SEM 1102 / MAT 200–Calculus and Analytic Geometry IImdvsamson.work/mat200/docs.pdf · SEM 1102 / MAT 200 Homework 1 13 January 2020 Name: dP ID: 3 9 0 0 0 §4.9 #49 Given that the

SEM 1102 / MAT 200–Calculus and Analytic Geometry II

Spring 2020

Prerequisites: SEM 1101 or MAT 150 or MAT 180

General Information

Class Schedule: Mondays and Wednesdays 9:00–11:10amClassroom: Edison (SR 2E)Instructor: Michael Daniel SamsonContact: [email protected], [email protected], +65 6577 1944Class Webpage: MoodleOffice Hours: Mondays and Wednesdays 11:10am–12:10pm

By appointment (through email): Fridays 1:00–7:00pm

Description

This course builds on the introduction to calculus in MAT 150. Topics in integration include applicationsof the integral in physics and geometry, and techniques of integration. The course also covers sequencesand series of real numbers, power series and Taylor series, and calculus of transcendental functions.Further topics may include a basic introduction to concepts in multivariable and vector calculus.

Course Objectives and Learning Outcomes

Upon completing this course students should be able to:

• Understand the concept of definite integral as a limit of Riemann sums.• Find indefinite and improper integrals using different integration techniques.• Perform standard operations with convergent power series, find Taylor and Maclaurin represen-

tations.• Use integrals to solve applied problems and analyze graphs of curves.

Textbooks

CALCULUS Early Transcendentals 8e, International Metric Version, James Stewart, Cengage Learning, ISBN-10 1-305-27237-4, ISBN-13 978-1-305-27237-8

Optional Textbooks

CALCULUS Early Vectors, James Stewart, Brooks/Cole Cengage Learning, ISBN-10 0-534-34941-2, ISBN-13 978-0-534-49348-6

1

Outline and Tentative Dates

The following schedule is subject to change.

Integration and Some TechniquesJanuary 6, 8: Summations, Fundamental Theorem of CalculusJanuary 13, 15: Substitution Rule, Integration by Parts, quizJanuary 20, 22: Trigonometric Integrals, Trigonometric Substitution,January 27: Lunar New YearJanuary 29: Partial Fractions, ⋆ Improper Integrals

Approximation of Definite Integrals as Infinite SumsFebruary 3: Founder’s DayFebruary 5: quiz, Approximation of IntegralsFebruary 10, 12: Sequences, Series, Tests of Series Convergence, Truncation ErrorFebruary 17–21: Study BreakFebruary 24: Mid-Term Examination (discussion on February 26)March 2, 4: Power Series, Taylor and Maclaurin Series, quiz

Geometric and Physical Applications of IntegrationMarch 9, 11: Areas Between Curves, Volumes, Cylindrical ShellsMarch 16, 18: Work, ⋆ Average Value of a Function, quizMarch 23, 25: Arc Length, Areas of Surfaces of RevolutionMarch 30, April 1: Applications in Other Fields, quizApril 6–17: Final Examination (venue and time to be announced by DigiPen Administration)

Grading Policy

The examination on week fifteen is optional. You must inform the instructor of your decision to not takethe final exam by week fourteen.

The relative weights of homework, quizzes and exams are:

10% Homework (at least ten)30% Quizzes (drop the lowest)30% Mid-Term Examination30% Final Examination

Grades will be computed out of 40 points. Letter grades will be computed subject to:

35 = at least A30 = at least B20 = at least C- (passing)

To pass the course, you need to

have a passing examination average and the course total should be greater than or equal to 20.

Late Policy

Late assignments will not be accepted. There will be no make-up quizzes or exams, unless authorizedby the instructor.

Attendance Policy

The duration of this semester is fifteen (15) weeks. During the first fourteen (14) weeks of the semester,the class will meet bi-weekly, for a total of twenty-eight (28) class sessions. There will be a three-hourlong final exam scheduled for the final week of the semester.

SEM 1102 / MAT 200 Spring 2020 Syllabus Page 2 of 3

Attendance is mandatory. You will be penalized for unexcused absences from class according to thefollowing scale:

• Three (3) or more absences will result in a 10% reduction of your overall course grade.• Six (6) or more absences will result in a 20% reduction of your overall course grade.• Eight (8) or more absences will result in a 30% reduction of your overall course grade.• Twelve (12) or more absences will result in your automatic failure in the course, irrespective of your

performance on homework, assignments, quizzes, and exams.

Medical leave and family emergencies—both accompanied by appropriate documents—will be the onlyexceptions to this policy. Sleeping, studying for another class and/or exam, working on your game, etc.,are not valid reasons for absence from the class.

On Use of Calculators

Calculator use is discouraged for this course, and will not be allowed during examinations. More sophisti-cated computing devices now regularly dispense as output the details taught in the course, without thebenefit of understanding the result. Calculators can be useful for doing away with the tedious clericalnature of computation, but this course will not evaluate students on their arithmetic—for most part,in-examination computations will be allowed to be left unsimplified without penalty.

Last Day to Withdraw

The final date to withdraw from this course is 3 March 2020. Scores for five (5) homework submissions,two (2) quizzes and one (1) examination should be available before this date. In order to withdrawfrom a course, in accordance with policy, contact your advisor or the Registrar to begin the withdrawalprocess—it is not sufficient simply to stop attending class or to inform the instructor. The last day forwithdrawal from this course is cited in the official catalog.

Academic Integrity Policy

Academic dishonesty in any form will not be tolerated in this course. Cheating, copying, plagiarizing,or any other form of academic dishonesty (including doing someone else’s individual assignments) willresult in, at the very minimum, a zero on the assignment in question, and could result in a failing gradein the course or even expulsion from DigiPen.

External Preparation

It is expected that the students in this class spend eight (8) hours on average per week for outside class-room activities through the semester, including, but not limited to, homework, reading assignments,project implementation, group discussions, preparation of examinations, etc.

Disability Support Service

Students who have special needs or medical conditions and require formal accommodations in orderto fully participate or effectively demonstrate learning in this class should contact the Student Life &Advising Office ([email protected]) at the beginning of each semester. A Student Life& Advising Officer will meet with the student privately to discuss how the accommodations will beimplemented.

SEM 1102 / MAT 200 Spring 2020 Syllabus Page 3 of 3

SEM 1102 / MAT 200 Homework 1 13 January 2020

Name: dP ID: 3 9 0 0 0

§4.9 #49 Given that the graph of f passes through the point (2, 5) and that the slope of its tangent line at(x, f(x)) is 3− 4x, find f(1).

From the given, f(2) = 5 and f ′(x) = 3 − 4x. It is indicated that f(x) = ax2 + bx + c, a quadraticfunction: f ′(x) = 2ax + b = 3 − 4x, so b = 3 and a = −2, so f(x) = −2x2 + 3x + c, andf(2) = −2(2)2 + 3(2) + c = c− 2 = 5, so c = 7. Thus, f(1) = −2(1)2 + 3(1) + 7 = 8.

§4.9 #79 A high-speed bullet train accelerates and decelerates at the rate of 1.2 m/s2. Its maximum cruisingspeed is 145 km/h.

(a) What is the maximum distance the train can travel if it accelerates from rest until it reachescruising speed and then runs at that speed for 15 minutes?

Note that 145 kilometers per hour is725

18= 40.27̄ meters per second. From the given, a(t) =

6

5

for 0 ≤ t ≤ T , where v(T ) =725

18, v(0) = 0 and v(t) <

725

18for 0 ≤ t < T ; a(t) = 0 for

T < t ≤ T +900; s(0) = 0. Thus, v(t) =6

5t for 0 ≤ t ≤ T , and v(t) =

725

18for T ≤ t < T +900:

thus6

5T =

725

18or T =

3625

108= 33.56481 seconds. Therefore, s(t) =

3

5t2 for 0 ≤ t ≤ T and

s(t) = s(T )+725

18(t−T ) for T ≤ t ≤ T +900. Thus, the maximum distance the train can travel

is

s(T + 900) =3

5

(

3625

108

)2

+725

18(900) =

143568125

3888≈ 36925.958,

or about 36.9 km.

(b) Suppose that the train starts from rest and must come to a complete stop in 15 minutes. Whatis the maximum distance it can travel under these conditions?

Here, the train starts the same way as above, with a(t) =6

5for 0 ≤ t ≤ T =

3625

108, v(0) = 0

and s(0) = 0, but a(t) = 0 only for T < t < T ∗, while a(t) = −6

5for T ∗ ≤ t ≤ 900 such

that v(900) = 0, while v(t) =725

18for T < t < T ∗. As above, v(t) =

6

5t for 0 ≤ t ≤ T , but

v(t) =725

18−

6

5(t−T ∗) for T ∗ ≤ t ≤ 900, so that v(900) = 0 =

725

18−

6

5(900− T ∗), which gives

900− T ∗ =3625

108= T , so T ∗ = 900− T =

93575

108= 866.43518 seconds. Therefore, s(t) =

3

5t2

for 0 ≤ t ≤ T , s(t) = s(T )+725

18(t−T ) for T ≤ t ≤ T ∗ and s(t) = s(T ∗)+

725

18(t−T ∗)−

3

5(t−T ∗)2

for T ∗ ≤ t ≤ 900. Thus, the maximum distance the train can travel is

s(900) =3

5

(

3625

108

)2

+725

18

(

93575

108−

3625

108

)

+725

18

(

3625

108

)

−3

5

(

3625

108

)2

=67841875

1944≈ 34898,

or about 34.9 km.

(c) Find the minimum time that the train takes to travel between two consecutive stations thatare 72 km apart.

