘₯)𝑑π‘₯ = βˆ’5

10

8, then which of the following

statements is true?

a. ∫ π‘₯𝑓(π‘₯)6

3𝑑π‘₯ = 4π‘₯

b. ∫ 2𝑓(π‘₯)10

3𝑑π‘₯ = βˆ’6

c. ∫ (𝑓(π‘₯)10

8βˆ’ 3)𝑑π‘₯ = βˆ’8

d. ∫ 5𝑓(π‘₯)8

3𝑑π‘₯ = 5

e. ∫ 𝑓(π‘₯)10

3𝑑π‘₯ = βˆ’9

8. The area of the region bounded by the curves 𝑦 = 5 βˆ’ π‘₯2, 𝑦 = π‘₯2 βˆ’ 5, π‘₯ = 1, π‘Žπ‘›π‘‘ π‘₯ = 2 is

a. 16

3

b. 29

3

c. 34

3

d. 92

3

e. 32

9. The volume of solid formed by the region bound by 𝑦 = π‘₯2, 𝑦 = 0, π‘Žπ‘›π‘‘ π‘₯ = 1 revolved

about the x-axis is

a. πœ‹

b. πœ‹

2

c. πœ‹

3

d. πœ‹

4

e. πœ‹

5

10. The volume of the solid formed by the region bound by 𝑦 = π‘₯2, 𝑦 = βˆ’2π‘₯ + 3, and the x-axis

revolved around 𝑦 = βˆ’1 is

a. Ο€ ∫ (x2 βˆ’ 1)21

0dx + Ο€ ∫ (βˆ’2x + 2)2dx

1.5

1

b. Ο€ ∫ ((x2)2 βˆ’ 1)1

0dx + Ο€ ∫ ((βˆ’2x + 3)2 βˆ’ 1)dx

1.5

1

c. Ο€ ∫ ((x2 + 1)2 βˆ’ 1)1

0dx + Ο€ ∫ ((βˆ’2x + 4)2 βˆ’ 1)dx

1.5

1

d. Ο€ ∫ ((x2 βˆ’ 1)2 βˆ’ 1)1

0dx + Ο€ ∫ ((βˆ’2x + 2)2 βˆ’ 1)dx

1.5

1

e. Ο€ ∫ (x2)21

0dx + Ο€ ∫ (βˆ’2x + 2)2dx

1.5

1

A

D

A

E

C

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11. The area of the region bounded by the curves 𝑦2 = π‘₯ π‘Žπ‘›π‘‘ 𝑦 = βˆ’π‘₯ + 4 is

a. 11.682

b. 10.600

c. 6.486

d. 5.796

e. 5.408

12. The volume of solid formed by the region bound by 𝑦 = π‘₯2, 𝑦 = 0, π‘Žπ‘›π‘‘ π‘₯ = 2 revolved

about the x=2 is

a. 25.133

b. 16.755

c. 8.378

d. 6.283

e. 5.924

13. The volume of the solid formed by the region bound by 𝑦 = π‘₯2 + 1, 𝑦 = βˆ’π‘₯2 + 3, π‘Žπ‘›π‘‘ π‘₯ = 0

revolved around 𝑦 = 4 is

a. 3.161

b. 5.194

c. 11.854

d. 13.433

e. 16.755

14. Use substitution to evaluate ∫(π‘₯2 + 2)√π‘₯3 + 6π‘₯ βˆ’ 5𝑑π‘₯

a. 2

3(

1

4π‘₯4 + 3π‘₯2 βˆ’ 5π‘₯)

3/2+ 𝐢

b. 2

3(π‘₯3 + 6π‘₯ βˆ’ 5)3/2 + 𝐢

c. (π‘₯3 + 6π‘₯ βˆ’ 5)3/2 + 𝐢

d. 2

9(π‘₯3 + 6π‘₯ βˆ’ 5)3/2 + 𝐢

e. 2

9(3π‘₯2 + 6)3/2 + 𝐢

15. Use substitution to evaluate ∫ π‘₯(2π‘₯2 βˆ’ 7)5𝑑π‘₯3

1

a. 1,755,936

b. 73,164

c. 438,984

d. 292,656

e. 26,321

A

C

E

D

B

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Extended Response

1. Use the graph to answer the following:

a. ∫ 𝑓(π‘₯)𝑑π‘₯5

0

b. ∫ 𝑓(π‘₯)𝑑π‘₯βˆ’4

0

c. 𝑓(0)

d. 𝑓′(0)

2. The table represents the fuel consumption of a car at given times.

a. Approximate the fuel consumption using MRAM for 0 ≀ 𝑑 ≀ 8 and 4 rectangles.

b. What is the approximate rate of change in the fuel consumption at 𝑑 = 1?

c. If the maximum rate of fuel consumption occurs at 𝑑 = 5 min, what is the rate of

change in the fuel consumption at 𝑑 = 5? Explain.

time(min) 0 1 2 3 4 5 6 7 8

gal/min 2 3 4 2 3 5 3 4 2

πŸ’

βˆ’πŸ“

𝟐

𝟎

28 gallons

1 gal/min per min

0; because it is a max

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3. The graph of a velocity function, 𝑓′(π‘₯), is shown

a. How far does the particle travel for 3 ≀ π‘₯ ≀ 7?

b. What is the particles acceleration at π‘₯ = 5?

c. Is particle speed increasing or decreasing at π‘₯ = 6? Explain.

4. Region R is bound by 𝑦 = π‘₯ + 3, 𝑦 = 9 βˆ’ π‘₯2 and π‘₯ = 1.

a. find the area of R

b. Find the volume of the solid created by rotating R about π‘₯ = βˆ’1 using the

Washer Method

c. Find the volume of the solid created by rotating R about π‘₯ = 4 using the Shell

Method

πŸπŸ–. πŸπŸ–

𝟎

Decreasing; the slope is negative

πŸπŸ‘

πŸ”β‰ˆ 𝟐. πŸπŸ”πŸ•

πŸ”πŸ

πŸ”β‰ˆ 𝟏𝟎𝟎. πŸ‘πŸ’

πŸπŸ‘

πŸβ‰ˆ πŸ‘πŸ”. πŸπŸπŸ–