contents 0.1 mth.121.05.04 ...faculty.essex.edu/~bannon/m121.r/notes/mth.121.review.04.pdfron bannon...

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Calculus IMTH-121

Essex County CollegeDivision of Mathematics

AssessmentsCourse Overview

Contents

0.1 mth.121.05.04 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20.2 mth.121.05.05 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 80.3 mth.121.05.06 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100.4 mth.121.06.01 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 200.5 mth.121.06.02 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 240.6 mth.121.06.03 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 270.7 mth.121.06.04 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29

The question presented here are representative of problems students will seeonline or from the textbook. Errors should be reported to

Ron Bannon [email protected].

This document may be shared with students and instructors of MTH 121 atEssex County College, Newark, NJ.

The Final Exam will have questions taken from homework sections mth.121.05.04 throughmth.121.06.04, and some questions from older material.

[email protected] LATEX 2ε

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0.1 mth.121.05.04

1. Find a formula for the area of the function f (x) = 8x+ 3 with the lower limit a = 2.

2. Let G (x) =

∫ x

1

(t2 − 3

)dt.

(a) G (1) =

Solution:

G (1) =

∫ 1

1

(t2 − 3

)dt = 0

(b) G′ (1) =

Solution:

G (x) =

∫ x

1

(t2 − 3

)dt

G′ (x) = x2 − 3

G′ (1) = −2

(c) G′ (2) =

Solution:

G (x) =

∫ x

1

(t2 − 3

)dt

G′ (x) = x2 − 3

G′ (2) = 1

(d) Find a formula for G (x).

Solution:

G (x) =

∫ x

1

(t2 − 3

)dt

=

(t3

3− 3t

)∣∣∣∣x1

=x3 − 9x+ 8

3

3. Let F (x) =

∫ x

−6

(1

t2 + 7

)dt.

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(a) F (−6) =

Solution:

F (−6) =

∫ −6−6

(1

t2 + 7

)dt = 0

(b) F ′ (−6) =

Solution:

F (x) =

∫ x

−6

(1

t2 + 7

)dt

F ′ (x) =1

x2 + 7

F ′ (−6) =1

43

4. Find the formula for the function represented by the integral.∫ √x3

1

6t+ 2dt.

Solution: ∫ √x3

1

6t+ 2dt. =

(ln |6t+ 2|

6

)∣∣∣∣√x

3

=1

6ln

(3√x+ 1

10

)

5. The antiderivative F (x) of f (x) =x+ 3

x2 + 5satisfying the initial condition F (5) = 0 is

given by F (x) =

∫ b

a

t+ 3

t2 + 5dt. Find a and b.

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Solution: a = 0 and b = x. Here’s why.

F ′ (x) = f (x) =x+ 3

x2 + 5∫x+ 3

x2 + 5dx = F (x) + C∫ x

a

t+ 3

t2 + 5dt = F (x)− F (a)∫ x

5

t+ 3

t2 + 5dt = F (x)− F (5) = F (x)

6. Let A (x) =

∫ x

0

f (t) dt for f (x) in the figure (Figure 1, page 5).

(a) A (2) =

Solution:

A (2) =

∫ 2

0

f (t) dt = 4

(b) A (3) =

Solution:

A (3) =

∫ 3

0

f (t) dt =13

2

(c) A′ (2) =

Solution:

A (x) =

∫ x

0

f (t) dt

A′ (x) = f (x)

A′ (2) = f (2) = 2

(d) A′ (3) =

Solution:

A (x) =

∫ x

0

f (t) dt

A′ (x) = f (x)

A′ (3) = f (3) = 3

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588 C H A P T E R 5 THE INTEGRAL

19.d

dx

! x

0(t3 ! t) dt

SOLUTION By FTC II,d

dx

! x

0(t3 ! t) dt = x3 ! x .

20.d

dx

! x

1sin(t2) dt

SOLUTION By FTC II,d

dx

! x

1sin(t2) dt = sin x2.

21.ddt

! t

100cos 5x dx

SOLUTION By FTC II,ddt

! t

100cos(5x) dx = cos 5t .

22.dds

! s

!2tan

"1

1 + u2

#du

SOLUTION By FTC II,dds

! s

!2tan

$ 1

1 + u2

%du = tan

$ 1

1 + s2

%.

23. Sketch the graph of A(x) =! x

0f (t) dt for each of the functions shown in Figure 10.

x

y

4321

(A)

2

1

0

!1

x

y

4321

2

1

0

!1

(B)

FIGURE 10

SOLUTION

• Remember that A"(x) = f (x). It follows from Figure 10(A) that A"(x) is constant and consequently A(x) is linearon the intervals [0, 1], [1, 2], [2, 3] and [3, 4]. With A(0) = 0, A(1) = 2, A(2) = 3, A(3) = 2 and A(4) = 2, weobtain the graph shown below at the left.