Here, the train travels in the same way as above, but for an unknown total time T̄ . Thus,

a(t) =6

5for 0 ≤ t ≤ T =

3625

108, a(t) = 0 for T < t < T̄−T , and a(t) = −

6

5for T̄−T ≤ t ≤ T̄ . It

follows that v(t) =6

5t for 0 ≤ t ≤ T , v(t) =

725

18for T < t < T̄−T , and v(t) =

725

18−6

5(t−T̄+T )

for T̄ − T ≤ t ≤ T̄ . Therefore, s(t) =3

5t2 for 0 ≤ t ≤ T , s(t) = s(T ) +

725

18(t − T ) for

T ≤ t ≤ T̄ − T and s(t) = s(T̄ − T ) +725

18(t− T̄ + T )−

3

5(t− T̄ + T )2 for T̄ − T ≤ t ≤ T̄ , with

s(T̄ ) = 72000. Thus,

s(T̄ ) = 72000 =3

5

(

3625

108

)2

+725

18

(

T̄ −3625

54

)

+725

18

(

3625

108

)

−3

5

(

3625

108

)2

Michael Daniel Samson Approximation of Areas Page 1 of 4


gives

=⇒725

18T̄ =

142596125

1944=⇒ T̄ =

5703845

3132≈ 1821.151,

or about 30 minutes and 21.151 seconds.

(d) The trip from one station to the next takes 37.5 minutes. How far apart are the stations?

Here, the train travels in the same way as above. Thus, a(t) =6

5for 0 ≤ t ≤ T =

3625

108,

a(t) = 0 for T < t < 2250 − T =239375

108, and a(t) = −

6

5for 2250 − T ≤ t ≤ 2250. It follows

that v(t) =6

5t for 0 ≤ t ≤ T , v(t) =

725

18for T < t < 2250−T , and v(t) =

725

18−6

5(t−2250+T )

for 2250 − T ≤ t ≤ 2250. Therefore, s(t) =3

5t2 for 0 ≤ t ≤ T , s(t) = s(T ) +

725

18(t − T ) for

T ≤ t ≤ 2250 − T and s(t) = s(2250 − T ) +725

18(t − 2250 + T ) −

3

5(t − 2250 + T )2 for

2250− T ≤ t ≤ 2250. Thus,

s(2250) =3

5

(

3625

108

)2

+725

18

(

2250−3625

54

)

+725

18

(

3625

108

)

−3

5

(

3625

108

)2

=173546875

1944≈ 89273,

or about 89.3 km.

§5.1 #5 (a) Estimate the area under the graph of f(x) = 1+x2 from x = −1 to x = 2 using three rectanglesand right endpoints. Then improve your estimate by using six rectangles. Sketch the curveand the approximating rectangles.

For three rectangles, A ≈ 1 · 1 + 1 · 2 + 1 · 5 = 8. For six rectangles, A ≈1

2·5

4+

1

2· 1 +

1

2·

5

4+

1

2· 2 +

1

2·13

4+

1

2· 5 =

55

8= 6.875.

1

2

3

4

5

-1 1 2x

y

(b) Repeat part (a) using left endpoints.

For three rectangles, A ≈ 1 · 2+ 1 · 1+ 1 · 2 = 5. For six rectangles, A ≈1

2· 2+

1

2·5

4+

1

2· 1+

1

2·5

4+

1

2· 2 +

1

2·13

4=

43

8= 5.375.

1

2

3

4

5

-1 1 2x

y



(c) Repeat part (a) using midpoints.

For three rectangles, A ≈ 1 ·5

4+ 1 ·

5

4+ 1 ·

13

4=

23

4= 5.75. For six rectangles, A ≈

1

2·25

16+

1

2·17

16+

1

2·17

16+

1

2·25

16+

1

2·41

16+

1

2·65

16=

95

16= 5.9375.

1

2

3

4

5

-1 1 2x

y

(d) From your sketches in parts (a)–(c), which appears to be the best estimate?

As using right endpoints tended to overestimate, and using left endpoints tended to underes-timate, the best estimates appear to be those using midpoints. As can be determined, thearea is 6 square units.

§5.1 #27 Let A be the area under the graph of an increasing continuous function f from a to b, and let Ln

and Rn be the approximations to A with n subintervals using left and right endpoints, respectively.

(a) How are A, Ln and Rn related?

Since f is increasing throughout the interval, the left endpoints are minimums over the subin-tervals, with the right endpoints are maximums over the subintervals, so Ln < A < Rn.

(b) Show that Rn − Ln =b− a

n[f(b) − f(a)]. Then draw a diagram to illustrate this equation

by showing that the n rectangles representing Rn − Ln can be reassembled to form a singlerectangle whose area is the right side of the equation.

If the partition points are a = x0, x1, . . . , xn−1, xn = b, with ∆x = xi − xi−1 =b− a

n, for

1 ≤ i ≤ n, then Rn = ∆x[f(x1) + · · · + f(xn)] and Ln = ∆x[f(x0) + · · · + f(xn−1)], so

Rn − Ln = ∆x[f(xn)− f(x0)] =b− a

n[f(b)− f(a)].

In the diagram, for each subinterval [xi−1, xi], the difference between the approximationsfor the interval is a rectangle whose area is ∆x[f(xi) − f(xi−1)]. If these rectangles are

stacked on top of each other, their area would be ∆xn∑

i=1

[f(xi) − f(xi−1)], which telescopes

to ∆x[f(xn) − f(x0)]—that is, it forms a rectangle whose width is ∆x, and whose length isf(xn)− f(x0).

(c) Deduce that Rn −A <b− a

n[f(b)− f(a)].

Since Ln < A < Rn, Rn −A < Rn − Ln =b− a

n[f(b)− f(a)].

§5.3 #85 A manufacturing company owns a major piece of equipment that depreciates at the (continuous)rate of f = f(t), where t is the time measured in months since its last overhaul. Because a fixed costA is incurred each time the machine is overhauled, the company wants to determine the optimaltime T (in months) between overhauls.

(a) Explain why

∫

t

0

f(s) ds represents the loss in value of the machine over the period of time t

since the last overhaul.

Since f(t) is the rate at which the equipment loses value, the total loss in value that themachine accrues over time t, where t = 0 is when the last overhaul occurs, then, by the



Fundamental Theorem of Calculus, g(t) =∫

t

0

f(s) ds represents a function corresponding to

loss of value, whose rate of change is g′(t) = f(t), and since g(0) = 0, this is the uniquefunction satisfying the conditions.

(b) Let C = C(t) =1

t

[

A+

∫

t

0

f(s) ds

]

. What does C represent and why would the company

want to minimize C?

Since g(t) corresponds to the loss of value of the equipment over time t > 0, and A is the costof overhauling the equipment, the total cost incurred when the equipment is overhauled attime t is A+ g(t), and C(t) = [A+ g(t)]/t is the average cost for overhauling the equipment, ifthe equipment is overhauled at time t (and not earlier)—minimizing the average cost of eachoverhaul is ideal for minimizing over time the costs associated with the equipment.

(c) Show that C has minimum value at the numbers t = T where C(T ) = f(T ).

As C(t) is continuous over t > 0 where f(t) is continuous, by Fermat’s Theorem, it minimizes

at a critical point: C ′(t) =tf(t)−A− g(t)

t2=

f(t)− C(t)

t, so when t > 0, C ′(t) = 0 only when

C(t) = f(t).By definition, f(t) ≥ 0 (since f(t) < 0 implies that the value of the equipment is increasingover some interval), so A + g(t) is an increasing function on t > 0, with lim

t→0+C(t) = ∞.

In addition, the loss on the equipment is capped by the cost of the equipment, say E—thatis, g(t) ≤ E, so f(t), C(t) → 0 as t → ∞. As such, if t = T is the smallest such T thatf(T ) = C(T ), then C(t) > f(t) for 0 < t < T , and C ′(t) < 0 for 0 < t < T—the smallest sucht = T gives a minimum value for C(t).It is unclear if the other values of t such that f(t) = C(t) also provides a minimum: if C ′(t) > 0for some interval T < t, since both are continuous, if f(t) = C(t) afterward, it will occur at amaximal value of C(t).

Source: James Stewart, Calculus Early Transcendentals, 8e, International Metric EditionReproduced without explicit permission, for use in SEM 1102 / MAT 200 class, DigiPen Institute of Technology, Singapore, January

2020.


SEM 1102 / MAT 200 Worksheet 1 Quiz 1 on January 15

Name: dP ID: 0 0 0

§5.5 #78 Evaluate

∫

1

0

x√

1− x4 dx by making a substitution and interpreting the resulting integral in terms

of an area.

Let u = x2: thendu = 2x dx and

∫

1

0

x√

1− x4 dx =1

2

∫

1

0

√

1− u2 du. This is half the area of the

portion of the unit circle in the first quadrant, thus

∫

1

0

x√

1− x4 dx =1

2

[

1

4π

]

=π

8.

§5.5 #94 (a) If f is continuous, prove that

∫ π/2

0

f(cosx) dx =

∫ π/2

0

f(sinx) dx.

Using the substitution u =π

2− x, x =

π

2− u, anddu = − dx, so

∫ π/2

0

f(cosx) dx =

∫

0

π/2

−f[

cos(π

2− u

)]

du =

∫ π/2

0

f(sinu) du =

∫ π/2

0

f(sinx) dx.

(b) Use part (a) to evaluate

∫ π/2

0

cos2 x dx and

∫ π/2

0

sin2 x dx.

Since∫ π/2

0

cos2 x dx+

∫ π/2

0

sin2 x dx =

∫ π/2

0

(cos2 x+ sin2 x) dx =

∫ π/2

0

dx =π

2,

and f(x) = x2 is continuous, from above,

∫ π/2

0

cos2 x dx =

∫ π/2

0

sin2 x dx =1

2

[π

2

]

=π

4.

§7.1 #32 Evaluate

∫

2

1

(lnx)2

x3dx.

Let u = (lnx)2: thendu =2 lnx dx

x,dv =

dx

x3, so v = −

1

2x2and

∫

2

1

(lnx)2

x3dx =

[

−1

2

(

lnx

x

)2]2

1

+

∫

2

1

lnx

x3dx.

Let U = lnx: thendU =dx

x,dV =

dx

x3, so V = −

1

2x2and

∫

2

1

(lnx)2

x3dx =

[

−(lnx)2 + lnx

2x2

]2

1

+1

2

∫

2

1

dx

x3=

[

−2(lnx)2 + 2 lnx+ 1

4x2

]2

1

=1

4−

2(ln 2)2 + 2 ln 2 + 1

16=

3− ln 4− 2(ln 2)2

16≈ 0.0408.

