Riemann Sums Class Work - NJCTL

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

Riemann Sums – Class Work

1. Consider the region between 𝑦 = 9𝑥 − 𝑥3and the x-axis for 0 ≤ 𝑥 ≤ 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.

𝟏𝟗. 𝟔𝟖𝟕𝟓

𝟏𝟗. 𝟔𝟖𝟕𝟓

𝟐𝟎. 𝟓𝟑𝟏𝟐𝟓

LRAM: 𝟐𝟎. 𝟏𝟎𝟗𝟑𝟕𝟓

RRAM: 𝟐𝟎. 𝟏𝟎𝟗𝟑𝟕𝟓

MRAM: 𝟐𝟎. 𝟑𝟐𝟎𝟑𝟏𝟐𝟓

About 𝟐𝟎. 𝟐𝟓


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(𝑥) = √𝑥, [0,10], 𝑛 = 5

7. 𝑓(𝑥) = 𝑥3, [1,3], 𝑛 = 4

8. 𝑓(𝑥) = cos 𝑥, [0,2𝜋], 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: 𝟏𝟕. 𝟑𝟖𝟒

RRAM: 𝟐𝟑. 𝟕𝟎𝟖𝟖

MRAM: 𝟐𝟏. 𝟐𝟐𝟖

LRAM: 𝟏𝟒

RRAM: 𝟐𝟕

MRAM: 𝟏𝟗. 𝟕𝟓

LRAM: 𝟎

RRAM: 𝟎

MRAM: 𝟎

MRAM: 𝟔

Approximate number of gallons consumed

Gallons


Reimann Sums – Homework


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.


𝟒. 𝟑𝟕𝟓

𝟒. 𝟑𝟕𝟓

𝟒. 𝟑𝟕𝟓

LRAM: 𝟒. 𝟒𝟔𝟖𝟕𝟓

RRAM: 𝟒. 𝟒𝟔𝟖𝟕𝟓

MRAM: 𝟒. 𝟓𝟏𝟓𝟔𝟐𝟓

About 𝟒. 𝟓


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(𝑥) = 𝑥2, [1,9], 𝑛 = 4

18. 𝑓(𝑥) = √𝑥3 [0,8], 𝑛 = 4

19. 𝑓(𝑥) = sin 𝑥, [0, 𝜋], 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.



LRAM: 𝟏𝟔𝟖

RRAM: 𝟑𝟐𝟖

MRAM: 𝟐𝟒𝟎

LRAM: 𝟗. 𝟑𝟐𝟗

RRAM: 𝟏𝟑. 𝟑𝟐𝟗

MRAM: 𝟏𝟐. 𝟏𝟑

LRAM: 𝟏. 𝟖𝟗𝟔

RRAM: 𝟏. 𝟖𝟗𝟔

MRAM: 𝟐. 𝟎𝟓𝟐

MRAM: 𝟐𝟏𝟏𝟓𝟎

Approximate number of downloads

Downloads


Trapezoid Rule – Class Work


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.


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(𝑥) =1

𝑥, [2,4], 𝑛 = 4

27. 𝑓(𝑥) = 𝑥 − 𝑥3, [1,3], 𝑛 = 4

28. 𝑓(𝑥) = sin 𝑥, [0,2𝜋], 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.



𝟑. 𝟕𝟓

𝟑. 𝟗𝟑𝟕𝟓

About 𝟒

𝟎. 𝟔𝟗𝟕

−𝟏𝟔. 𝟓

𝟎

𝟏𝟎𝟖. 𝟕𝟓

Approximate number of miles driven

Miles


Trapezoid Rule – Homework

32. Consider the region between 𝑦 = 8 − 4𝑥 and the x-axis for 0 ≤ 𝑥 ≤ 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.


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(𝑥) = 𝑥2 − 4, [1,3], 𝑛 = 6

36. 𝑓(𝑥) = √4 − 𝑥[0,4], 𝑛 = 4

37. 𝑓(𝑥) = cos 𝑥, [0, 𝜋], 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.



𝟖

𝟖

Exactly 𝟖

𝟏𝟗

𝟐𝟕

𝟓. 𝟏𝟒𝟔

𝟎

𝟐𝟒𝟗𝟑𝟎

Approximate number of words typed

Words


Accumulation Functions – Class Work

Use the graph of 𝑓′(𝑥) to answer the following questions. 𝑓(0) = 2

41. ∫ 𝑓′(𝑥)𝑑𝑥3

0

42. ∫ 𝑓′(𝑥)𝑑𝑥5

0

43. ∫ 𝑓′(𝑥)𝑑𝑥0

−4

44. ∫ 𝑓′(𝑥)𝑑𝑥0

3

45. ∫ 𝑓′(𝑥)𝑑𝑥−4

5

46. 𝑓′(2)

47. 𝑓(2)

48. 𝑓"(2)

49. When is 𝑓"(𝑥) > 0?

𝟑

𝟐. 𝟓

𝟎

−𝟑

−𝟐. 𝟓

𝟏

𝟒

𝟎

[−𝟐, −𝟏]𝒂𝒏𝒅 [𝟒, 𝟓]


Homework

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

50. ∫ 𝑓′(𝑥)𝑑𝑥3

0

51. ∫ 𝑓′(𝑥)𝑑𝑥5

0

52. ∫ 𝑓′(𝑥)𝑑𝑥0

−4

53. ∫ 𝑓′(𝑥)𝑑𝑥−4

0

54. ∫ 𝑓′(𝑥)𝑑𝑥−4

5

55. 𝑓′(3)

56. 𝑓(4)

57. 𝑓"(−2)

58. When is 𝑓(𝑥) increasing?

𝟐. 𝟓

𝟒. 𝟓

𝟔. 𝟐𝟖

−𝟔. 𝟐𝟖

−𝟏𝟎. 𝟕𝟖

−𝟐

𝟑

𝟎

[𝟎, 𝟓]


Anti-Derivatives – Class Work

∫ 𝑓(𝑥)𝑑𝑥 = 42

−2

, ∫ 𝑔(𝑥)𝑑𝑥 = −32

−2

, ∫ 𝑓(𝑥)𝑑𝑥 = 8,5

2

∫ 𝑓(𝑥)𝑑𝑥 = 32

0

, ∫ 𝑓(𝑥)𝑑𝑥 = 25

8

59. ∫ (𝑓(𝑥) + 𝑔(𝑥))𝑑𝑥2

−2 60. ∫ (𝑓(𝑥) − 𝑔(𝑥))𝑑𝑥

2

−2 61. ∫ (3𝑓(𝑥) + |𝑔(𝑥)|)𝑑𝑥

2

−2

62. ∫ 𝑓(𝑥)𝑑𝑥5

−2 63. ∫ 4𝑓(𝑥)𝑑𝑥

8

2 64. ∫ 𝑓(𝑥)𝑑𝑥

8

−2 65. ∫ 𝑓(𝑥)𝑑𝑥

0

−2

Find the value of following definite integrals.