• Since the graph of y = f (x) in Figure 10(B) lies above the x-axis for x # [0, 4], it follows that A(x) is increasingover [0, 4]. For x # [0, 2], area accumulates more rapidly with increasing x , while for x # [2, 4], area accumulatesmore slowly. This suggests A(x) should be concave up over [0, 2] and concave down over [2, 4]. A sketch of A(x)

is shown below at the right.

1 2 3 4

1

2

3

4

x

y

1 2 3 4

0.51

1.52

2.53

x

y

24. Let A(x) =! x

0f (t) dt for f (x) shown in Figure 11. Calculate A(2), A(3), A"(2), and A"(3). Then find a formula

for A(x) (actually two formulas, one for 0 $ x $ 2 and one for 2 $ x $ 4) and sketch the graph of A(x).

4321

2

3

4

1x

y

y = f (x)

FIGURE 11Figure 1: Partial graph of A (x).

7. Calculate the derivative.d

dx

∫ x4

x2

√3t dt

Solution:

d

dx

∫ x4

x2

√3t dt =

√3x4 · 4x3 −

√3x2 · 2x

= 4√

3x5 − 2√

3x |x|

8. Let A (x) =

∫ x

0

f (t) dt and B (x) =

∫ x

2

f (t) dt as in figure (Figure 2, page 6).

(a) Find a formula for A (x) that is valid on [0, 6].

Solution:

A (x) =

∫ x

0

f (t) dt

A (x) = −x 0 ≤ x < 1

A (x) = x2 − 3x+ 1 1 ≤ x < 2

A (x) = x− 3 2 ≤ x < 4

A (x) = −x2 + 9x− 19 4 ≤ x < 5

A (x) = 4− x 5 ≤ x ≤ 6

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(b) Find a formula for B (x) that is valid on [2, 6].

Solution:

B (x) =

∫ x

2

f (t) dt

B (x) = x− 2 2 ≤ x < 4

B (x) = −x2 + 9x− 18 4 ≤ x < 5

B (x) = 7− x 5 ≤ x ≤ 6

S E C T I O N 5.4 The Fundamental Theorem of Calculus, Part II 591

SOLUTION Let

G(x) =! x2

!x

tan t dt =! x2

0tan t dt "

! !x

0tan t dt.

Applying the Chain Rule combined with FTC twice, we have

G#(x) = tan(x2) · 2x " tan(!

x) · 12

x"1/2 = 2x tan(x2) " tan(!

x)

2!

x.

36.d

du

! 3u+9

"u

"x2 + 1 dx

SOLUTION Let

F(u) =! 3u+9

"u

"x2 + 1 dx =

! 3u+9

0

"x2 + 1 dx "

! "u

0

"x2 + 1 dx .

Applying the Chain Rule combined with FTC,

F #(u) =#

(3u + 9)2 + 1 · 3 "#

("u)2 + 1 · ("1) = 3#

(3u + 9)2 + 1 +"

u2 + 1.

In Exercises 37–38, let A(x) =! x

0f (t) dt and B(x) =

! x

2f (t) dt, with f (x) as in Figure 13.

x

y

63 4 521

2

1

0

!1

!2

y = f (x)

FIGURE 13

37. Find the min and max of A(x) on [0, 6].SOLUTION The minimum values of A(x) on [0, 6] occur where A#(x) = f (x) goes from negative to positive. Thisoccurs at one place, where x = 1.5. The minimum value of A(x) is therefore A(1.5) = "1.25. The maximum values ofA(x) on [0, 6] occur where A#(x) = f (x) goes from positive to negative. This occurs at one place, where x = 4.5. Themaximum value of A(x) is therefore A(4.5) = 1.25.

38. Find formulas for A(x) and B(x) valid on [2, 4].

SOLUTION On the interval [2, 4], A#(x) = B #(x) = f (x) = 1. A(2) =! 2

0f (t) dt = "1 and B(2) =

! 2

2f (t) dt =

0. Hence A(x) = (x " 2) " 1 and B(x) = (x " 2).

39. Let A(x) =! x

0f (t)dt , with f (x) as in Figure 14.

(a) Does A(x) have a local maximum at P?(b) Where does A(x) have a local minimum?(c) Where does A(x) have a local maximum?(d) True or false? A(x) < 0 for all x in the interval shown.

x

y

SR

Q

Py = f (x)

FIGURE 14 Graph of f (x).