§7.1 #70 If f(0) = g(0) = 0 and f ′′ and g′′ are continuous, show that∫ a

0

f(x)g′′(x) dx = f(a)g′(a)− f ′(a)g(a) +

∫ a

0

f ′′(x)g(x) dx.

Let u = f(x): thendu = f ′(x) dx,dv = g′′(x) dx, so v = g′(x) and∫ a

0

f(x)g′′(x) dx = [f(x)g′(x)]a0−

∫ a

0

f ′(x)g′(x) dx = f(a)g′(a)−

∫ a

0

f ′(x)g′(x) dx.

Likewise,∫ a

0

f ′′(x)g(x) dx = [f ′(x)g(x)]a0−

∫ a

0

f ′(x)g′(x) dx = f ′(a)g(a)−

∫ a

0

f ′(x)g′(x) dx,

thus

∫ a

0

f ′(x)g′(x) dx = f ′(a)g(a)−

∫ a

0

f ′′(x)g(x) dx, and the statement follows by substitution.

Michael Daniel Samson Substitution Rule and Integration by Parts Page 1 of 2

SEM 1102 / MAT 200 Worksheet 1 Quiz 1 on January 15

Source: James Stewart, Calculus Early Transcendentals, 8e, International Metric Edition

Reproduced without explicit permission, for use in SEM 1102 / MAT 200 class, DigiPen Institute of Technology, Singapore, January

2020.

Michael Daniel Samson Substitution Rule and Integration by Parts Page 2 of 2


Name: dP ID: 3 9 0 0 0

§5.5 #77 Evaluate

∫

2

−2

(x+3)√

4− x2 dx by writing it as a sum of two integrals and interpreting one of those

integrals in terms of an area.∫

2

−2

(x+3)√

4− x2 dx = 3

∫

2

−2

√

4− x2 dx+

∫

2

−2

x√

4− x2 dx. The first integral is thrice the area of

the semicircle above the x-axis x2 + y2 = 4, which is 6π; using u = 4− x2 with the second interval

givesdu = −2x dx and

∫

2

−2

x√

4− x2 dx = −1

2

∫

0

0

√u du = 0. Thus, the integral is equal to 6π.

§5.5 #91 If a and b are positive numbers, show that

∫

1

0

xa(1− x)b dx =

∫

1

0

xb(1− x)a dx.

Using y = 1− x, givesdy = − dx and

∫

1

0

xa(1− x)b dx = −∫

0

1

(1− y)ayb dy =

∫

1

0

yb(1− y)a dy =∫

1

0

xb(1− x)a dx, which results from changing the dummy variable of the integral from y to x.

§7.1 #53 Use integration by parts to prove

∫

tann x dx =tann−1 x

n− 1−

∫

tann−2 x dx, (n 6= 1).

Noting

∫

tanx dx = ln | secx| + C, consider for n ≥ 2, that tann x = tann−2 x(sec2 x − 1), so∫

tann x dx =

∫

tann−2 x sec2 x dx−∫

tann−2 x dx. For the first integral, letting u = tanx setsdu =

sec2 x dx, such that∫

tann x dx =

∫

un−2 du−∫

tann−2 x dx =un−1

n− 1−∫

tann−2 x dx =tann−1 x

n− 1−∫

tann−2 x dx.

§7.1 #67 The Fresnel function S(x) =

∫

x

0

sin

(

1

2πt2

)

dt was discussed in Example 5.3.3 and is used exten-

sively in the theory of optics. Find

∫

S(x) dx. [Your answer will involve S(x).]

Let u = S(x): thendv = dx gives v = x anddu = sin

(

1

2πx2

)

dx, so

∫

S(x) dx = xS(x)−∫

x sin

(

1

2πx2

)

dx.

Let w =πx2

2: thendw = πx dx, so

∫

S(x) dx = xS(x)− 1

π

∫

sinw dw = xS(x) +1

πcosw + C = xS(x) +

1

πcos

(

1

2πx2

)

+ C.

§7.1 #71 Suppose that f(1) = 2, f(4) = 7, f ′(1) = 5, f ′(4) = 3, and f ′′ is continuous. Find the value of∫

4

1

xf ′′(x) dx.

Letting u = x setsdv = f ′′(x) dx, such thatdu = dx, v = f ′(x) and

∫

4

1

xf ′′(x) dx = xf ′(x)]x=4

x=1−

∫

4

1

f ′(x) dx = 4f ′(4)− f ′(1)− f(4) + f(1) = 12− 5− 7 + 2 = 2.



2020.

Michael Daniel Samson Substitution Rule and Integration by Parts One page only

SEM 1102 / MAT 200 Trigonometric Integrals 20 January 2020

Find

∫

sinm x cosn x dx, m,n ≥ 0, integers.

• If m is odd

1. Replace sinm−1 x = (1− cos2 x)(m−1)/2:

∫

sinm x cosn x dx =

∫

(1− cos2 x)(m−1)/2 cosn x sinx dx =

∫

p(cosx) sinx dx,

where p is a polynomial.

2. Use the substitution u = cosx and integrate: du = − sinx dx and

∫

sinm x cosn x dx = −

∫

p(u) du = −P (u) + C = −P (cosx) + C,

where P is a polynomial such that P ′ = p.

• If n is odd

1. Replace cosn−1 x = (1− sin2 x)(n−1)/2:

∫

sinm x cosn x dx =

∫

sinm x(1− sin2 x)(n−1)/2 cosx dx =

∫

p(sinx) cosx dx,


2. Use the substitution u = sinx and integrate: du = cosx dx and

∫

sinm x cosn x dx =

∫

p(u) du = P (u) + C = P (sinx) + C,


• Otherwise (m and n are both even)

1a. If m ≥ n: replace sinn x cosn x =

(

sin 2x

2

)n

and sinm−n x =

(

1− cos 2x

2

)(m−n)/2

∫

sinm x cosn x dx =1

2(m+n)/2

∫

sinn 2x(1− cos 2x)(m−n)/2 dx.

1b. If m < n: replace sinm x cosm x =

(

sin 2x

2

)m

and cosn−m x =

(

1 + cos 2x

2

)(n−m)/2

∫

sinm x cosn x dx =1

2(m+n)/2

∫

sinm 2x(1 + cos 2x)(n−m)/2 dx.

2. Use substitution u = 2x and perform termwise integration (return to top): du = 2 dx,

∫

sinm x cosn x dx =1

2(m+n+2)/2

|n−m|/2∑

k=0

sign(n−m)k(

|n−m|/2

k

)∫

sinmin(n,m) u cosk u du.

Since the power of sinx will still be even, terms where k is odd can be solved by the secondcase above, but this case will be revisited when k is even.

Find

∫

sinm x

cosn xdx, m,n ≥ 0, integers. (Analogous process for

∫

cosm x

sinn xdx, using cofunctions.)

• If m > n

Michael Daniel Samson Algorithms Page 1 of 2

SEM 1102 / MAT 200 Trigonometric Integrals 20 January 2020

1. Replace sinm x = (1− cos2 x)⌊m/2⌋ sinm mod 2 x∫

sinm x

cosn xdx =

∫

(1− cos2 x)⌊m/2⌋ sinm mod 2 x

cosn xdx.

2. Perform termwise integration (go to next cases or use previous algorithm)

∫

sinm x

cosn xdx =

⌊m/2⌋∑

k=0

(−1)k(

⌊m/2⌋

k

)∫

sinm mod 2 x cos2k−n x dx.

If 2k − n ≥ 0, the previous algorithm is used, otherwise the next cases are used.

• If m = n,sinn x

cosn x= tann x: Use

∫

tanx dx =

∫

sinx dx

cosx= −

∫

d(cosx)

cosx= ln | secx|+ C, and,

∫

tann x dx =

∫

tann−2 x sec2 x dx−

∫

tann−2 x dx =1

n− 1tann−1 x−

∫

tann−2 x dx, for n > 1.

• Otherwise

(

n > m,sinm x

cosn x= tanm x secn−m x

)

– If n−m is even

1. Replace secn−m−2 x = (1 + tan2 x)(n−m−2)/2:∫

sinm x

cosn xdx =

∫

tanm x(1 + tan2 x)(n−m−2)/2 sec2 x dx =

∫

p(tanx) sec2 x dx,


2. Use the substitution u = tanx and integrate: du = sec2 x dx and∫

sinm x

cosn xdx =

∫

p(u) du = P (u) + C = P (tanx) + C,


– If m is odd

1. Replace tanm−1 x = (sec2 x− 1)(m−1)/2:∫

sinm x

cosn xdx =

∫

(sec2 x− 1)(m−1)/2 secn−m x tanx dx =

∫

p(secx) secx tanx dx,


2. Use the substitution u = secx and integrate: du = secx tanx dx and∫

sinm x

cosn xdx =

∫

p(u) du = P (u) + C = P (secx) + C,


– Otherwise (m is even and n is odd)

1. Replace tanm x = (sec2 x− 1)m/2:

∫

sinm x

cosn xdx =

∫

(sec2 x− 1)m/2 secn−m x dx =

m/2∑

k=0

(−1)k(

m/2

k

)∫

secn−2k x dx.

2. Use

∫

secx dx =

∫

secx(secx+ tanx)

secx+ tanxdx =

∫

d(secx+ tanx)

secx+ tanx= ln | secx+tanx|+C,

and, for n > 0, by integration by parts (u = sec2n−1 x and dv = sec2 x dx):∫

sec2n+1 x dx = sec2n−1 x tanx− (2n− 1)

∫

sec2n−1 x tan2 x dx

= sec2n−1 x tanx− (2n− 1)

∫

sec2n+1 x dx+ (2n− 1)

∫

sec2n−1 x dx

=1

2nsec2n−1 x tanx+

2n− 1

2n

∫

sec2n−1 x dx.

Michael Daniel Samson Algorithms Page 2 of 2


Name: dP ID: 3 9 0 0 0

§7.2 #65 A particle moves on a straight line with velocity function v(t) = sinωt cos2 ωt. Find its positionfunction s = f(t) if f(0) = 0.