66. ∫ 3𝑑𝑥4

1 67. ∫ 𝑥𝑑𝑥

5

2 68. ∫ 4𝑥3𝑑𝑥

3

−2

69. ∫1

𝑥

5

1𝑑𝑥 70. ∫ 𝑒𝑥𝑑𝑥

6

0 71. ∫ (3𝑥2 + 6𝑥 − 5)𝑑𝑥

1

−2

72. ∫1

𝑥2 𝑑𝑥2

1 73. ∫ 𝑠𝑒𝑐2𝑥

𝜋

40

𝑑𝑥 74. ∫ sin 𝑥 𝑑𝑥2𝜋

0

75. ∫1

1+𝑥2 𝑑𝑥1

0

𝟏 𝟕 𝟏𝟓

𝟏𝟐 𝟐𝟒 𝟏𝟎 𝟏

𝟗 𝟏𝟎. 𝟓 𝟔𝟓

𝐥𝐧 𝟓 ≈ 𝟏. 𝟔𝟎𝟗 𝒆𝟔 − 𝟏 ≈ 𝟒𝟎𝟐. 𝟒𝟑 −𝟏𝟓

𝟏

𝟐 𝟏 𝟎

𝝅

𝟒


Anti-Derivatives – Homework

∫ 𝑓(𝑥)𝑑𝑥 = 52

−2

, ∫ 𝑔(𝑥)𝑑𝑥 = 92

−2

, ∫ 𝑓(𝑥)𝑑𝑥 = −6,5

2

∫ 𝑓(𝑥)𝑑𝑥 = −12

0

, ∫ 𝑓(𝑥)𝑑𝑥 = −75

8

76. ∫ (𝑓(𝑥) + 2𝑔(𝑥))𝑑𝑥2

−2 77. ∫ (2𝑓(𝑥) − 2𝑔(𝑥))𝑑𝑥

2

−2 78. ∫ (3𝑓(𝑥) + |𝑔(𝑥)|)𝑑𝑥

2

−2

79. ∫ (𝑓(𝑥) + 1)𝑑𝑥5

2 80. ∫ 3𝑓(𝑥)𝑑𝑥

8

2 81. ∫ 𝑓(𝑥)𝑑𝑥

8

−2 82. ∫ 𝑓(𝑥)𝑑𝑥

0

−2

Find the value of following definite integrals.

83. ∫ 4𝑑𝑥4

2 84. ∫ 2𝑥𝑑𝑥

5

2 85. ∫ 𝑥3𝑑𝑥

6

−1

86. ∫2

𝑥

6

1𝑑𝑥 87. ∫ (4𝑒𝑥 + 1)𝑑𝑥

5

0 88. ∫ (3𝑥2 + 6𝑥 − 5)𝑑𝑥

1

−2

89. ∫6

𝑥3 𝑑𝑥2

1 90. ∫ 𝑠𝑒𝑐𝑥𝑡𝑎𝑛𝑥

𝜋

40

𝑑𝑥 91. ∫ cos 𝑥 𝑑𝑥2𝜋

0

92. ∫1

√1−𝑥2𝑑𝑥

1

20

𝟐𝟑 −𝟖 𝟐𝟒

−𝟑 𝟑 𝟔 𝟔

𝟖 𝟐𝟏 𝟑𝟐𝟑. 𝟕𝟓

𝟐 𝐥𝐧 𝟔 ≈ 𝟑. 𝟓𝟖 𝟒𝒆𝟓 + 𝟏 ≈ 𝟓𝟗𝟒. 𝟔𝟓 𝟒

𝟐. 𝟐𝟓 √𝟐 − 𝟏 ≈ 𝟎. 𝟒𝟏𝟒 𝟎

𝝅

𝟔


Fundamental Theorem of Calculus – Class Work

Find 𝑑𝑦

𝑑𝑥

93. 𝑦 = ∫ (4𝑡 − 2)𝑑𝑡𝑥

1 94. 𝑦 = ∫ (3𝑢2 − 4𝑢)𝑑𝑢

2𝑥

2

95. 𝑦 = ∫ 𝑙𝑛(𝑣)𝑑𝑣4

𝑥 96. 𝑦 = ∫ (4𝑡3 − 2𝑡)𝑑𝑡

0

𝑥2

97. 𝑦 = ∫ (7𝑢)𝑑𝑢2𝑥

3𝑥2 98. 𝑦 = ∫ 𝑒𝑣𝑑𝑣𝑥 𝑙𝑛 𝑥

𝑙𝑛 𝑥

99. Let 𝐹(𝑥) = ∫ 𝑓(𝑡)𝑑𝑡,𝑥

0 where 𝑓(𝑡) is defined by the graph.

a. 𝑓(2)

b. 𝐹(2)

c. 𝐹′(2)

d. 𝑓′(2)