SOLUTION

(a) In order for A(x) to have a local maximum, A#(x) = f (x) must transition from positive to negative. As this doesnot happen at P , A(x) does not have a local maximum at P .(b) A(x) will have a local minimum when A#(x) = f (x) transitions from negative to positive. This happens at R, soA(x) has a local minimum at R.

Figure 2: Graph related to A (x) and B (x).

9. Determine f (x), assuming that

∫ x

1

f (t) dt = 3x2 − 5x+ 2.

Solution: ∫ x

1

f (t) dt = 3x2 − 5x+ 2

F (x)− F (1) = 3x2 − 5x+ 2

d

dx[F (x)− F (1)] =

d

dx

[3x2 − 5x+ 2

]F ′ (x) = 6x− 5

f (x) = 6x− 5

10. Let f (x) = x2 − 13x+ 40 and F (x) =

∫ x

0

f (t) dt.

(a) Find the critical points of F (x) and determine whether they are local minima ormaxima.

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Solution:

F ′ (x) = x2 − 13x+ 40

= (x− 5) (x− 8)

The critical points are x = 5 and x = 8; simple sign-analysis indicates that a localmaximum occurs at x = 5 and a local minimum occurs at x = 8.

(b) Find the points of inflection of F (x) and determine whether the concavity changesfrom up to down or down to up.

Solution:

F ′ (x) = x2 − 13x+ 40

F ′′ (x) = 2x− 13

The point-of-inflection occurs at are x = 13/2; concavity changes from down toup.

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0.2 mth.121.05.05

1. A population of insects increases at a rate 190 + 6t + 0.9t2 insects per day. Find theinsect population after 3 days, assuming that there are 50 insects at t = 0.1

Solution:

P ′ (t) = 190 + 6t+ 0.9t2

P (t) = 190t+ 3t2 + 0.3t3 + C

P (0) = C = 50

P (t) = 190t+ 3t2 + 0.3t3 + 50

P (3) = 190 · 3 + 3 · 32 + 0.3 · 33 + 50 = 655.1

You may also do this. ∫ 3

0

P ′ (t) dt = P (3)− P (0)

605.1 = P (3)− 50

655.1 = P (3)

655 insects

2. Find the displacement of a particle moving in a straight line with velocity v (t) = 12t−3m/s over the time interval [2, 5].

Solution: ∫ 5

2

v (t) dt = 117 m

3. Find the displacement over the time interval [1, 9] of a helicopter whose vertical velocityat time t is v (t) = 0.04t2 + t m/s.2

Solution: ∫ 9

1

v (t) dt =3728

75≈ 49.71 m

1Round your answer to the nearest insect.2Round your answer to two decimal places.

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4. A projectile is released with an initial (vertical) velocity of 90 m/s. Use the formulav (t) = 90 − 9.8t for velocity to determine the distance traveled during the first 16seconds.3

Solution: ∫ 16

0

|90− 9.8t| dt =

∫ 90/9.8

0

90− 9.8t dt+

∫ 16

90/9.8

9.8t− 90 dt

=157028

245≈ 640.93 m

5. Find the net change in velocity over [4, 5] of an object with the acceleration a (t) =23t− 2t2 (in m/s2).4

Solution: ∫ 5

4

a (t) dt =

(23t2

3− 2t3

3

)∣∣∣∣54

=377

6≈ 62.83 m/s

3Round your answer to two decimal places.4Round your answer to two decimal places.

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0.3 mth.121.05.06

1. Evaluate the indefinite integral. Use substitution.5∫x(x2 + 6

)5dx, u = x2 + 6

Solution:

u = x2 + 6

du = 2x dx∫x(x2 + 6

)5dx =

∫ ∗ u52

du

=u6

12+ C∗

=(x2 + 6)

6

12+ C

2. Evaluate the indefinite integral. Use substitution.6∫t√

24t2 + 5 dt, u = 24t2 + 5

Solution:

u = 24t2 + 5

du = 48t dt∫t√

24t2 + 5 dt, =

∫ ∗ u1/248

du

=u3/2

72+ C∗

=(24t2 + 5)

3/2

72+ C

=(24t2 + 5)

√24t2 + 5

72+ C

5Use C for the constant of integration.6Use C for the constant of integration.