Since s(t) =

∫

t

0

v(x) dx, as s(0) = 0, using u = cosωx givesdu = −ω sinωx dx and

s(t) =

∫

t

0

sinωx cos2 ωx dx = − 1

ω

∫

cosωt

1

u2 du =

[

u3

3ω

]1

cosωt

=1− cos3 ωt

3ω,

assuming ω 6= 0, otherwise v(t) = s(t) ≡ 0.

§7.3 #13 Evaluate

∫

√x2 − 9

x3dx.

Using x = 3 sec θ givesdx = 3 sec θ tan θ dθ and√x2 − 9 = 3 tan θ and, since sin2 θ =

1

2(1−cos 2θ),

∫

√x2 − 9

x3dx =

∫

3 tan θ(3 sec θ tan θ dθ)

(3 sec θ)3dx =

1

3

∫

sin2 θ dθ =1

6

∫

dθ − 1

6

∫

cos 2θ dθ

=θ

6− sin 2θ

12+ C =

1

6sec−1

(x

3

)

−√x2 − 9

2x2+ C.

Check:d

dx

[

1

6sec−1

(x

3

)

−√x2 − 9

2x2+ C

]

=1

6x√

(x/3)2 − 1− 2x2(x/

√x2 − 9)− 4x

√x2 − 9

4x4=

1

2x√x2 − 9

− x3 − 2x(x2 − 9)

2x4√x2 − 9

=x3 − x3 + 2x3 − 18x

2x4√x2 − 9

=x2 − 9

x3√x2 − 9

=

√x2 − 9

x3.

§7.3 #29 Evaluate

∫

x√

1− x4 dx.

Letting u = x2 = sin θ givesdu = 2x dx = cos θ dθ,√1− u2 =

√1− x4 = cos θ and

∫

x√

1− x4 dx =

1

2

∫

cos2 θ dθ =1

4

∫

dθ +1

4

∫

cos 2θ dθ =θ

4+

sin 2θ

8+ C =

1

4sin−1(x2) +

x2√1− x4

4+ C.

Check:d

dx

[

1

4sin−1(x2) +

x2√1− x4

4+ C

]

=x

2√1− x4

+x√1− x4

2− x5

2√1− x4

=x+ x(1− x4)− x5

2√1− x4

=

x(1− x4)√1− x4

= x√

1− x4.

§7.3 #31 (a) Use trigonometric substitution to show that

∫

dx√x2 + a2

= ln(x+√

x2 + a2) + C.

Using x = a tan θ givesdx = a sec2 θ dθ,√x2 + a2 = a sec θ and

∫

dx√x2 + a2

=

∫

a sec2 θ dθ

a sec θ=

∫

sec θ dθ = ln | sec θ + tan θ|+ C = ln

∣

∣

∣

∣

∣

√x2 + a2

a+

x

a

∣

∣

∣

∣

∣

+ C = ln(x+√

x2 + a2) + C.

(b) Use the hyperbolic substitution x = a sinh t to show that

∫

dx√x2 + a2

= sinh−1

(x

a

)

+ C.

Using x = a sinh t givesdx = a cosh t dt,√x2 + a2 = a cosh t and

∫

dx√x2 + a2

=

∫

a cosh t dt

a cosh t=

∫

dt = t+ C = sinh−1

(x

a

)

+ C.

These formulas are connected by sinh−1 x = ln(x+√x2 + 1), x ∈ R.

Note sinh−1

(x

a

)

= ln

(

x

a+

√

[x

a

]2

+ 1

)

= ln

∣

∣

∣

∣

∣

x

a+

√x2 + a2

a

∣

∣

∣

∣

∣

.

Michael Daniel Samson Integrals with Trigonometric Functions Page 1 of 2




2020.

Michael Daniel Samson Integrals with Trigonometric Functions Page 2 of 2

SEM 1102 / MAT 200 Homework 4 5 February 2020

Name: dP ID: 0 0 0

37. Evaluate

∫

x2 − 3x+ 7

(x2 − 4x+ 6)2dx.

Let u = x − 2 =√2 tan θ: thendu = dx =

√2 sec2 θ dθ,

√x2 − 4x+ 4 + 2 =

√

u2 + (√2)2 =

√2√tan2 θ + 1 =

√2 sec θ and

∫

x2 − 3x+ 7

(x2 − 4x+ 6)2dx =

∫

dx

x2 − 4x+ 6dx+

1

2

∫

2x− 4

(x2 − 4x+ 6)2dx+ 3

∫

dx

(x2 − 4x+ 6)2

=

∫

√2 sec2 θ dθ

(√2 sec θ)2

+1

2

∫

d(x2 − 4x+ 6)

(x2 − 4x+ 6)2+ 3

∫

√2 sec2 θ dθ

(√2 sec θ)4

=

√2

2tan−1

√2(x− 2)

2−

1

2(x2 − 4x+ 6)+

3√2

4

∫

cos2 θ dθ

=

√2

2tan−1

√2(x− 2)

2−

1

2(x2 − 4x+ 6)+

3√2

8tan−1

√2(x− 2)

2+

3√2

8

∫

cos 2θ dθ

=7√2

8tan−1

√2(x− 2)

2−

1

2(x2 − 4x+ 6)+

3√2

8sin θ cos θ + C

=7√2

8tan−1

√2(x− 2)

2−

1

2(x2 − 4x+ 6)+

3

4

x− 2

x2 − 4x+ 6+ C

=7√2

8tan−1

√2(x− 2)

2+

3x− 8

4(x2 − 4x+ 6)+ C.

43. Make a substitution to express the integrand as a rational function and then evaluate

∫

x3 dx3√x2 + 1

.

Let u = 3√x2 + 1: thus, u3 − 1 = x2, 3u2 du = 2x dx and

∫

x3 dx3√x2 + 1

=1

2

∫

x2(2x dx)3√x2 + 1

=1

2

∫

(u3 − 1)(3u2 du)

u=

3

2

∫

(u4 − u) du =3

2

(

u5

5−

u2

2

)

+ C

=3(2x2 − 3)

203

√

(x2 + 1)2 + C.

59. The German mathematician Karl Weierstrass (1815–1897) noticed that the substitution t = tan(x/2)will convert any rational function of sinx and cosx into an ordinary rational function of t.

(a) If t = tan(x/2), −π < x < π, sketch a right triangle or use trigonometric identities to show

that cos(x

2

)

=1

√1 + t2

and sin(x

2

)

=t

√1 + t2

.

Since t = tan(x

2

)

, t2 + 1 = tan2(x

2

)

+ 1 = sec2(x

2

)

, so cos2(x

2

)

=1

t2 + 1, sin2

(x

2

)

= 1−

cos2(x

2

)

=t2

t2 + 1, and the above follow, since cos

(x

2

)

≥ 0 and sin(x

2

)

= tan(x

2

)

cos(x

2

)

.

(b) Show that cosx =1− t2

1 + t2and sinx =

2t

1 + t2.

These follow, since cosx = cos2(x

2

)

− sin2(x

2

)

and sinx = 2 sin(x

2

)

cos(x

2

)

.

(c) Show thatdx =2

1 + t2dt.

Given that sinx =2t

1 + t2, cosx dx =

1− t2

1 + t2dx =

2(1 + t2)− (2t)2

(1 + t2)2dt, thus

dx =1 + t2

1− t22t+ 2t2 − 4t2

(1 + t2)2dt =

2(1− t2) dt

(1− t2)(1 + t2)=

2 dt

1 + t2.

Michael Daniel Samson Integrals with Rational Functions Page 1 of 2


63. Use the substitution in Exercise 59 to transform the integrand into a rational function of t and then

evaluate

∫ π/2

0

sin 2x dx

2 + cosx.

Using Weierstrass substitution:

∫ π/2

0

sin 2x dx

2 + cosx= 8

∫

1

0

t(1− t2) dt

2(1 + t2)3 + (1− t2)(1 + t2)2= 8

∫

1

0

(t− t3) dt

(3 + t2)(1 + t2)2.

The integrand can be given in partial fractions as

t− t3

(3 + t2)(1 + t2)2=

At+B

t2 + 3+Ct+D

t2 + 1+

Et+ F

(t2 + 1)2=

(At+B)(t2 + 1)2 + (Ct+D)(t2 + 3)(t2 + 1) + (Et+ F )(t2 + 3)

(3 + t2)(1 + t2)2.

The numerators, evaluated at the given values of t:

t = 0 : B + 3D + 3F = 0

t = 1 : 4A+ 4B+ 8C+ 8D+ 4E+ 4F = 0

t = −1 : −4A+ 4B− 8C+ 8D− 4E+ 4F = 0

t = 2 : 50A+ 25B+ 70C+ 35D+ 14E+ 7F = −6

t = −2 : −50A+ 25B− 70C+ 35D− 14E+ 7F = 6

t = 3 : 300A+ 100B+ 360C+ 120D+ 36E+ 12F = −24

These produce the values:

B + 3D + 3F = 0

8B + 16D + 8F = 0

50B + 70D + 14F = 0

=⇒

B = 0

D = 0

F = 0

,

4A+ 8C + 4E = 0

50A+ 70C + 14E = −6

300A+ 360C + 36E = −24

=⇒

A = 1

C = −1

E = 1

These indicate

∫ π/2

0

sin 2x dx

2 + cosx= 4

∫

1

0

2t dt

3 + t2− 4

∫

1

0

2t dt

1 + t2+ 4

∫

1

0

2t dt

(1 + t2)2=

[

4 ln

∣

∣

∣

∣

t2 + 3

t2 + 1

∣

∣

∣

∣

−4

t2 + 1

]1

0

=[

4 ln∣

∣

∣1 + 2 cos2

x

2

∣

∣

∣− 4 cos2

x

2

]π/2

0

= 4 ln 2− 2− 4 ln 3 + 4 = 2− 4 ln3

2≈ 1.9729719975561.

71. The rational number22

7has been used as an approximation to the number π since the time of

Archimedes. Show that

∫

1

0

x4(1− x)4

1 + x2dx =

22

7− π.