100. ∫ (5𝑢 − 6)𝑑𝑢 + 𝐾𝑥

−2= ∫ (5𝑢 − 6)𝑑𝑢

𝑥

4, find K

𝒅𝒚

𝒅𝒙= 𝟒𝒙 − 𝟐

𝒅𝒚

𝒅𝒙= 𝟐𝟒𝒙𝟐 − 𝟏𝟔𝒙

𝒅𝒚

𝒅𝒙= − 𝐥𝐧 𝒙

𝒅𝒚

𝒅𝒙= −𝟖𝒙𝟕 + 𝟒𝒙𝟑

𝒅𝒚

𝒅𝒙= 𝟐𝟖𝒙 − 𝟏𝟐𝟔𝒙𝟑 𝒅𝒚

𝒅𝒙= 𝐥𝐧 𝒙 𝒆𝒙 𝐥𝐧 𝒙 + 𝒆𝒙 𝐥𝐧 𝒙 −

𝒆𝐥𝐧 𝒙

𝒙

𝒇(𝟐) = 𝟒

𝑭(𝟐) = 𝟗

𝑭′(𝟐) = 𝟒

𝒇′(𝟐) = −𝟏

𝟐

𝑲 = 𝟔


Fundamental Theorem of Calculus – Homework

Find 𝑑𝑦

𝑑𝑥

101. 𝑦 = ∫ 𝑒𝑢𝑑𝑢𝑥

2 102. 𝑦 = ∫ 𝑡2𝑑𝑡

√𝑥

3

103. 𝑦 = ∫ 𝑠𝑖𝑛 𝑣 𝑑𝑣𝜋

𝑥 104. 𝑦 = ∫ √5 − 𝑡 𝑑𝑡

5

4−𝑥

105. 𝑦 = ∫ (𝑢2 − 4𝑢 + 2)𝑑𝑢7𝑥

2𝑥 106. 𝑦 = ∫ √𝑣

𝑥2

1

𝑥

𝑑𝑣

107. Let 𝐹(𝑥) = ∫ 𝑓(𝑡)𝑑𝑡,𝑥

0 where 𝑓(𝑡) is defined by the graph.

a. 𝑓(2)

b. 𝐹(2)

c. 𝐹′(2)

d. 𝑓′(2)

108. ∫ (3𝑢2 + 2𝑢 + 1)𝑑𝑢 + 𝐾𝑥

1= ∫ (3𝑢2 + 2𝑢 + 1)𝑑𝑢

𝑥

3, find K

𝒅𝒚

𝒅𝒙= 𝒆𝒙 𝒅𝒚

𝒅𝒙=

√𝒙

𝟐

𝒅𝒚

𝒅𝒙= − 𝐬𝐢𝐧 𝒙

𝒅𝒚

𝒅𝒙= √𝟏 + 𝒙

𝒅𝒚

𝒅𝒙= 𝟑𝟑𝟓𝒙𝟐 − 𝟏𝟖𝟎𝒙 + 𝟏𝟎 𝒅𝒚

𝒅𝒙= 𝟐𝒙𝟐 +

√𝒙

𝒙𝟑

𝒇(𝟐) = 𝟒

𝑭(𝟐) = 𝟔

𝑭′(𝟐) = 𝟒

𝒇′(𝟐) = 𝒖𝒏𝒅𝒆𝒇𝒊𝒏𝒆𝒅

𝑲 = −𝟑𝟔


Substitution Method – Class Work

Evaluate the indefinite integral using the Substitution Method

109. ∫ 2𝑥√𝑥2 + 1𝑑𝑥 110. ∫𝑥3

(𝑥4+1)4 𝑑𝑥

111. ∫ sin5 𝑥 cos 𝑥 𝑑𝑥 112. ∫ 𝑥 cos(𝑥2) 𝑑𝑥

113. ∫ 𝑥𝑒−𝑥2𝑑𝑥 114. ∫

𝑑𝑥

𝑥 ln 𝑥

Evaluate the definite integral

115. ∫𝑥

(𝑥2+1)3 𝑑𝑥1

0 116. ∫ (𝑥 + 1)(𝑥2 + 2𝑥)3𝑑𝑥

2

1

117. ∫ 𝑥 tan(𝑥2) 𝑑𝑥1

0 118. ∫ √5𝑥 + 6

2

−1𝑑𝑥

𝟐

𝟑(𝒙𝟐 + 𝟏)𝟑/𝟐 + 𝑪 −

𝟏

𝟏𝟐(𝒙𝟒+𝟏)𝟑 + 𝑪

𝟏

𝟔𝐬𝐢𝐧𝟔 𝒙 + 𝑪

𝟏

𝟐𝐬𝐢𝐧(𝒙𝟐) + 𝑪

−𝟏

𝟐𝒆−𝒙𝟐

+ 𝑪 𝐥𝐧(𝐥𝐧 𝒙) + 𝑪

𝟑

𝟏𝟔

𝟒𝟎𝟏𝟓

𝟖= 𝟓𝟎𝟏. 𝟖𝟕𝟓

𝟏

𝟑(𝟐√𝟐 − 𝟏) ≈ 𝟎. 𝟔𝟎𝟗 𝟖. 𝟒


Substitution Method – Homework

Evaluate the indefinite integral using the Substitution Method

119. ∫ 𝑥2(𝑥3 + 1)4𝑑𝑥 120. ∫1

(𝑥+2)2 𝑑𝑥

121. ∫2𝑥2+𝑥

(4𝑥3+3𝑥2)5 𝑑𝑥 122. ∫ sin(2𝑥 − 4) 𝑑𝑥

123. ∫𝑥

√𝑥2+9𝑑𝑥 124. ∫ sec2 𝑥 (4 tan3 𝑥 − 3 tan2 𝑥)𝑑𝑥

Evaluate the definite integral

125. ∫ 𝑥√𝑥2 + 94

0𝑑𝑥 126. ∫

𝑥+3

(𝑥2+6𝑥+1)3 𝑑𝑥2

0

127. ∫ (𝑥 − 9)−2/3𝑑𝑥17

10 128. ∫ tan2 𝑥 sec2 𝑥 𝑑𝑥

𝜋/4

0

𝟏

𝟏𝟓(𝒙𝟑 + 𝟏)𝟓 + 𝑪 −

𝟏

(𝒙+𝟐)+ 𝑪

−𝟏

𝟐𝟒(𝟒𝒙𝟑+𝟑𝒙𝟐)𝟒 + 𝑪 −

𝟏

𝟐𝐜𝐨𝐬(𝟐𝒙 − 𝟒) + 𝑪

√𝒙𝟐 + 𝟗 + 𝑪 𝐭𝐚𝐧𝟒 𝒙 − 𝐭𝐚𝐧𝟑 𝒙 + 𝑪

𝟗𝟖

𝟑≈ 𝟑𝟐. 𝟔𝟕

𝟕𝟐

𝟐𝟖𝟗≈ 𝟎. 𝟐𝟒𝟗

𝟑 𝟏

𝟑


Area Between Curves – Class Work

Find the total area between the functions.