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3. Evaluate the indefinite integral. Use substitution.7∫22x2√x+ 10 dx, u = x+ 10

Solution:

u = x+ 10

du = dx∫22x2√x+ 10 dx = 22

∫ ∗(u− 10)2 u1/2 du

= 22

∫ ∗u5/2 − 20u3/2 + 100u1/2 du

= 22

(2u7/2

7− 8u5/2 +

200u3/2

3

)+ C∗

=44 (x+ 10)3/2 (3x2 − 24x+ 160)

21+ C

4. Evaluate the indefinite integral. Use substitution.8∫sin9 θ cos θ dθ, u = sin θ

Solution:

u = sin θ

du = cos θ dθ∫sin9 θ cos θ dθ =

∫ ∗u9 du

=u10

10+ C∗

=sin10 θ

10+ C

5. Evaluate the indefinite integral. Use substitution.9∫6 sec2 x tanx dx, u = tanx

7Use C for the constant of integration.8Use C for the constant of integration.9Use C for the constant of integration.

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Solution:

u = tanx

du = sec2 x dx∫6 sec2 x tanx dx =

∫ ∗6u du

= 3u2 + C∗

= 3 tan2 x+ C

6. Evaluate the indefinite integral. Use substitution.10∫(arctanx)6

x2 + 1dx, u = arctanx

Solution:

u = arctanx

du =1

x2 + 1dx∫

(arctanx)6

x2 + 1dx =

∫ ∗u6 du

=u7

7+ C∗

=(arctanx)7

7+ C

7. Evaluate the indefinite integral.11 ∫x6 cos

(x7)

dx


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Solution:

u = x7

du = 7x6 dx∫x6 cos

(x7)

dx =

∫ ∗ cosu

7du

=sinu

7+ C∗

=sin (x7)

7+ C

8. Evaluate the indefinite integral.12∫cos (9x) · cos [sin (9x)] dx

Solution:

u = sin 9x

du = 9 cos 9x dx∫cos (9x) · cos [sin (9x)] dx =

∫ ∗ cosu

9du

=sinu

9+ C∗

=sin (sin 9x)

9+ C

9. Evaluate the indefinite integral.13 ∫1

(2x− 5)5dx


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Solution:

u = 2x− 5

du = 2 dx∫1

(2x− 5)5dx =

∫ ∗ u−52

du

= − 1

8u4+ C∗

= − 1

8 (2x− 5)4+ C

10. Evaluate the indefinite integral.14 ∫10x+ 1

(10x2 + 2x)3dx

Solution:

u = 10x2 + 2x

du = 2 (10x+ 1) dx∫10x+ 1

(10x2 + 2x)3dx =

∫ ∗ u−32

du

= − 1

4u2+ C∗

= − 1

4 (10x2 + 2x)2+ C

11. Evaluate the indefinite integral.15 ∫x√x2 + 5

dx


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Solution:

u = x2 + 5

du = 2x dx∫x√x2 + 5

dx =

∫ ∗ u−1/22

du

=√u+ C∗

=√x2 + 5 + C

12. Evaluate the indefinite integral.16 ∫sin (8− 8θ) dθ

Solution:

u = 8− 8θ

du = −8 dθ∫sin (8− 8θ) dθ =

∫ ∗−sinu

8du

=cosu

8+ C∗

=cos (8− 8θ)

8+ C

13. Evaluate the indefinite integral.17∫14θ sin

(14θ2

)dθ,


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Solution:

u = 14θ2

du = 28θ dθ∫14θ sin

(14θ2

)dθ =

∫ ∗ sinu

2du

= −cosu

2+ C∗

= −cos (14θ2)

2+ C

14. Evaluate the indefinite integral.18 ∫sec2 (4x+ 6) dx,

Solution:

u = 4x+ 6

du = 4 dx∫sec2 (4x+ 6) dx =

∫ ∗ sec2 u

4du

=tanu

4+ C∗

=tan (4x+ 6)

4+ C

15. Evaluate the indefinite integral.19∫(x+ 5) ex

2+10x dx,


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Solution:

u = x2 + 10x

du = 2 (x+ 5) dx∫(x+ 5) ex

2+10x dx, =

∫ ∗ eu2

du

=eu

2+ C∗

=ex

2+10x

2+ C

16. Evaluate the indefinite integral.20 ∫ex

(ex + 5)3dx,

Solution:

u = ex + 5

du = ex dx∫ex

(ex + 5)3dx =

∫ ∗u−3 du

= − 1

2u2+ C∗

= − 1

2 (ex + 5)2+ C

17. Evaluate the indefinite integral.21 ∫1

x ln 4xdx


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Solution:

u = ln 4x

du =1

xdx∫

1

x ln 4xdx =

∫ ∗ 1

udu

= ln |u|+ C∗

= ln |ln 4x|+ C

18. Evaluate the definite integral. ∫ 4

3

x (x+ 2)5 dx

Solution:

u = x+ 2

du = x dx∫ 4

3

x (x+ 2)5 dx =

∫ 6

5

(u− 2)u5 du

=

∫ 6

5

u6 − 2u5 du

=u7

7− u6

3

∣∣∣∣65

=388216

21


−5x√x+ 6 dx

Solution:

u = x+ 6

du = x dx∫ 19

−5x√x+ 6 dx =

∫ 25

1

(u− 6)u1/2 du

=

∫ 25

1

u3/2 − 6u1/2 du

=2u2/5

5− 4u3/2

∣∣∣∣251

=3768

5

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0

x

(x2 + 1)3dx

Solution:

u = x2 + 1

du = 2x dx∫ 1

0

x

(x2 + 1)3dx =

1

2

∫ 2

1

u−3 du

= − 1

4u2

∣∣∣∣21

=3

16


0

θ5 tan(θ6)

dθ

Solution:

u = θ6

du = 6θ5 dθ∫ 1

0

θ5 tan(θ6)

dθ =1

6

∫ 1

0

tanu du

= − ln |cosu|6

∣∣∣∣10

= − ln (cos 1)

6

22. Evaluate the definite integral. ∫ π/2

0

sin8 x cosx dx

Solution:

u = sin x

du = cos x dx∫ π/2

0

sin8 x cosx dx =

∫ 1

0

u8 du

=u9

9

∣∣∣∣10

=1

9

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0.4 mth.121.06.01

1. Find the area (Figure 3, page 20) of the region between y = 3x2 + 12 and y = 4x + 4over [−3, 3].

6 APPLICATIONS OFTHE INTEGRAL

6.1 Area Between Two Curves

Preliminary Questions

1. What is the area interpretation of! b

a

"f (x) ! g(x)

#dx if f (x) " g(x)?

SOLUTION Because f (x) " g(x),$ b

a ( f (x) ! g(x)) dx represents the area of the region bounded between the graphsof y = f (x) and y = g(x), bounded on the left by the vertical line x = a and on the right by the vertical line x = b.

2. Is! b

a

"f (x) ! g(x)

#dx still equal to the area between the graphs of f and g if f (x) " 0 but g(x) # 0?

SOLUTION Yes. Since f (x) " 0 and g(x) # 0, it follows that f (x) ! g(x) " 0.

3. Suppose that f (x) " g(x) on [0, 3] and g(x) " f (x) on [3, 5]. Express the area between the graphs over [0, 5] as asum of integrals.

SOLUTION Remember that to calculate an area between two curves, one must subtract the equation for the lower curvefrom the equation for the upper curve. Over the interval [0, 3], y = f (x) is the upper curve. On the other hand, over theinterval [3, 5], y = g(x) is the upper curve. The area between the graphs over the interval [0, 5] is therefore given by

! 3

0( f (x) ! g(x)) dx +

! 5

3(g(x) ! f (x)) dx .

4. Suppose that the graph of x = f (y) lies to the left of the y-axis. Is! b

af (y) dy positive or negative?

SOLUTION If the graph of x = f (y) lies to the left of the y-axis, then for each value of y, the corresponding value of x

is less than zero. Hence, the value of$ b

a f (y) dy is negative.

Exercises1. Find the area of the region between y = 3x2 + 12 and y = 4x + 4 over [!3, 3] (Figure 8).

50

25

y

x

y = 3x2 + 12

y = 4x + 4

3!1

!2

!3 1 2

FIGURE 8

SOLUTION As the graph of y = 3x2 + 12 lies above the graph of y = 4x + 4 over the interval [!3, 3], the areabetween the graphs is

! 3

!3

%(3x2 + 12) ! (4x + 4)

&dx =

! 3

!3(3x2 ! 4x + 8) dx =

%x3 ! 2x2 + 8x

&'''3

!3= 102.

2. Compute the area of the region in Figure 9(A), which lies between y = 2 ! x2 and y = !2 over [!2, 2].

y

x2!2 !2

!2 !2

y

x1

y = 2 ! x2

y = !2

y = 2 ! x2

y = x

(A) (B)

FIGURE 9

Figure 3: Area of interest.

Solution: ∫ −3−3

(3x2 + 12

)− (4x+ 4) dx =

∫ 3

−33x2 − 4x+ 8 dx

=[x3 − 2x2 + 8x

]∣∣∣∣3−3

= 102

2. Consider the following.

f (x) = 8x− 10

g (x) = x2 − 4x+ 10

(a) Find the points of intersection of the graphs.