It can be shown that x4(1− x)4 = (x6 − 4x5 + 5x4 − 4x2 + 4)(x2 + 1)− 4, thus

∫

1

0

x4(1− x)4

1 + x2dx =

∫

1

0

(x6 − 4x5 + 5x4 − 4x2 + 4) dx− 4

∫

1

0

dx

1 + x2

=

[

x7

7−

2x6

3+ x5 −

4x3

3+ 4x− 4 tan−1 x

]1

0

=1

7−

2

3+ 1−

4

3+ 4− π =

22

7− π.

Source: James Stewart, Calculus Early Transcendentals, 8e, International Metric Edition, Section 7.4Reproduced without explicit permission, for use in SEM 1102 / MAT 200 class, DigiPen Institute of Technology, Singapore, January

2020.

Michael Daniel Samson Integrals with Rational Functions Page 2 of 2

SEM 1102 / MAT 200 Worksheet 2 Quiz 2 on February 5

Name: dP ID: 0 0 0

§7.2 #56 Evaluate

∫

sinx cosx dx by four methods:

(a) the substitution u = cosx

This givesdu = − sinx dx, thus

∫

sinx cosx dx = −∫

u du = −u2

2+ C = −cos2 x

2+ C.

(b) the substitution u = sinx

This givesdu = cosx dx, thus

∫

sinx cosx dx =

∫

u du =u2

2+ C =

sin2 x

2+ C.

(c) the identity sin 2x = 2 sinx cosx

Letting u = 2x givesdu = 2dx, thus

∫

sinx cosx dx =1

4

∫

sinu du = −cosu

4+C = −cos 2x

4+

C.

(d) integration by parts

Letting u = cosx setsdv = sinx dx, thusdu = − sinx dx, v = − cosx and

∫

sinx cosx dx =

− cos2 x−∫

sinx cosx dx, so

∫

sinx cosx dx = −cos2 x

2+ C.

Letting u = sinx setsdv = cosx dx, thusdu = cosx dx, v = sinx and

∫

sinx cosx dx =

sin2 x−∫

sinx cosx dx, so

∫

sinx cosx dx =sin2 x

2+ C.

Explain the different appearances of the answers.

From the identities cos 2x = cos2 x − sin2 x and sin2 x + cos2 x = 1, − cos 2x = 1 − 2 cos2 x =

2 sin2 x − 1, so the arbitrary constant C absorbs the constant ±1

4in the equivalent values for

−cos 2x

4, and the expressions are equivalent.

§7.2 #70 A finite Fourier sine series is given by the sum f(x) =N∑

n=1an sinnx = a1 sinx + a2 sin 2x + · · · +

aN sinNx. Show that the mth coefficient am is given by the formula am =1

π

∫ π

−π

f(x) sinmx dx.

First, it must be shown that

∫ π

−π

sinmx sinnx dx =

{

0 if m 6= n,

π if m = n,where m and n are positive

integers: 2 sinmx sinnx = cos[(m− n)x]− cos[(m+ n)x], so, if m 6= n,

∫ π

−π

sinmx sinnx dx =1

2

∫ π

−π

(cos[(m− n)x]− cos[(m+ n)x]) dx =

[

sin[(m− n)x]

2(m− n)− sin[(m+ n)x]

2(m+ n)

]π

−π

,

all of whose terms are zero. But, if m = n,

∫ π

−π

sin2 nx dx =1

2

∫ π

−π

dx− 1

2

∫ π

−π

cos 2nx dx =

[

x

2− sin 2nx

4n

]π

−π

= π.

Then,

∫ π

−π

f(x) sinmx dx =

∫ π

−π

(

N∑

n=1

an sinnx sinmx

)

dx =

N∑

n=1

[

an

∫ π

−π

sinnx sinmx dx

]

= 0 + · · ·+ 0 + am

∫ π

−π

sinmx sinmx dx+ 0 + · · · = amπ,

and the conclusion follows from dividing both sides by π.

Michael Daniel Samson Integration with Trigonometric Functions Page 1 of 3


§7.3 #32 Evaluate

∫

x2

(x2 + a2)3/2dx

(a) by trigonometric substitution

Using x = a tan θ givesdx = a sec2 θ dθ,√x2 + a2 = a sec θ and

∫

x2

(x2 + a2)3/2dx =

∫

a2 tan2 θ(a sec2 θ dθ)

a3 sec3 θ=

∫

sin2 θ dθ

cos θ=

∫

(1− cos2 θ) dθ

cos θ

=

∫

sec θ dθ −∫

cos θ dθ = ln | sec θ + tan θ| − sin θ + C

= ln

∣

∣

∣

∣

∣

√x2 + a2

a+

x

a

∣

∣

∣

∣

∣

− x/a√x2 + a2/a

+ C = ln∣

∣

∣x+

√

x2 + a2∣

∣

∣− x√

x2 + a2+ C.

(b) by the hyperbolic substitution x = a sinh t

Using x = a sinh t givesdx = a cosh t dt,√x2 + a2 = a cosh t and

∫

x2

(x2 + a2)3/2dx =

∫

a2 sinh2 t(a cosh t dt)

a3 cosh3 t=

∫

tanh2 t dt =

∫

(1− sech2t) dt,

since dividing both sides of the identity cosh2 t−sinht = 1 by cosh2 t gives 1−tanh2 t = sech2t,

∫

x2

(x2 + a2)3/2dx =

∫

dt−∫

sech2t dt = t− tanh t+ C = sinh−1(x

a

)

− x/a√x2 + a2/a

+ C,

sinced

dttanh t =

cosh2 t− sinh2 t

cosh2 t= sech2t, and

∫

x2

(x2 + a2)3/2dx == ln

∣

∣

∣

∣

∣

x

a+

√

(x

a

)2

+ 1

∣

∣

∣

∣

∣

− x√x2 + a2

+C = ln∣

∣

∣x+

√

x2 + a2∣

∣

∣− x√

x2 + a2+C,

since y = sinh−1 x means x = sinh y =ey − e−y

2and e2y − 2xey − 1 = 0: by the quadratic

formula, ey =2x

2±√

(−2x)2 + 4

4= x+

√x2 + 1 ≥ 0, thus y = ln |x+

√x2 + 1| = sinh−1 x.

§7.3 #44 A water storage tank has the shape of a cylinder with diameter 10 m. It is mounted so that thecircular cross-sections are vertical. If the depth of the water is 7 m, what percentage of the totalcapacity is being used?

The percentage of the total capacity occupied by the water in the tank is the same as the area of

any vertical circular cross-section that is in water: if such a cross-section has a cartesian grid with

units of length one meter, the origin in the center of the circle, and whose positive x-axis is pointing

downward, the percentage to be determined is the area of the portion of the circle x2 + y2 = 25to the right of the line x = −2. By symmetry with respect to the x-axis, the percentage is the area

under the curve y =√25− x2 from x = −2 to x = 5, divided by the area of the semicircle, which

is25π

2. Thus, using x = 5 sin θ, givingdx = 5 cos θ dθ and

√25− x2 = 5 cos θ, the percentage is

2

25π

∫ 5

−2

√

25− x2 dx =2

π

∫ π/2

sin−1(−2/5)

cos2 θ dθ =1

π

∫ π/2

sin−1(−2/5)

dθ +1

π

∫ π/2

sin−1(−2/5)

cos 2θ dθ

=1

2π[2θ + sin 2θ]

π/2

sin−1(−2/5)=

1

2− 1

πsin−1

(

−2

5

)

+ 0− 1

2πsin

[

2 sin−1

(

−2

5

)]

,

where cos

[

sin

(

−2

5

)]

=

√21

5gives the last term as 2

(

−2

5

)√21

5=

−4√21

25; thus the percentage

is25π + 4

√21

50π− 1

πsin−1

(

−2

5

)

≈ 0.7476842 or about 74.8%.





2020.



Name: dP ID: 0 0 0

Evaluate the integral.

26.

∫ 1

0

(3x2 + 1) dx

x3 + x2 + x+ 1

Since the denominator can be factored into (x2 + 1)(x+ 1), the integrand can be split into

∫ 1

0

3x2 + 1

x3 + x2 + x+ 1dx =

∫ 1

0

Adx

x+ 1+

∫ 1

0

Bx+ C

x2 + 1dx =

[

A ln |x+ 1|+ B

2ln(x2 + 1) + C tan−1 x

]1

0

=2A+B

Bln 2 +

Cπ

4, where 3x2 + 1 = A(x2 + 1) + (Bx+ C)(x+ 1).

The unknowns A, B and C are defined by a system: A + B = 3, B + C = 0 and A + C = 1, so

A = 2, B = 1 and C = −1, so

∫ 1

0

(3x2 + 1) dx

x3 + x2 + x+ 1=

5

2ln 2− π

4.

Check :d

dx

[

ln(

(x+ 1)2√

x2 + 1)

− tan−1 x]

=

2(x+ 1)√x2 + 1 +

x(x+ 1)2√x2 + 1

(x+ 1)2√x2 + 1

− 1

x2 + 1

=2(x2 + 1) + x(x+ 1)− (x+ 1)

(x+ 1)(x2 + 1)=

2x2 + 2 + x2 + x− x− 1

(x+ 1)(x2 + 1)=

3x2 + 1

x3 + x2 + x+ 1

52.

∫

dx

x(x4 + 1)

Since the denominator can be factored into x(x2+√2+1)(x2−

√2+1), the integrand can be split

into∫

dx

x(x4 + 1)dx =

∫

Adx

x+

∫

Bx+ C

[x+ (1/√2)]2 + (1/2)

dx+

∫

Dx+ E

[x− (1/√2)]2 + (1/2)

dx

= A ln |x|+∫

Bu+ − (B/√2) + C

u2+ + (1/

√2)2

dx+

∫

Du− + (D/√2) + E

u2− + (1/

√2)2

dx

= A ln |x|+ B

2ln(x2 +

√2x+ 1) + (

√2C −B) tan−1(

√2x+ 1) +

D

2ln(x2 −

√2x+ 1)

+ (√2E +D) tan−1(

√2x− 1), where u± = x± 1√

2, and the unknowns satisfy

1 = A(x4 + 1) + (Bx+ C)x(x2 −√2x+ 1) + (Dx+ E)x(x2 +

√2x+ 1).

