SEM 1102 / MAT 200–Calculus and Analytic Geometry IImdvsamson.work/mat200/docs.pdf · SEM 1102 /...
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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
SEM 1102 / MAT 200 Homework 1 13 January 2020
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
Michael Daniel Samson Approximation of Areas Page 2 of 4
SEM 1102 / MAT 200 Homework 1 13 January 2020
(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
Michael Daniel Samson Approximation of Areas Page 3 of 4
SEM 1102 / MAT 200 Homework 1 13 January 2020
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.
Michael Daniel Samson Approximation of Areas Page 4 of 4
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
SEM 1102 / MAT 200 Homework 2 20 January 2020
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.
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 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,
where p is a polynomial.
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,
where P is a polynomial such that P ′ = p.
• 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,
where p is a polynomial.
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,
where P is a polynomial such that P ′ = p.
– 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,
where p is a polynomial.
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,
where P is a polynomial such that P ′ = p.
– 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
SEM 1102 / MAT 200 Homework 3 29 January 2020
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
SEM 1102 / MAT 200 Homework 3 29 January 2020
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 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
SEM 1102 / MAT 200 Homework 4 5 February 2020
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
SEM 1102 / MAT 200 Worksheet 2 Quiz 2 on February 5
§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%.
Michael Daniel Samson Integration with Trigonometric Functions Page 2 of 3
SEM 1102 / MAT 200 Worksheet 2 Quiz 2 on February 5
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 Integration with Trigonometric Functions Page 3 of 3
SEM 1102 / MAT 200 Worksheet 3 Quiz 2 on February 5
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
SEM 1102 / MAT 200 Worksheet 3 Quiz 2 on February 5
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
Reproduced without explicit permission, for use in SEM 1102 / MAT 200 class, DigiPen Institute of Technology, Singapore, January
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.
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, 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
SEM 1102 / MAT 200 Newton-Cotes Formulas 5 February 2020
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)
]
.
Michael Daniel Samson Approximating Definite Integrals Page 2 of 3
SEM 1102 / MAT 200 Newton-Cotes Formulas 5 February 2020
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)|}.
Michael Daniel Samson Approximating Definite Integrals Page 3 of 3
SEM 1102 / MAT 200 Homework 5 10 February 2020
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
SEM 1102 / MAT 200 Homework 5 10 February 2020
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
]
,
approximately 1.518362.
(c) S10 =1
30
[
√
662
125+
√
364
125+
√
2394
125+
√
872
125+√38 +
√
1548
125+
√
7826
125+
√
2416
125+
√
11718
125+
√7
]
,
approximately 1.511519.
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
]
,
approximately 2.681046.
(c) S6 =1
6
[
4
√
ln3
2+ 2
√ln 2 + 4
√
ln5
2+ 2
√ln 3 + 4
√
ln7
2+√2 ln 2
]
, approximately 2.631976.
wolframalpha.com approximates the integral as 2.66142.
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)
]
, approximately 8.799212.
(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.
Michael Daniel Samson Newton-Cotes Formulas Page 2 of 3
SEM 1102 / MAT 200 Homework 5 10 February 2020
wolframalpha.com approximates the integral as 8.80423.
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
]
,
approximately 0.902333.
• 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
)]
, approximately 0.905620.
(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
Reproduced without explicit permission, for use in SEM 1102 / MAT 200 class, DigiPen Institute of Technology, Singapore, February
2020.
Michael Daniel Samson Newton-Cotes Formulas Page 3 of 3
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)
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, February
2020.
Michael Daniel Samson Approximation of Functions by Infinite Polynomials One page only
SEM 1102 / MAT 200 Homework 6 21 February 2020
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.
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, February
2020.
Michael Daniel Samson Convergence of Series One page only