129. 𝑦 = 𝑥4 − 6𝑥2 𝑎𝑛𝑑 𝑦 = 6 − 𝑥2 130. 𝑦 = cos 𝑥 𝑎𝑛𝑑 𝑦 = 2cos 𝑥 𝑓𝑜𝑟 0 ≤ 𝑥 ≤𝜋

2

131. 𝑦 = 𝑥3 − 3𝑥 𝑎𝑛𝑑 𝑦 = 𝑥 𝑓𝑜𝑟 𝑥 ≥ 0 132. 𝑥 = 𝑦2 − 6𝑦 𝑎𝑛𝑑 𝑥 + 𝑦 = 6

133. 𝑥 = 𝑦2 𝑎𝑛𝑑 𝑥 = 𝑦 134. 𝑦 = cos 𝑥 𝑎𝑛𝑑 𝑦 = 𝑥2 − 𝜋2

Area Between Curves – Homework

Find the total area between the functions.

135. 𝑦 = 2 + 3𝑥2 𝑎𝑛𝑑 𝑦 = 6 − 2𝑥2 136. 𝑦 = sin 𝑥 𝑎𝑛𝑑 𝑦 = cos 𝑥 𝑓𝑜𝑟 0 ≤ 𝑥 ≤ 𝜋

137. 𝑦 = −2𝑥3 + 10𝑥 𝑎𝑛𝑑 𝑦 = −8𝑥 𝑓𝑜𝑟 𝑥 ≤ 0 138. 𝑥 = −𝑦2 + 9 𝑎𝑛𝑑 𝑥 + 𝑦 = 3

139. 𝑥 = 𝑦2 𝑎𝑛𝑑 𝑥 = 𝑦2

3⁄ 140. 𝑦 = sin 𝑥 𝑎𝑛𝑑 𝑦 = 𝑥2 − 𝜋𝑥

𝟖𝟖

𝟓√𝟔 ≈ 𝟒𝟑. 𝟏𝟏𝟏 𝟏

𝟒 𝟑𝟒𝟑

𝟔≈ 𝟓𝟕. 𝟏𝟕

𝟏

𝟔 𝟒𝟏. 𝟓

𝟏𝟔

𝟑≈ 𝟓. 𝟑𝟑 𝟐√𝟐 ≈ 𝟐. 𝟖𝟐𝟖

𝟒𝟎. 𝟓 𝟏𝟐𝟓

𝟔≈ 𝟐𝟎. 𝟖𝟑

𝟒

𝟏𝟓≈ 𝟎. 𝟐𝟔𝟕 𝟐 +

𝟏

𝟔𝝅𝟑 ≈ 𝟕. 𝟏𝟕


Volume: Disk Method – Class Work

Find the volume of the solid.

141. 𝑦 = 6𝑥 − 𝑥2 revolved about the x-axis

142. Area between 𝑦 = 6𝑥 − 𝑥2 and the x-axis revolved about 𝑦 = −5

143. 𝑦 = 𝑥, 𝑦 = −2𝑥 + 4, 𝑎𝑛𝑑 𝑦 = 0 revolved about the x-axis

144. 𝑦 = 𝑥, 𝑦 = −2𝑥 + 4, 𝑎𝑛𝑑 𝑥 = 0 revolved about the y-axis

𝝅 ∫ (𝟔𝒙 − 𝒙𝟐)𝟐𝒅𝒙𝟔

𝟎

𝟏𝟐𝟗𝟔

𝟓𝝅 ≈ 𝟖𝟏𝟒. 𝟑

𝝅 ∫ (𝟓 + 𝟔𝒙 − 𝒙𝟐)𝟐𝒅𝒙𝟔

𝟎

𝟑𝟖𝟒𝟔

𝟓𝝅 ≈ 𝟐𝟒𝟏𝟔. 𝟓𝟏

𝝅 ∫ (𝒙)𝟐𝒅𝒙𝟒/𝟑

𝟎+ 𝝅 ∫ (𝟒 − 𝟐𝒙)𝟐𝒅𝒙

𝟐

𝟒/𝟑

𝟑𝟐

𝟐𝟕𝝅 ≈ 𝟑. 𝟕𝟐𝟑

𝝅 ∫ (𝒚)𝟐𝒅𝒚𝟒/𝟑

𝟎+ 𝝅 ∫ (𝟐 −

𝟏

𝟐𝒚)

𝟐

𝒅𝒚𝟒

𝟒/𝟑

𝟔𝟒

𝟐𝟕𝝅 ≈ 𝟕. 𝟒𝟓


145. 𝑦 = 2𝑥, 𝑥 = 3, 𝑎𝑛𝑑 𝑥 − 𝑎𝑥𝑖𝑠 revolved about the x-axis

146. 𝑦 = 2𝑥, 𝑥 = 3, 𝑎𝑛𝑑 𝑥 − 𝑎𝑥𝑖𝑠 revolved about the 𝑥 = 3

147. 𝑦 = 6 − 𝑥, 𝑦 = 𝑥2, 𝑎𝑛𝑑 𝑦 = 0 revolved about the x-axis

148. 𝑦 = 6 − 𝑥, 𝑦 = 𝑥2, 𝑎𝑛𝑑 𝑥 = 0 revolved about the y-axis

𝝅 ∫ (𝟐𝒙)𝟐𝒅𝒙𝟑

𝟎

𝟑𝟔𝝅 ≈ 𝟏𝟏𝟑. 𝟎𝟗𝟕

𝝅 ∫ (𝟑 −𝟏

𝟐𝒚)

𝟐

𝒅𝒚𝟔

𝟎

𝟏𝟖𝝅 ≈ 𝟓𝟔. 𝟓𝟒𝟗

𝝅 ∫ (𝒙𝟐)𝟐𝒅𝒙𝟐

𝟎+ 𝝅 ∫ (𝟔 − 𝒙)𝟐𝒅𝒙

𝟔

𝟐

𝟒𝟏𝟔

𝟏𝟓𝝅 ≈ 𝟖𝟕. 𝟏𝟐𝟕

𝝅 ∫ (√𝒚)𝟐

𝒅𝒚𝟒

𝟎+ 𝝅 ∫ (𝟔 − 𝒚)𝟐𝒅𝒚

𝟔

𝟒

𝟒𝟒𝟖

𝟑𝝅 ≈ 𝟒𝟔𝟗. 𝟏𝟒𝟓


Volume: Disk Method – Homework

Find the volume of the solid.