Solution: (2, 6) and (10, 70).

(b) Compute the area of the region below the graph of f and above the graph of g.

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Solution: ∫ 10

2

8x− 10−(x2 − 4x+ 10

)dx = 6x2 − 20x− x3

3

∣∣∣∣102

=256

3≈ 85.333

3. Find the area of the region between y = sinx and y = cosx over[π4, π].

Solution: ∫ π

π/4

sinx− cosx dx = − cosx− sinx

∣∣∣∣ππ/4

=√

2 + 1

≈ 2.414

4. Find the area A between y = ex and y = e2x over [0, 1].

Solution: ∫ 1

0

e2x − ex dx =e2x

2− ex

∣∣∣∣10

=e2 − 2e+ 1

2≈ 1.476

5. Find the area of the region bounded by

y =3√

1− x2and y = − 3√

1− x2for − 1

5≤ x ≤ 1

5.

Solution: ∫ 1/5

−1/5

3√1− x2

−(− 3√

1− x2

)dx = 6

∫ 1/5

−1/5

1√1− x2

dx

= 6 arcsinx

∣∣∣∣1/5−1/5

= 12 arcsin 1/5

≈ 2.416

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6. Find the area between the graphs x = sin (y) and x = 1 − cos (y) over the interval0 ≤ y ≤ π/2 in figure (Figure 4, page 22).

676 C H A P T E R 6 APPLICATIONS OF THE INTEGRAL

In Exercises 21–22, find the area between the graphs of x = sin y and x = 1 ! cos y over the given interval (Figure 14).

x = 1 ! cos y

x = sin y

x

y

!

2

2

FIGURE 14

21. 0 " y " "2

SOLUTION As shown in the figure, the graph on the right is x = sin y and the graph on the left is x = 1 ! cos y.Therefore, the area between the two curves is given by

! "/2

0(sin y ! (1 ! cos y)) dy = (! cos y ! y + sin y)

""""/2

0=

#!"

2+ 1

$! (!1) = 2 ! "

2.

22. !"2

" y " "2

SOLUTION The shaded region in the figure shows the area between the graphs from y = 0 to y = "2 . It is bounded on

the right by x = sin y and on the left by x = 1 ! cos y. Therefore, the area between the graphs from y = 0 to y = "2 is

! "/2

0(sin y ! (1 ! cos y)) dy = (! cos y ! y + sin y)

""""/2

0=

#!"

2+ 1

$! (!1) = 2 ! "

2.

The graphs cross at y = 0. Since x = 1 ! cos y lies to the right of x = sin y on the interval [!"2 , 0] along the y-axis,

the area between the graphs from y = !"2 to y = 0 is

! 0

!"/2((1 ! cos y) ! sin y) dy = (y ! sin y + cos y)

"""0

!"/2= 1 !

#!"

2+ 1

$= "

2.

The total area between the graphs from y = !"2 to y = "

2 is the sum

! "/2

0(sin y ! (1 ! cos y)) dy +

! 0

!"/2((1 ! cos y) ! sin y) dy = 2 ! "

2+ "

2= 2.

23. Find the area of the region lying to the right of x = y2 + 4y ! 22 and the left of x = 3y + 8.

SOLUTION Setting y2 + 4y ! 22 = 3y + 8 yields

0 = y2 + y ! 30 = (y + 6)(y ! 5),

so the two curves intersect at y = !6 and y = 5. The area in question is then given by

! 5

!6

#(3y + 8) ! (y2 + 4y ! 22)

$dy =

! 5

!6

#!y2 ! y + 30

$dy =

%

! y3

3! y2

2+ 30y

&"""""

5

!6

= 13316

.

24. Find the area of the region lying to the right of x = y2 ! 5 and the left of x = 3 ! y2.

SOLUTION Setting y2 + 5 = 3 ! y2 yields 2y2 = 8 or y = ±2. The area of the region enclosed by the two graphs isthen

! 2

!2

#(3 ! y2) ! (y2 + 5)

$dy =

! 2

!2

#8 ! 2y2

$dy =

'8y ! 2

3y3

(""""2

!2= 64

3.

25. Calculate the area enclosed by x = 9 ! y2 and x = 5 in two ways: as an integral along the y-axis and as an integralalong the x-axis.

Figure 4: Area of interest.