The unknowns A, B, C, D and E are defined by a system: A+B+D = 0, −√2B+C+

√2D+E = 0,

B−√2C +D+

√2E = 0, C +E = 0 and A = 1, so B = −1

2, C = − 1

2√2

, D = −1

2and E =

1

2√2

,

so

∫

dx

x(x4 + 1)= ln |x| − 1

4ln(x4 + 1) + C.

Check :d

dx

[

ln

∣

∣

∣

∣

x4√x4 + 1

∣

∣

∣

∣

]

=4√x4 + 1

x

4√x4 + 1− x4

4

√

(x4 + 1)3√x4 + 1

=x4 + 1− x4

x(x4 + 1)=

1

x(x4 + 1)

56.

∫

dx√x+ x

√x

Using u2 = x gives 2u du = dx and

∫

dx√x+ x

√x=

∫

2u du

u+ u3= 2 tan−1 u+ 1 = 2 tan−1

√x+ C.

Check :d

dx

[

2 tan−1√x]

=2

x+ 1

1

2√x=

1√x+ x

√x

Michael Daniel Samson Strategy for Integration One page only


72.

∫

ln(x+ 1) dx

x2

Using u = ln(x+ 1) leavesdv =dx

x2, thusdu =

dx

x+ 1and v = − 1

x, and

∫

ln(x+ 1) dx

x2= − ln(x+ 1)

x+

∫

dx

x(x+ 1)= − ln(x+ 1)

x+

∫

dx

x−∫

dx

x+ 1= ln

∣

∣

∣

∣

x

x+ 1

∣

∣

∣

∣

− ln(x+ 1)

x+C.

Check :d

dx

[

ln

∣

∣

∣

∣

x

x+ 1

∣

∣

∣

∣

− ln(x+ 1)

x

]

=x+ 1

x

1

(x+ 1)2−

x

x+ 1− ln(x+ 1)

x2=

1

x(x+ 1)− 1

x(x+ 1)+

ln(x+ 1)

x2

Source: James Stewart, Calculus Early Transcendentals, 8e, International Metric Edition, Section 7.5


2020.

Michael Daniel Samson Strategy for Integration One page only

SEM 1102 / MAT 200 Integration Strategy 12 February 2020

A noncomprehensive guide to antidifferentiation and integration

0. Preprocessing: likely only need to be performed once

• Definite integrals: §5.3

∫

a

a

f(x) dx = 0,

∫

b

a

f(x) dx =

∫

c

a

f(x) dx+

∫

b

c

f(x) dx,

∫

a

b

f(x) dx = −∫

b

a

f(x) dx.

• If f(x) is odd (f(−x) = −f(x)),

∫

a

−a

f(x) dx = 0; §5.3

if f(x) is even (f(−x) = f(x)),

∫

a

−a

f(x) dx = 2

∫

a

0

f(x) dx.

• If p(x) = q(x)d(x) + r(x), all polynomials,

∫

p(x)

d(x)dx =

∫(

q(x) +r(x)

d(x)

)

dx. §7.4

1. Known Antiderivative: constants of integration are omitted below §7.5

∫

xn dx =xn+1

n+ 1, n 6= −1;

∫

dx

x= ln |x|;

∫

bx dx =bx

ln b;

∫

sinhx dx = coshx;∫

secx dx = ln | secx+ tanx|;∫

cscx dx = ln | cscx− cotx|;∫

coshx dx = sinhx

2. Integral Linearity:

∫ n∑

k=1

ckfk(x) dx =

n∑

k=1

ck

∫

fk(x) dx, go to 1 §5.3

3. Reduction Formulas: some examples, for n ≥ 1, go to 1 after §7.2

∫

tann+2 x dx =tann+1 x

n+ 1−∫

tann x dx,

∫

sec2n+1 x dx =sec2n−1 x tanx

2n+

2n− 1

2n

∫

sec2n−1 x dx,

∫

cotn+2 x dx = −cotn+1 x

n+ 1−∫

cotn x dx,

∫

csc2n+1 x dx = −csc2n−1 x cotx

2n+

2n− 1

2n

∫

csc2n−1 x dx.

4. Direct Substitution:

∫

f(g(x))g′(x) dx =

∫

f(g(x)) d(g(x)) =

∫

f(u) du: if a definite integral, §5.5

limits become from u = g(a) to u = g(b), when before from x = a to x = b, go to 0

5. Indirect Substitution: go to 0 after

• Using u = f−1(x) gives f(u) = x so f ′(u) du = dx.

These include: rationalizing substitutions, e.g., u =√x+ a gives u2 = x+ a and 2u du = dx; §7.4

trigonometric substitutions, §7.3

u = sin−1 x+ a

bgives b sinu = x+ a, b cosu du = dx and

√

b2 − (x+ a)2 = b cosu,

u = tan−1 x+ a

bgives b tanu = x+ a, b sec2 u du = dx and

√

b2 + (x+ a)2 = b secu,

u = sec−1 x+ a

bgives b secu = x+ a, b secu tanu du = dx and

√

(x+ a)2 − b2 = b tanu;

hyperbolic substitutions, §7.3

u = sinh−1 x+ a

bgives b sinhu = x+ a, b coshu du = dx and

√

(x+ a)2 + b2 = b coshu,

u = cosh−1 x+ a

bgives b coshu = x+ a, b sinhu du = dx and

√

(x+ a)2 − b2 = b sinhu.

• Weierstrass substitution: t = tanx

2gives dx =

2 dt

1 + t2, cosx =

1− t2

1 + t2, sinx =

2t

1 + t2. §7.4

6. Trigonometric Identities: go to 1 after §7.2

Michael Daniel Samson Heuristics Page 1 of 2

SEM 1102 / MAT 200 Integration Strategy 12 February 2020

• From the unit circle: cos2 x+ sin2 x ≡ 1, 1 + tan2 x = secx, cot2 x+ 1 = csc2 x

• sin(u±v) = sinu cos v±cosu sin v and cos(u±v) = cosu cos v∓sinu sin v. These also give rise to

the double- and half-angle formulas: cos2 x =1 + cos 2x

2and sin2 x =

1− cos 2x

2, as well as

sin ax cos bx =sin[(a+ b)x]− sin[(a− b)x]

2, cos ax cos bx =

cos[(a+ b)x] + cos[(a− b)x]

2and

sin ax sin bx =cos[(a− b)x]− cos[(a+ b)x]

2.

• By definition and differentiation: §3.3

d

dxtanx =

d

dx

sinx

cosx= sec2 x,

d

dxcotx =

d

dx

cosx

sinx= − csc2 x,

d

dxsecx =

d

dx(cosx)−1 = secx tanx,

d

dxcscx =

d

dx(sinx)−1 = − cscx cotx.

7. Partial Fractions: with p(x), q(x) polynomials, deg p < deg q,

∫

p(x)

q(x)dx =

∫[

∑ pi(x)

qi(x)

]

dx, §7.4

where qi(x) is a factor of q(x): qi(x) = (x + a)n sets pi(x) ≡ b; qi(x) = [(x + a)2 + b2]n, b > 0, setspi(x) = 2c(x+ a) + d, go to 1

8. Integration by Parts: go to 1 after §7.1

∫

f(x)g′(x) dx =

∫

f(x) d(g(x)) = f(x)g(x)−∫

g(x) d(f(x)) = f(x)g(x)−∫

f ′(x)g(x) dx.

In particular, if xf ′(x) is integrable,

∫

f(x) dx = xf(x)−∫

xf ′(x) dx, and, if F ′(x) = f(x), using

indirect substitution,

∫

f−1(x) dx =

∫

uf ′(u) du = uf(u)−∫

f(u) du = xf−1(x)−F (f−1(x))+C.


Reproduced without explicit permission, for use in SEM 1102 / MAT 200 class, DigiPen Institute of Technology, Singapore, February

2020.

Michael Daniel Samson Heuristics Page 2 of 2

SEM 1102 / MAT 200 Newton-Cotes Formulas 5 February 2020

Approximate the area bound by y = f(x), x = a, x = b and the x-axis, given by

∫ b

a

f(x) dx, partitioning

the interval [a, b] into n intervals of equal width, [xk−1, xk], 1 ≤ k ≤ n, with a = x0 and b = xn and

h =b− a

n= xk − xk−1.

Trapezoidal Rule

Approximate f(x) with a piecewise-linear function g(x),

g(x) = gk(x) = akx+ bk on [xk−1, xk], 1 ≤ k ≤ n, f(xm) = g(xm), 0 ≤ m ≤ n.

The approximation of the definite integral is, thus,

∫ b

a

f(x) dx ≈

∫ b

a

g(x) dx =n∑

k=1

∫ xk

xk−1

g(x) dx =n∑

k=1

∫ xk

xk−1

gk(x) dx.

It can be determined, using Lagrange basis functions, that

gk(x) = f(xk−1)x− xk

xk−1 − xk+ f(xk)

x− xk−1

xk − xk−1=

f(xk)(x− xk−1) + f(xk−1)(xk − x)

h,

so

∫ xk

xk−1

gk(x) dx =f(xk)

h

∫ xk

xk−1

(x− xk−1) dx+f(xk−1)

h

∫ xk

xk−1

(xk − x) dx

=f(xk)

h

[

(x− xk−1)2

2

]xk

xk−1

−f(xk−1)

h

[

(xk − x)2

2

]xk

xk−1

=f(xk)

h

[

h2

2

]

+f(xk−1)

h

[

h2

2

]

=h

2[f(xk−1) + f(xk)],

which is the area of the trapezoid made by y = gk(x), x = xk−1, x = xk and the x-axis. Thus, theTrapezoidal Rule uses the approximation

∫ b

a

f(x) dx ≈

n∑

k=1

∫ xk

xk−1

gk(x) dx =

n∑

k=1

h

2[f(xk−1) + f(xk)] = h

[

f(a) + f(b)

2+

n−1∑

k=1

f(xk)

]

.

The error bound for the approximation is

∣

∣

∣

∣

∣

∫ b

a

f(x) dx− h

[

f(a) + f(b)

2+

n−1∑

k=1

f(xk)

]∣

∣

∣

∣

∣

<M(b− a)3

12n2=

Mh2(b− a)

12where M = max

a≤x≤b{|f ′′(x)|}.