149. 𝑦 = 8𝑥 − 2𝑥2 revolved about the x-axis

150. 𝑦 = 8𝑥 − 2𝑥2 and the x-axis revolved about 𝑦 = −3

151. 𝑦 = 3𝑥, 𝑦 = −𝑥2 + 4, 𝑎𝑛𝑑 𝑦 = 0 revolved about the x-axis

152. 𝑦 = 3𝑥, 𝑦 = −𝑥2 + 4, 𝑎𝑛𝑑 𝑥 = 0 revolved about the y-axis

𝝅 ∫ (𝟖𝒙 − 𝟐𝒙𝟐)𝟐𝒅𝒙𝟒

𝟎

𝟐𝟎𝟒𝟖

𝟏𝟓𝝅 ≈ 𝟒𝟐𝟖. 𝟗𝟑𝟐

𝝅 ∫ (𝟑 + 𝟖𝒙 − 𝟐𝒙𝟐)𝟐𝒅𝒙𝟒

𝟎

𝟒𝟓𝟎𝟖

𝟏𝟓𝝅 ≈ 𝟗𝟒𝟒. 𝟏𝟓𝟑

𝝅 ∫ (𝟑𝒙)𝟐𝒅𝒙𝟏

𝟎+ 𝝅 ∫ (𝟒 − 𝒙𝟐)𝟐𝒅𝒙

𝟐

𝟏

𝟗𝟖

𝟏𝟓𝝅 ≈ 𝟐𝟎. 𝟓𝟐𝟓

𝝅 ∫ (𝟏

𝟑𝒚)

𝟐

𝒅𝒚𝟑

𝟎+ 𝝅 ∫ (√𝟒 − 𝒚)

𝟐𝒅𝒚

𝟒

𝟑

𝟑

𝟐𝝅 ≈ 𝟒. 𝟕𝟏𝟐


153. 𝑦 = 3𝑥, 𝑥 = 4, 𝑎𝑛𝑑 𝑥 − 𝑎𝑥𝑖𝑠 revolved about the x-axis

154. 𝑦 = 3𝑥, 𝑥 = 4, 𝑎𝑛𝑑 𝑥 − 𝑎𝑥𝑖𝑠 revolved about the 𝑥 = 4

155. 𝑦 = 𝑥2 + 2, 𝑦 = 10 − 𝑥2, 𝑦 = 0, 𝑥 = 0, 𝑎𝑛𝑑 𝑥 = 3 revolved about the x-axis

156. 𝑦 = 𝑥2 + 2, 𝑦 = 10 − 𝑥2, 𝑥 = 0, 𝑎𝑛𝑑 𝑥 = 3 revolved about the x-axis

𝝅 ∫ (𝟑𝒙)𝟐𝒅𝒙𝟒

𝟎

𝟏𝟗𝟐𝝅 ≈ 𝟔𝟎𝟑. 𝟏𝟖𝟔

𝝅 ∫ (𝟒 −𝟏

𝟑𝒚)

𝟐

𝒅𝒚𝟏𝟐

𝟎

𝟔𝟒𝝅 ≈ 𝟐𝟎𝟏. 𝟎𝟔

𝝅 ∫ (𝒙𝟐 + 𝟐)𝟐𝒅𝒙𝟐

𝟎+ 𝝅 ∫ (𝟏𝟎 − 𝒙𝟐)𝟐𝒅𝒙

𝟑

𝟐

𝟐𝟎𝟑

𝟓𝝅 ≈ 𝟏𝟐𝟕. 𝟓𝟓

𝝅 ∫ ((𝟏𝟎 − 𝒙𝟐) − (𝒙𝟐 + 𝟐))𝟐

𝒅𝒙𝟐

𝟎+ 𝝅 ∫ ((𝒙𝟐 + 𝟐) − (𝟏𝟎 − 𝒙𝟐))

𝟐

𝒅𝒙𝟑

𝟐

𝟒𝟗𝟐

𝟓𝝅 ≈ 𝟑𝟎𝟗. 𝟏𝟑𝟑


Volume: Washer Method – Class Work

Find the solid created by rotating the region 𝑦 = 𝑥2, 𝑦 = 𝑥, 𝑥 = 1, 𝑎𝑛𝑑 𝑥 = 4 about

157. 𝑦 = 0

158. 𝑦 = 1

159. 𝑦 = 100

160. 𝑦 = −2

161. 𝑥 = 0

𝝅 ∫ (𝒙𝟐)𝟐 − (𝒙)𝟐𝒅𝒙𝟒

𝟏

𝟗𝟏𝟖

𝟓𝝅 ≈ 𝟓𝟕𝟔. 𝟕𝟗𝟔

𝝅 ∫ (𝒙𝟐 − 𝟏)𝟐 − (𝒙 − 𝟏)𝟐𝒅𝒙𝟒

𝟏

𝟕𝟖𝟑

𝟓𝝅 ≈ 𝟒𝟗𝟏. 𝟗𝟕𝟑

𝝅 ∫ (𝟏𝟎𝟎 − 𝒙)𝟐 − (𝟏𝟎𝟎 − 𝒙𝟐)𝟐𝒅𝒙𝟒

𝟏

𝟏𝟐𝟓𝟖𝟐

𝟓𝝅 ≈ 𝟕𝟗𝟎𝟓. 𝟓𝟎𝟒

𝝅 ∫ (𝟐 + 𝒙𝟐)𝟐 − (𝟐 + 𝒙)𝟐𝒅𝒙𝟒

𝟏

𝟏𝟏𝟖𝟖

𝟓𝝅 ≈ 𝟕𝟒𝟔. 𝟒𝟒𝟐

𝝅 ∫ (𝒚)𝟐 − (√𝒚)𝟐

𝒅𝒚𝟒

𝟏+ 𝝅 ∫ (𝟒)𝟐 − (√𝒚)

𝟐𝒅𝒚

𝟏𝟔

𝟒

𝟏𝟕𝟏

𝟐𝝅 ≈ 𝟐𝟔𝟖. 𝟔𝟎𝟔


162. 𝑥 = −2

163. 𝑥 = 1

164. 𝑥 = 4

165. 𝑥 = 10

𝝅 ∫ (𝟐 + 𝒚)𝟐 − (𝟐 + √𝒚)𝟐

𝒅𝒚𝟒

𝟏+ 𝝅 ∫ (𝟔)𝟐 − (𝟐 + √𝒚)