Solution: ∫ π/2

0

sin y − (1− cos y) dy = sin y − cos y − y∣∣∣∣π/20

= 2 +π

2≈ 3.571

7. Find the area of the region lying to the right of x = y2− 9 and to the left of x = 5− y2.

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Solution: ∫ √7−√7

5− y2 −(y2 − 9

)dy = 14y − 2y3

3

∣∣∣∣√7

−√7

=56√

7

3≈ 49.387

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0.5 mth.121.06.02

1. Find the volume of the wedge in figure (Figure 5, page 24 by integrating the area ofvertical cross sections. Assume that a = 6, b = 8 and c = 3.

S E C T I O N 6.2 Setting Up Integrals: Volume, Density, Average Value 691

5. Find the volume of liquid needed to fill a sphere of radius R to height h (Figure 18).

R

y

h

FIGURE 18 Sphere filled with liquid to height h.

SOLUTION The radius r at any height y is given by r =!

R2 ! (R ! y)2. Thus, the volume of the filled portion ofthe sphere is

!" h

0r2 dy = !

" h

0

#R2 ! (R ! y)2

$dy = !

" h

0(2Ry ! y2) dy = !

%

Ry2 ! y3

3

&'''''

h

0

= !%

Rh2 ! h3

3

&

.

6. Find the volume of the wedge in Figure 19(A) by integrating the area of vertical cross sections.

68

(A) (B)

4

ba

c

xx

FIGURE 19

SOLUTION Cross sections of the wedge taken perpendicular to the x-axis are right triangles. Using similar triangles,we find the base and the height of the cross sections to be 3

4 (8 ! x) and 12 (8 ! x), respectively. The volume of the wedge

is then

316

" 8

0(8 ! x)2 dx = 3

16

" 8

0

#64 ! 16x + x2

$dx = 3

16

(64x ! 8x2 + 1

3x3

)''''8

0= 32.

7. Derive a formula for the volume of the wedge in Figure 19(B) in terms of the constants a, b, and c.

SOLUTION The line from c to a is given by the equation (z/c) + (x/a) = 1 and the line from b to a is given by(y/b) + (x/a) = 1. The cross sections perpendicular to the x-axis are right triangles with height c(1 ! x/a) and baseb(1 ! x/a). Thus we have

" a

0

12

bc (1 ! x/a)2 dx = !16

abc#

1 ! xa

$3''''a

0= 1

6abc.

8. Let B be the solid whose base is the unit circle x2 + y2 = 1 and whose vertical cross sections perpendicular to thex-axis are equilateral triangles. Show that the vertical cross sections have area A(x) =

"3(1 ! x2) and compute the

volume of B.

SOLUTION At the arbitrary location x , the side of the equilateral triangle cross section that lies in the base of the solid

extends from the top half of the unit circle (with y =*

1 ! x2) to the bottom half (with y = !*

1 ! x2). The equilateraltriangle therefore has sides of length s = 2

*1 ! x2 and an area of

A(x) = s2"3

4=

"3(1 ! x2).

Finally, the volume of the solid is

"3

" 1

!1

#1 ! x2

$dx =

"3(

x ! 13

x3)''''

1

!1= 4

"3

3.

In Exercises 9–14, find the volume of the solid with given base and cross sections.

9. The base is the unit circle x2 + y2 = 1 and the cross sections perpendicular to the x-axis are triangles whose heightand base are equal.

Figure 5: Wedge

Solution: ∫ 0

−6

1

2

(x2

+ 3)(4x

3+ 8

)dx =

∫ 0

−6

x2

3+ 4x+ 12 dx

=

[x3

9+ 2x2 + 12x

]∣∣∣∣0−6

= 24

2. Find the volume V of the solid whose base is the semicircle y =√

16− x2, where−4 ≤ x ≤ 4 and whose cross sections perpendicular to the x-axis are squares.

Solution: ∫ −4−4

(√16− x2

)2dx =

∫ 4

−416− x2 dx

=

[16x− x3

3

]∣∣∣∣4−4

=256

3

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3. Calculate the average over the given interval.

f (x) = x5, [0, 6]

Solution:

1

6− 0

∫ 6

0

x5 dx =x6

36

∣∣∣∣60

= 1296

4. Calculate the average over the given interval.

f (x) = cos x,[0,

π

6

]

Solution:

1

π/6− 0

∫ π/6

0

cosx dx =6 sinx

π

∣∣∣∣π/60

=3

π

5. Calculate the average over the given interval.22

f (x) = e−3x, [−1, 9]

Solution:

1

9 + 1

∫ 9

−1e−3x dx = −e

−3x

30

∣∣∣∣9−1

=e30 − 1

30e27

≈ 0.6695

22Round your answer to three decimal places.

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6. The temperature (in centigrade) at time t (in hours) in an art museum varies accordingto

T (t) = 18 + 5 cos( π

12t).