Simpson’s 1 / 3 Rule

Let n be even. Approximate f(x) with a piecewise-parabolic function g(x),

g(x) = gk(x) = akx2 + bkx+ ck on [x2k−2, x2k], 1 ≤ k ≤ n/2, f(xm) = g(xm), 0 ≤ m ≤ n.

The approximation of the definite integral is, thus,

∫ b

a

f(x) dx ≈

∫ b

a

g(x) dx =

n/2∑

k=1

∫ x2k

x2k−2

g(x) dx =

n/2∑

k=1

∫ xk

x2k−2

gk(x) dx.

Michael Daniel Samson Approximating Definite Integrals Page 1 of 3


It can be determined, using Lagrange basis functions, that

gk(x) = f(x2k−2)(x− x2k)(x− x2k−1)

(x2k−2 − x2k)(x2k−2 − x2k−1)+ f(x2k−1)

(x− x2k)(x− x2k−2)

(x2k−1 − x2k)(x2k−1 − x2k−2)

+ f(x2k)(x− x2k−1)(x− x2k−2)

(x2k − x2k−1)(x2k − x2k−2)

gk(x) =f(x2k−2)

2h2(x2k−1 + h− x)(x2k−1 − x) +

f(x2k−1)

h2(x2k−1 + h− x)(x− x2k−1 + h)

+f(x2k)

2h2(x− x2k−1)(x− x2k−1 + h),

so∫ x2k

x2k−2

gk(x) dx =f(x2k−2)

2h2

∫ x2k−1+h

x2k−1−h

[x2 − (2x2k−1 + h)x+ x2k−1(x2k−1 + h)] dx

−f(x2k−1)

h2

∫ x2k−1+h

x2k−1−h

[x2 − 2x2k−1x+ x22k−1 − h2] dx

+f(x2k)

2h2

∫ x2k−1+h

x2k−1−h

[x2 − (2x2k−1 − h)x+ x2k−1(x2k−1 − h)] dx

=f(x2k−2)

2h2

[

x3

3−

(2x2k−1 + h)x2

2+ x2k−1(x2k−1 + h)x

]x2k−1+h

x2k−1−h

−f(x2k−1)

h2

[

x3

3− x2k−1x

2 + (x22k−1 − h2)x

]x2k−1+h

x2k−1−h

+f(x2k)

2h2

[

x3

3−

(2x2k−1 − h)x2

2+ x2k−1(x2k−1 − h)x

]x2k−1+h

x2k−1−h

=f(x2k−2)− 2f(x2k−1) + f(x2k)

6h2[(x2k−1 + h)3 − (x2k−1 − h)3]

−f(x2k−2)(2x2k−1 + h)− 4f(x2k−1)x2k−1 + f(x2k)(2x2k−1 − h)

4h2[(x2k−1 + h)2 − (x2k−1 − h)2]

+f(x2k−2)x2k−1(x2k−1 + h)− 2f(x2k−1)(x

22k−1 − h2) + f(x2k)x2k−1(x2k−1 − h)

h

=f(x2k−2)− 2f(x2k−1) + f(x2k)

3h(3x2

2k−1 + h2)

−f(x2k−2)(2x2k−1 + h)− 4f(x2k−1)x2k−1 + f(x2k)(2x2k−1 − h)

hx2k−1

+f(x2k−2)x2k−1(x2k−1 + h)− 2f(x2k−1)(x

22k−1 − h2) + f(x2k)x2k−1(x2k−1 − h)

h

=f(x2k−2)

3h[(3x2

2k−1 + h2)− 3(2x22k−1 + x2k−1h) + 3x2k−1(x2k−1 + h)]

+f(x2k−1)

3h[−2(3x2

2k−1 + h2) + 12x22k−1 − 6(x2

2k−1 − h2)]

+f(x2k)

3h[(3x2

2k−1 + h2)− 3(2x22k−1 − x2k−1h) + 3x2k−1(x2k−1 − h)]

=h

3f(x2k−2) +

4h

3f(x2k−1) +

h

3f(x2k) =

h

3[f(x2k−2) + 4f(x2k−1) + f(x2k)].

Thus, Simpson’s 1 / 3 Rule uses the approximation

∫ b

a

f(x) dx ≈

n/2∑

k=1

∫ x2k

x2k−2

gk(x) dx =

n/2∑

k=1

h

3[f(x2k−2) + 4f(x2k−1) + f(x2k)]

≈h

3

[

f(a) + f(b) +

n−1∑

k=1

[3− (−1)k]f(xk)

]

.



The error bound for the approximation is given by

∣

∣

∣

∣

∣

∫ b

a

f(x) dx−h

3

[

f(a) + f(b) +n−1∑

k=1

[3− (−1)k]f(xk)

]∣

∣

∣

∣

∣

<M(b− a)5

180n4=

Mh4(b− a)

180

where M = maxa≤x≤b

{|f (4)(x)|}.



Name: dP ID: 0 0 0

Round your answers to six decimal places. Let ∆x =b− a

nand xi = a+ i∆x.

5. Use (a) the Midpoint Rule

∫

b

a

f(x) dx ≈ Mn = ∆x[f(x1) + f(x2) + · · ·+ f(xn)] where xi =1

2(xi−1 + xi),

and (b) Simpson’s Rule

∫

b

a

f(x) dx ≈ Sn =∆x

3[f(x0)+4f(x1)+2f(x2)+4f(x3)+· · ·+2f(xn−2)+4f(xn−1)+f(xn)] (n even),

to approximate

∫

π

0

x2 sinx dx with n = 8. Compare your results to the actual value to determine

the error in each approximation.

Given f(x) = x2 sinx, a = 0, b = π and n = 8, then ∆x =π

8, and xi =

iπ

8, 0 ≤ i ≤ 8. By integration

by parts, first using u = x2 anddv = sinx dx, second using U = x anddV = cosx dx,

∫

b

a

f(x) dx =[

−x2 cosx]π

0+2

∫

π

0

x cosx dx = π2+[2x sinx]π

0−2

∫

π

0

sinx dxπ2+[2 cosx]π

0= π2−4.

(a) M8 =π

8

8∑

i=1

(

(2i− 1)π

16

)2

sin

(

(2i− 1)π

16

)

=π3

2048

4∑

i=1

[

(2i− 1)2 + (17− 2i)2]

sin

(

(2i− 1)π

16

)

,

where cosπ

8=

√

1

2

(

1 + cosπ

4

)

=

√

2 +√2

2and cos

3π

8=

√

1

2

(

1 + cos3π

4

)

=

√

2−√2

2,

so

sinπ

16=

√

1

2

(

1− cosπ

8

)

=

√

2−√

2 +√2

2= cos

7π

16,

cosπ

16=

√

1

2

(

1 + cosπ

8

)

=

√

2 +√

2 +√2

2= sin

7π

16,

sin3π

16=

√

1

2

(

1− cos3π

8

)

=

√

2−√

2−√2

2= cos

5π

16,

cos3π

16=

√

1

2

(

1 + cos3π

8

)

=

√

2 +√

2−√2

2= sin

5π

16.

Thus, M8 =π3

2048

[

113

√

2−√

2 +√2 + 89

√

2−√

2−√2 + 73

√

2 +√

2−√2 + 65

√

2 +√

2 +√2

]

,

approximately 5.932957, with

∣

∣

∣

∣

∣

∫

b

a

f(x) dx−M8

∣

∣

∣

∣

∣

≈ 0.063352.

(b) S8 =π

24

[

π2

16sin

π

8+

π2

8sin

π

4+

9π2

16sin

3π

8+

π2

2sin

π

2+

25π2

16sin

5π

8+

9π2

8sin

3π

4+

49π2

16sin

7π

8+ π sinπ

]

,

where sinπ

8=

√

1

2

(

1− cosπ

4

)

=

√

2−√2

2= sin

7π

8and sin

3π

8=

√

1

2

(

1− cos3π

4

)

=

√

2 +√2

2= sin

5π

8. Thus, S8 =

π3

384

[

25√

2−√2 + 10

√2 + 17

√

2 +√2 + 8

]

, approximately

5.869247, with

∣

∣

∣

∣

∣

∫

b

a

f(x) dx− S8

∣

∣

∣

∣

∣

≈ 0.000358.

Michael Daniel Samson Newton-Cotes Formulas Page 1 of 3


7. Use (a) the Trapezoidal Rule

∫

b

a

f(x) dx ≈ Tn =∆x

2[f(x0) + 2f(x1) + 2f(x2) + · · ·+ 2f(xn−1) + f(xn)],

(b) the Midpoint Rule, and (c) Simpson’s Rule to approximate

∫

2

1

√

x3 − 1 dx with n = 10.

Given f(x) =√x3 − 1, a = 1, b = 2 and n = 10, then ∆x =

1

10, and xi =

10 + i

10, 0 ≤ i ≤ 10.

(a) T10 =1

20

[

√

331

250+

√

364

125+

√

1197

250+

√

872

125+

√

19

2+

√

1548

125+

√

3913

250+

√

2416

125+

√

5859

250+√7

]

,

approximately 1.506361.

(b) M10 =1

10

10∑

i=1

√

(19 + 2i)3 − 8000

8000=

√5

2000

10∑

i=1

√8i3 + 228i2 + 2166i− 1141. Thus, M10 =

√5

2000

[√1261 + 3

√463 + 5

√305 +

√11683 + 3

√1871 +

√21791 +

√27937 + 15

√155 +

√42653 +

√51319

]

,


(c) S10 =1

30

[

√

662

125+

√

364

125+

√

2394

125+

√

872

125+√38 +

√

1548

125+

√

7826

125+

√

2416

125+

√

11718

125+

√7

]

,


wolframalpha.com approximates the integral as 1.51593.

9. Use (a) the Trapezoidal Rule, (b) the Midpoint Rule, and (c) Simpson’s Rule to approximate

∫

4

1

√lnx dx

with n = 6.

Given f(x) =√lnx, a = 1, b = 4 and n = 6, then ∆x =

1

2, and xi =

2 + i

2, 0 ≤ i ≤ 6.

(a) T6 =1

4

[

2

√

ln3

2+ 2

√ln 2 + 2

√

ln5

2+ 2

√ln 3 + 2

√

ln7

2+

√2 ln 2

]

, approximately 2.591334.