𝟐𝒅𝒚

𝟏𝟔

𝟒

𝟐𝟕𝟗

𝟐𝝅 ≈ 𝟒𝟑𝟖. 𝟐𝟓𝟐

𝝅 ∫ (𝒚 − 𝟏)𝟐 − (√𝒚 − 𝟏)𝟐

𝒅𝒚𝟒

𝟏+ 𝝅 ∫ (𝟓)𝟐 − (√𝒚 − 𝟏)

𝟐𝒅𝒚

𝟏𝟔

𝟒

𝟓𝟎𝟏

𝟐𝝅 ≈ 𝟕𝟖𝟔. 𝟗𝟔𝟗

𝝅 ∫ (𝟒 − √𝒚)𝟐

− (𝟒 − 𝒚)𝟐𝒅𝒚𝟒

𝟏+ 𝝅 ∫ (𝟒 − √𝒚)

𝟐𝒅𝒚

𝟏𝟔

𝟒

𝟒𝟓

𝟐𝝅 ≈ 𝟕𝟎. 𝟔𝟖𝟔

𝝅 ∫ (𝟏𝟎 − √𝒚)𝟐

− (𝟏𝟎 − 𝒚)𝟐𝒅𝒚𝟒

𝟏+ 𝝅 ∫ (𝟏𝟎 − √𝒚)

𝟐− (𝟔)𝟐𝒅𝒚

𝟏𝟔

𝟒

𝟑𝟔𝟗

𝟐𝝅 ≈ 𝟓𝟕𝟗. 𝟔𝟐𝟒


Volume: Washer Method – Homework

Find the solid created by rotating the region 𝑦 = −𝑥3, 𝑦 = −𝑥, 𝑥 = 1, 𝑎𝑛𝑑 𝑥 = 3 about

166. 𝑦 = 0

167. 𝑦 = 10

168. 𝑦 = −1

169. 𝑦 = −50

170. 𝑥 = 0

𝝅 ∫ (−𝒙𝟑)𝟐 − (−𝒙)𝟐𝒅𝒙𝟑

𝟏

𝟔𝟑𝟕𝟔

𝟐𝟏𝝅 ≈ 𝟗𝟓𝟑. 𝟖𝟒𝟕

𝝅 ∫ (𝟏𝟎 + 𝒙𝟑)𝟐 − (𝟏𝟎 + 𝒙)𝟐𝒅𝒙𝟑

𝟏

𝟏𝟑𝟎𝟗𝟔

𝟐𝟏𝝅 ≈ 𝟏𝟗𝟓𝟗. 𝟏𝟓𝟕

𝝅 ∫ (𝟏 − 𝒙𝟑)𝟐 − (𝟏 − 𝒙)𝟐𝒅𝒙𝟑

𝟏

𝟓𝟕𝟎𝟒

𝟐𝟏𝝅 ≈ 𝟖𝟓𝟑. 𝟑𝟏𝟔

𝝅 ∫ (𝟓𝟎 − 𝒙)𝟐 − (𝟓𝟎 − 𝒙𝟑)𝟐𝒅𝒙𝟑

𝟏

𝟐𝟕𝟐𝟐𝟒

𝟐𝟏𝝅 ≈ 𝟒𝟎𝟕𝟐. 𝟕𝟎𝟏

𝝅 ∫ (𝒚)𝟐 − (𝒚𝟏/𝟑)𝟐

𝒅𝒚𝟑

𝟏+ 𝝅 ∫ (𝟑)𝟐 − (𝒚𝟏/𝟑)

𝟐𝒅𝒚

𝟐𝟕

𝟑

𝟏𝟏𝟗𝟐

𝟏𝟓𝝅 ≈ 𝟐𝟒𝟗. 𝟔𝟓𝟐


171. 𝑥 = −2

172. 𝑥 = 1

173. 𝑥 = 3

174. 𝑥 = 10

𝝅 ∫ (𝟐 + 𝒚)𝟐 − (𝟐 + 𝒚𝟏/𝟑)𝟐

𝒅𝒚𝟑

𝟏+ 𝝅 ∫ (𝟓)𝟐 − (𝟐 + 𝒚𝟏/𝟑)

𝟐𝒅𝒚

𝟐𝟕

𝟑

𝟐𝟏𝟓𝟐

𝟏𝟓𝝅 ≈ 𝟒𝟓𝟎. 𝟕𝟏𝟒

𝝅 ∫ (𝒚 − 𝟏)𝟐 − (𝒚𝟏/𝟑 − 𝟏)𝟐

𝒅𝒚𝟑

𝟏+ 𝝅 ∫ (𝟐)𝟐 − (𝒚𝟏/𝟑 − 𝟏)

𝟐𝒅𝒚

𝟐𝟕

𝟑

𝟕𝟏𝟐

𝟏𝟓𝝅 ≈ 𝟏𝟒𝟗. 𝟏𝟐𝟏

𝝅 ∫ (𝟑 − 𝒚𝟏/𝟑)𝟐

− (𝟑 − 𝒚)𝟐𝒅𝒚𝟑

𝟏+ 𝝅 ∫ (𝟑 − 𝒚𝟏/𝟑)

𝟐𝒅𝒚

𝟐𝟕

𝟑

𝟐𝟒𝟖

𝟏𝟓𝝅 ≈ 𝟓𝟏. 𝟗𝟒𝟏

𝝅 ∫ (𝟏𝟎 − 𝒚𝟏/𝟑)𝟐

− (𝟏𝟎 − 𝒚)𝟐𝒅𝒚𝟑

𝟏+ 𝝅 ∫ (𝟏𝟎 − 𝒚𝟏/𝟑)

𝟐− (𝟕)𝟐𝒅𝒚

𝟐𝟕

𝟑

𝟑𝟔𝟎𝟖

𝟏𝟓𝝅 ≈ 𝟕𝟓𝟓. 𝟔𝟓𝟖


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 + 𝑥2, 𝑦 = 0, 𝑥 = 1, 𝑥 = 3 revolved about the y-axis