Find the average temperature over the time period [8, 24].23

Solution:

1

24− 8

∫ 24

8

18 + 5 cos( π

12t)

dt =

[9t

8+

15

4πsin

(πt

12

)]∣∣∣∣248

= 18− 15√

3

8π≈ 16.97 degrees centigrade

23Round your answer to two decimal places.

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0.6 mth.121.06.03

1. Calculate the volume of the solid obtained by revolving the region under the graph off (x) =

√x+ 9 about the x-axis over the interval [2, 7].

Solution: ∫ 7

2

π(√

x+ 9)2

dx = π

∫ 7

2

x+ 9 dx

= π

[x2

2+ 9x

]∣∣∣∣72

=135π

2

2. Find the volume of the solid obtained by rotating the region enclosed by the curvesy = x2 + 4 and y = 22− x2 about the x-axis.

Solution: ∫ 3

−3π(22− x2

)2 − π (x2 + 4)2

dx = π

∫ 3

−3468− 52x2 dx

= π

[468x− 52x3

3

]∣∣∣∣3−3

= 1872π

3. Find the volume of the solid obtained by rotating the region enclosed by the curvesy = 13− x, y = 3x+ 9 and x = −1 about the x-axis.

Solution: ∫ 1

−1π (13− x)2 − π (3x+ 9)2 dx = π

∫ 1

−188− 80x− 8x2 dx

= π

[88x− 40x3 − 8x3

3

]∣∣∣∣1−1

=512π

3

4. Find the volume V of the solid obtained by rotating the region enclosed by the graphsx =√y and x = 0 about the y-axis between y = 3 and y = 5.

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Solution: ∫ 5

3

π (√y)2 dy = π

∫ 5

3

y dy

=πy2

2

∣∣∣∣53

= 8π

5. Find the volume of the solid obtained by rotating the region enclosed by the curvesy = 12− x, x =

√y, x = 0 about y = −6.

Solution:∫ 3

0

π (18− x)2 − π(x2 + 6

)2dx = π

∫ 3

0

288− 36x− 11x2 − x4dx

= π

[288x− 18x2 − 11x3

3− x5

5

]∣∣∣∣30

=2772π

5

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0.7 mth.121.06.04

1. Use the shell method to compute the volume obtained by rotating the region enclosedby the graphs as indicated, about the y-axis.

f (x) = 3x− 10, g (x) = 6− x, x = 0

Solution: ∫ 4

0

2πx (6− x− 3x+ 10) dx = 8π

∫ 4

0

4x− x2 dx

= 8π

[2x2 − x3

3

]∣∣∣∣40

=256π

3

2. Use the shell method to compute the volume V of the solid obtained by rotating theregion enclosed by the graphs of the functions y = x8 and y = x1/8 about the y-axis.

Solution: ∫ 1

0

2πx(x1/8 − x8

)dx = 2π

∫ 1

0

x9/8 − x9 dx

= 2π

[8x17/8

17− x10

10

]∣∣∣∣10

=63π

85

3. Use the shell method to compute the volume V of the solid obtained by rotating theregion enclosed by the graphs of the functions y = x2, y = 32− x2, and x ≥ 1 about they-axis.

Solution: ∫ 4

1

2πx(32− x2 − x2

)dx = 4π

∫ 4

1

16x− x3 dx

= 4π

[8x2 − x4

4

]∣∣∣∣41

= 225π

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4. Use the shell method to compute the volume obtained by rotating the region enclosedby the graphs as indicated, about the y-axis.

y = 8− x3, y = 8− 4x, for x ≥ 0

Solution: ∫ 2

0

2πx(8− x3 − 8 + 4x

)dx = 2π

∫ 2

0

4x2 − x4 dx

= 2π

[4x3

3− x5

5

]∣∣∣∣20

=128π

15

5. Find the volume of rotation of the region enclosed by the graphs of y = x3+7, y = 9−x2,and x = 3 as shown in the figure (Figure 6, page 30 about the line x = 5.

0 1 2 3 4 5 6

5

10

15

20

25

30

35

Figure 6: Region of interest.

Solution:∫ 3

1

2π (5− x)(x3 + 7− 9 + x2

)dx = −2π

∫ 3

1

x4 − 4x3 − 5x2 − 2x+ 10 dx

= −2π

[x5

5− x4 − 5x3

3− x3 + 10x

]∣∣∣∣31

=1888π

15

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contents 0.1 mth.121.05.04 ...faculty.essex.edu/~bannon/m121.r/notes/mth.121.review.04.pdfron bannon...

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