(b) M6 =1

2

6∑

i=1

√

ln3 + 2i

2. Thus, M6 =

1

2

[

√

ln5

4+

√

ln7

4+

√

ln9

4+

√

ln11

4+

√

ln13

4+

√

ln15

4

]

,


(c) S6 =1

6

[

4

√

ln3

2+ 2

√ln 2 + 4

√

ln5

2+ 2

√ln 3 + 4

√

ln7

2+√2 ln 2

]



17. Use (a) the Trapezoidal Rule, (b) the Midpoint Rule, and (c) Simpson’s Rule to approximate, with

n = 8,

∫

4

0

ln(1 + ex) dx.

Given f(x) = ln(1 + ex), a = 0, b = 4 and n = 8, then ∆x =1

2, and xi =

i

2, 0 ≤ i ≤ 8.

(a) T8 =1

4ln[

2(1 +√e)2(1 + e)2(1 +

√e3)2(1 + e2)2(1 +

√e5)2(1 + e3)2(1 +

√e7)2(1 + e4)

]

, ap-

proximately 8.814278.

(b) M8 =1

2

8∑

i=1

ln

[

1 + exp

(

2i− 1

4

)]

=1

2ln

8∏

i=1

[

1 + exp

(

2i− 1

4

)]

. Thus, M8 =1

2ln [(1 + 4

√e)×

(1 +4√e3)(1 +

4√e5)(1 +

4√e7)(1 +

4√e9)(1 +

4√e11)(1 +

4√e13)(1 +

4√e15)

]


(c) S8 =1

6ln

[

2(1 +√e)4(1 + e)2(1 +

√e3)4(1 + e2)2(1 +

√e5)4(1 + e3)2(1 +

√e7)4(1 + e4)

]

, ap-

proximately 8.804229.




19. (a) Find the approximations T8 and M8 for the integral

∫

1

0

cos(x2) dx.

Given f(x) = cos(x2), a = 0, b = 1 and n = 8, then ∆x =1

8, and xi =

i

8, 0 ≤ i ≤ 8.

• T8 =1

8

[

1

2+ cos

(

1

64

)

+ cos

(

1

16

)

+ cos

(

9

64

)

+ cos

(

1

4

)

+ cos

(

25

64

)

+ cos

(

9

16

)

+ cos

(

49

64

)

+cos 1

2

]

,


• M8 =1

8

8∑

i=1

cos

[

(2i− 1)2

256

]

. Thus, M8 =1

8ln

[

cos

(

1

256

)

+ cos

(

9

256

)

+ cos

(

25

256

)

+

cos

(

49

256

)

+ cos

(

81

256

)

+ cos

(

121

256

)

+ cos

(

169

256

)

+ cos

(

225

256

)]


(b) Estimate the errors in the approximations of part (a).

Suppose |f ′′(x)| ≤ K for a ≤ x ≤ b. If ET and EM are the errors in the Trapzoidaland Midpoint Rules, then

|ET | ≤K(b− a)3

12n2and |EM | ≤ K(b− a)3

24n2.

Since f ′′(x) = −2 sin(x2) − 4x2 cos(x2), set K = 6, as |f ′′(x)| ≤ |2 + 4x2| for all x, giving

upper-bound error estimates of |ET | ≤1

128and |EM | ≤ 1

256.

Using a tighter bound of K = 3.85 makes |ET | ≤73

15360≈ 0.0047526 and |EM | ≤ 73

30720≈

0.0023763.

(c) How large do we have to choose n so that the approximations Tn and Mn to the integral inpart (a) are accurate to within 0.0001?

With the given estimates, |ET | ≤1

2n2<

1

10000when n ≥ 71, and |EM | ≤ 1

4n2<

1

10000when

n ≥ 50. Note that T71 ≈ 0.904496, M50 ≈ 0.904552 and wolframalpha.com approximates

the integral as 0.904524.

Using a tighter bound of K = 3.85 makes |ET | ≤73

240n2<

1

10000when n ≥ 56, and T56 ≈

0.904480 and |EM | ≤ 73

480n2<

1

10000when n ≥ 39, and M39 ≈ 0.904570.

Source: James Stewart, Calculus Early Transcendentals, 8e, International Metric Edition, Section 7.7


2020.


SEM 1102 / MAT 200 Series Convergence Tests and Error Estimates 12 February 2020

Given a sequence {an}∞n=0, determine properties of the series∞∑

n=0

an = limN→∞

sN , where the partial sum

sN is defined as sN =N∑

n=0

an. (⋆ indicates use in power series, Estimates after Tests for convergence)

§11.1 {sN}∞N=0is a sequence: if it converges, s = lim

N→∞

sN , converges to the series; otherwise, the series

diverges

– sN = f(N): limN→∞

sN = limx→∞

f(x)—both limits exist (converge) or do not exist (diverge)

– limN→∞

|sN | = 0: limN→∞

sN = 0

– |r| < 1: rn → 0; r = 1: rn → 1

– every monotone (increasing sN+1 ≥ sN or decreasing sN+1 ≤ sN ) bounded (there exist L, Rsuch that L ≤ sN ≤ R) sequence converges

§11.2–3 the convergence of some series can be determined by the sequence {an}∞n=0

Test/Formula an = arn:∞∑

n=0

arn =a

1− rif |r| < 1, otherwise the series diverges (geometric series)

Test an =1

np:

∞∑

n=1

1

npconverges if p > 1, otherwise the series diverges (p-series)

Test limn→∞

an 6= 0:∞∑

n=0

an diverges (divergence test)

§11.3–4∞∑

n=0

an can be compared with something known to converge or diverge

Test an = f(n), f(x) > 0, descending, continuous on x ≥ 1:∞∑

n=1

f(n) and

∫

∞

1

f(x) dx = limt→∞

∫ t

1

f(x) dx

both converge or diverge (integral test)

EstimateN∑

n=1

f(n) +

∫

∞

N+1

f(x) dx ≤∞∑

n=1

f(n) ≤N∑

n=1

f(n) +

∫

∞

N

f(x) dx (integral error estimate)

Test 0 ≤ an ≤ bn and∞∑

n=0

bn converges:∞∑

n=0

an converges (comparison test)

Test 0 ≤ bn ≤ an and∞∑

n=0

bn diverges:∞∑

n=0

an diverges (comparison test)

Test 0 ≤ an, bn and limn→∞

bn

an= c > 0:

∞∑

n=0

an and∞∑

n=0

bn both converge or diverge (limit comparison

test)

⋆ §11.5 if an = (−1)nbn, bn > 0 (alternating series)

Test bn+1 ≤ bn and limn→∞

bn = 0:∞∑

n=0

(−1)nbn converges (alternating series test)

Estimate

∣

∣

∣

∣

∞∑

n=N+1

(−1)nbn

∣

∣

∣

∣

=

∣

∣

∣

∣

∞∑

n=0

(−1)nbn −N∑

n=0

(−1)nbn

∣

∣

∣

∣

< bN+1 (alternating series error estimate)

§11.6 other trends of an as n → ∞ may determine convergence of∞∑

n=0

an

⋆ Test limn→∞

∣

∣

∣

∣

an+1

an

∣

∣

∣

∣

< 1:∞∑

n=0

an converges; limn→∞

∣

∣

∣

∣

an+1

an

∣

∣

∣

∣

> 1:∞∑

n=0

an diverges (ratio test)

Test limn→∞

n

√an < 1:

∞∑

n=0

an converges; limn→∞

n

√an > 1:

∞∑

n=0

an diverges (root test)



2020.

Michael Daniel Samson Approximation of Functions by Infinite Polynomials One page only


Name: dP ID: 0 0 0

§11.2 #89 The Cantor set, named after the German mathematician Georg Cantor (1845–1918), is constructed

as follows. We started with the closed interval [0, 1] and remove the open interval

(

1

3,2

3

)

. That

leaves two intervals

[

0,1

3

]

and

[

2

3, 1

]

and we remove the middle third of each. Four intervals

remain and again we remove the open middle third of each of them. We continue this processindefinitely, at each step removing the open middle third of every interval that remains from thepreceding step. The Cantor set consists of the numbers that remain in [0, 1] after all those intervalshave been removed.

(a) Show that the total length of all the intervals that are removed is 1. Despite that, the Cantorset contains infinitely many numbers. Give examples of some numbers in the Cantor set.

(b) The Sierpinski carpet is a two-dimensional counterpart of the Cantor set. It is constructed byremoving the center one-ninth of a square of side 1, then removing the centers of the eightsmaller remaining squares, and so on. Show that the sum of the areas of the removed squaresis 1. This implies that the Sierpinski carpet has area 0.

§11.5 #23 Show that∞∑

n=1

(−1)n+1

n6is convergent. How many terms of the series do we need to add in order

to find the sum to the accuracy |error| < 0.00005?

§11.5 #27 Approximate the sum of the series∞∑

n=1

(−1)n

(2n)!correct to four decimal places.

§11.6 #45 (a) Show that∞∑

n=0

xn

n!converges for all x.

(b) Deduce that limn→∞

xn

n!= 0 converges for all x.

§11.6 #47 (a) Find the partial sum s5 of the series∞∑

n=1

1

n2n. Use Exercise 46

Let∑

an be a series with positive terms and let rn =an+1

an. Suppose that lim

n→∞

rn =

L < 1, so∑

an converges by the Ratio Test.

• if {rn} is a decreasing sequence and rn+1 < 1,∞∑

k=n+1

ak ≤an+1

1− rn+1

• if {rn} is an increasing sequence,∞∑

k=n+1

ak ≤an+1

1− L

to estimate the error in using s5 as an approximation to the sum of the series.

(b) Find a value of n so that sn is within 0.00005 of the sum. Use this value of n to approximatethe sum of the series.



2020.

Michael Daniel Samson Convergence of Series One page only

Documents

SEM 1102 / MAT 200–Calculus and Analytic Geometry IImdvsamson.work/mat200/docs.pdf · SEM 1102 / MAT 200 Homework 1 13 January 2020 Name: dP ID: 3 9 0 0 0 §4.9 #49 Given that the