176. 𝑦 = 8 − 𝑥3, 𝑦 = 8 − 4𝑥 revolved about the y-axis

177. 𝑦 = 𝑥−4, 𝑦 = 0, 𝑥 = −3, 𝑥 = −1 revolved about 𝑥 = 4

178. 𝑦 = 𝑥2, 𝑦 = 8 − 𝑥2, 𝑥 = 0 revolved about 𝑥 = −3

𝟐𝝅 ∫ 𝒙(𝟏 + 𝒙𝟐)𝒅𝒙𝟑

𝟏

𝟒𝟖𝝅 ≈ 𝟏𝟓𝟎. 𝟕𝟗𝟔

𝟐𝝅 ∫ 𝒙 ((𝟖 − 𝒙𝟑) − (𝟖 − 𝟒𝒙)) 𝒅𝒙𝟐

𝟎

𝟏𝟐𝟖

𝟏𝟓𝝅 ≈ 𝟐𝟔. 𝟖𝟎𝟖

𝟐𝝅 ∫ (−𝒙 + 𝟒)(𝒙−𝟒)𝒅𝒙−𝟏

−𝟑

𝟐𝟖𝟎

𝟖𝟏𝝅 ≈ 𝟏𝟎. 𝟖𝟔

𝟐𝝅 ∫ (𝒙 + 𝟑) ((𝟖 − 𝒙𝟐) − (𝒙𝟐)) 𝒅𝒙𝟐

𝟎

𝟖𝟎𝝅 ≈ 𝟐𝟓𝟏. 𝟑𝟐𝟕


179. 𝑦 = 𝑥, 𝑥 = 0, 𝑥 = 1 revolved about the x-axis

180. 𝑦 = 𝑥1/3 − 2, 𝑦 = 0, 𝑥 = 8, 𝑥 = 27 revolved about 𝑦 = 4

181. 𝑦 = √−𝑥 − 2, 𝑦 = 0, 𝑥 = −6, 𝑥 = −2 revolved about 𝑦 = −1

182. 𝑦 = √𝑥 − 2, 𝑦 = √10 − 𝑥, 𝑦 = 0 revolved about 𝑦 = −6

𝟐𝝅 ∫ 𝒚(𝒚)𝒅𝒚𝟏

𝟎

𝟐

𝟑𝝅 ≈ 𝟐. 𝟎𝟗𝟒

𝟐𝝅 ∫ (𝟒 − 𝒚)(𝒚 + 𝟐)𝟑𝒅𝒚𝟏

𝟎

𝟓𝟓𝟑

𝟓𝝅 ≈ 𝟑𝟒𝟕. 𝟒𝟔

𝟐𝝅 ∫ (𝒚 + 𝟏)(−𝒚𝟐 − 𝟐)𝒅𝒚𝟎

𝟐

𝟖𝟖

𝟑𝝅 ≈ 𝟗𝟐. 𝟏𝟓𝟑

𝟐𝝅 ∫ (𝒚 + 𝟔) ((𝟏𝟎 − 𝒚𝟐) − (𝒚𝟐 + 𝟐)) 𝒅𝒚𝟐

𝟎

𝟏𝟒𝟒𝝅 ≈ 𝟒𝟓𝟐. 𝟑𝟖𝟗


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 − 2𝑥 + 3𝑥2 − 2𝑥3, 𝑦 = 0, 𝑥 = 0, 𝑥 = 1 revolved about the y-axis

184. 𝑦 = 9 − 𝑥2, 𝑦 = 9 − 3𝑥 revolved about the y-axis

185. 𝑦 = 𝑥−1/2, 𝑦 = 0, 𝑥 = 1, 𝑥 = 4 revolved about 𝑥 = −3

186. 𝑦 = sin(𝑥2) , 𝑦 = 0, 𝑥 = 0, 𝑥 = √𝜋 revolved about y-axis

𝟐𝝅 ∫ 𝒙(𝟏 − 𝟐𝒙 + 𝟑𝒙𝟐 − 𝟐𝒙𝟑)𝒅𝒙𝟏

𝟎

𝟏𝟏

𝟑𝟎𝝅 ≈ 𝟏. 𝟏𝟓𝟐

𝟐𝝅 ∫ 𝒙 ((𝟗 − 𝒙𝟐) − (𝟗 − 𝟑𝒙)) 𝒅𝒙𝟑

𝟎

𝟐𝟕

𝟐𝝅 ≈ 𝟒𝟐. 𝟒𝟏𝟐

𝟐𝝅 ∫ (𝒙 + 𝟑)(𝒙−𝟏/𝟐)𝒅𝒙𝟒

𝟏

𝟔𝟒

𝟑𝝅 ≈ 𝟔𝟕. 𝟎𝟐𝟏

𝟐𝝅 ∫ 𝒙 𝐬𝐢𝐧(𝒙𝟐 ) 𝒅𝒙√𝝅

𝟎

𝟐𝝅 ≈ 𝟔. 𝟐𝟖𝟑


187. 𝑦 = 3𝑥 − 1, 𝑦 = 2, 𝑥 = 1, 𝑥 = 3 revolved about the x-axis

188. 𝑦 = 𝑥3, 𝑦 = 𝑥 revolved about the x-axis

189. 𝑦 = √𝑥 − 2, 𝑦 = 0, 𝑥 = 1, 𝑥 = 4 revolved about 𝑦 = 5

190. 𝑦 = √𝑥 − 1, 𝑦 = 7 − 𝑥, 𝑦 = 0 revolved about 𝑦 = −10

𝟐𝝅 ∫ 𝒚 (𝟏

𝟑𝒚 +

𝟏

𝟑) 𝒅𝒚

𝟖

𝟐

𝟏𝟑𝟐𝝅 ≈ 𝟒𝟏𝟒. 𝟔𝟗

𝟐𝝅 ∫ 𝒚(𝒚𝟏/𝟑 − 𝒚)𝒅𝒚𝟏

𝟎

𝟒

𝟐𝟏𝝅 ≈ 𝟎. 𝟓𝟗𝟖

𝟐𝝅 ∫ (−𝒚 + 𝟑)(𝒚 + 𝟐)𝟐𝒅𝒚𝟎

−𝟏

𝟗𝟓

𝟔𝝅 ≈ 𝟒𝟗. 𝟕𝟒𝟐

𝟐𝝅 ∫ (𝒚 + 𝟏𝟎) ((𝟕 − 𝒚) − (𝒚𝟐 + 𝟏)) 𝒅𝒚𝟐

𝟎

𝟒𝟕𝟐

𝟑𝝅 ≈ 𝟒𝟗𝟒. 𝟐𝟕𝟕


Unit Review Multiple Choice

1. ∫4𝑥+2

4𝑥

4

1𝑑𝑥 =

a. ln 4

b. 1

2ln 4

c. 3 +1

2ln 4

d. 5 +1

2ln 4

e. 2

2. ∫ (𝑥2 − 4𝑥 + 7)𝑑𝑥2

−2

a. 0

b. 16

3

c. 8

3

d. 100

3

e. −100

3

3. The area under 𝑦 =1

𝑥 from 𝑥 = 1 to 𝑥 = 𝑒4 is split into two equal area regions by 𝑥 = 𝑘.

Find 𝑘.

a. 2

b. 2.5

c. 𝑒2

d. ln 4

e. 𝑒

4. 𝐹(𝑥) = ∫ (𝑡2 − 1)2𝑥

𝑥𝑑𝑡, 𝐹′(𝑥) =

a. 2𝑥2 − 𝑥

b. 4𝑥2 − 𝑥

c. 8𝑥2 − 𝑥

d. 3𝑥2 − 1

e. 7𝑥2 − 1

5. ∫ 𝑥3𝑑𝑥3

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


6. Using the trapezoid rule and 𝑛 = 8, approximate ∫ (𝑥2 − 6)𝑑𝑥4

2

a. 6.688

b. 3.839

c. 7.647

d. 6.667

e. 13.366

7. ∫ 𝑓(𝑥)𝑑𝑥6

3= 4, ∫ 𝑓(𝑥) = −8𝑑𝑥

10

6, 𝑎𝑛𝑑 ∫ 𝑓(𝑥)𝑑𝑥 = −5

10

8, then which of the following

statements is true?

a. ∫ 𝑥𝑓(𝑥)6

3𝑑𝑥 = 4𝑥

b. ∫ 2𝑓(𝑥)10

3𝑑𝑥 = −6

c. ∫ (𝑓(𝑥)10

8− 3)𝑑𝑥 = −8

d. ∫ 5𝑓(𝑥)8

3𝑑𝑥 = 5

e. ∫ 𝑓(𝑥)10

3𝑑𝑥 = −9

8. The area of the region bounded by the curves 𝑦 = 5 − 𝑥2, 𝑦 = 𝑥2 − 5, 𝑥 = 1, 𝑎𝑛𝑑 𝑥 = 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 = 𝑥2, 𝑦 = 0, 𝑎𝑛𝑑 𝑥 = 1 revolved

about the x-axis is

a. 𝜋

b. 𝜋

2

c. 𝜋

3

d. 𝜋

4

e. 𝜋

5

10. The volume of the solid formed by the region bound by 𝑦 = 𝑥2, 𝑦 = −2𝑥 + 3, and the x-axis

revolved around 𝑦 = −1 is

a. π ∫ (x2 − 1)21

0dx + π ∫ (−2x + 2)2dx

1.5

1

b. π ∫ ((x2)2 − 1)1

0dx + π ∫ ((−2x + 3)2 − 1)dx

1.5

1

c. π ∫ ((x2 + 1)2 − 1)1

0dx + π ∫ ((−2x + 4)2 − 1)dx

1.5

1

d. π ∫ ((x2 − 1)2 − 1)1

0dx + π ∫ ((−2x + 2)2 − 1)dx

1.5

1

e. π ∫ (x2)21

0dx + π ∫ (−2x + 2)2dx

1.5

1

A

D

A

E

C


11. The area of the region bounded by the curves 𝑦2 = 𝑥 𝑎𝑛𝑑 𝑦 = −𝑥 + 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 = 𝑥2, 𝑦 = 0, 𝑎𝑛𝑑 𝑥 = 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 = 𝑥2 + 1, 𝑦 = −𝑥2 + 3, 𝑎𝑛𝑑 𝑥 = 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 ∫(𝑥2 + 2)√𝑥3 + 6𝑥 − 5𝑑𝑥

a. 2

3(

1

4𝑥4 + 3𝑥2 − 5𝑥)

3/2+ 𝐶

b. 2

3(𝑥3 + 6𝑥 − 5)3/2 + 𝐶

c. (𝑥3 + 6𝑥 − 5)3/2 + 𝐶

d. 2

9(𝑥3 + 6𝑥 − 5)3/2 + 𝐶

e. 2

9(3𝑥2 + 6)3/2 + 𝐶

15. Use substitution to evaluate ∫ 𝑥(2𝑥2 − 7)5𝑑𝑥3

1

a. 1,755,936

b. 73,164

c. 438,984

d. 292,656

e. 26,321

A

C

E

D

B


Extended Response

1. Use the graph to answer the following:

a. ∫ 𝑓(𝑥)𝑑𝑥5

0

b. ∫ 𝑓(𝑥)𝑑𝑥−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 ≤ 𝑡 ≤ 8 and 4 rectangles.

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

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

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

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

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

𝟒

−𝟓

𝟐

𝟎

28 gallons

1 gal/min per min

0; because it is a max


3. The graph of a velocity function, 𝑓′(𝑥), is shown

a. How far does the particle travel for 3 ≤ 𝑥 ≤ 7?

b. What is the particles acceleration at 𝑥 = 5?

c. Is particle speed increasing or decreasing at 𝑥 = 6? Explain.

4. Region R is bound by 𝑦 = 𝑥 + 3, 𝑦 = 9 − 𝑥2 and 𝑥 = 1.

a. find the area of R

b. Find the volume of the solid created by rotating R about 𝑥 = −1 using the

Washer Method

c. Find the volume of the solid created by rotating R about 𝑥 = 4 using the Shell

Method

𝟏𝟖. 𝟐𝟖

𝟎

Decreasing; the slope is negative

𝟏𝟑

𝟔≈ 𝟐. 𝟏𝟔𝟕

𝟔𝟏

𝟔≈ 𝟏𝟎𝟎. 𝟑𝟒

𝟐𝟑

𝟐≈ 𝟑𝟔. 𝟏𝟐𝟖

Documents

Riemann Sums Class Work - NJCTL