Math 118: Honours Calculus II
Winter, 2005
List of Theorems
Lemma 5.1 (Partition Renement): If P and Q are partitions of [a, b] such that
Q P , then
L(P, f ) L(Q, f ) U(Q, f ) U(P, f ).
Lemma 5.2 (Upper Sums Bound Lower Sums): Let f be bounded on [a, b].
Math 118 Winter 2015 Lecture 1
Recall we have three methods to evaluate an integral
i. By denition;
Z
(Jan. 5, 2015)
b
f (x) dx.
a
ii. First check that f is integrable on [a; b], then try to nd a family of partitions fPn g
such that U (f ; Pn) (or L(f ; P
Math 118 Winter 2015 Homework 1 Solutions
Due Thursday Jan. 15 3pm in Assignment Box
Question 1. (5 pts) Let F (x) be dierentiable on (a, b) and let c (a, b). Assume that both
limxc+F (x), limxcF (x) exist and are nite. Prove that the two limits are equal
Math 118 Winter 2015 Lecture 12
Z
Integrals of the form
rational.
(Jan. 23, 2015)
a x + b n1/m1
a x + b nk/mk
R x;
; :;
dx where R(x; y) is
cx+d
cx+d
Note. We say a multi-variable function R(x1; :; xk) is rational if there are polynomials
P (x1; :; xk)
Math 118 Winter 2015 Lecture 10
The method of partial fractions: To evaluate
Z
(Jan. 21, 2015)
P (x)
dx where P ; Q are polynomials,
Q(x)
1. Check deg P < deg Q. If not perform P = Q P0 + R with P0; R polynomials and
deg R < deg Q and write
Z
Z
Z
P (x)
R(
Math 118 Winter 2015 Lecture 14
(Jan. 28, 2015)
Integration by parts for denite integrals.
Theorem 1. (Integration by parts) If u; v are continuous on [a; b] an dierentiable on
(a; b), and if u 0; v 0 are integrable on [a; b], then
Z b
Z b
u(x) v 0(x) dx
Math 118 Honors Calculus II Winter, 2003-2004
Solutions to Homework #2
Dr. Michael Li
#7.25 Show 1 + tan2 x = sec2 x,
1 + cot2 x = csc2 x.
Proof. For x such that cos x = 0, dividing cos2 x in the identity sin2 x +
cos2 x = 1, we have
1 + tan2 x = sec2 x.
Math 118 Honors Calculus II Winter, 2003-2004
Solutions to Homework #6
Dr. Michael Li
#9.1 Find the area of the shaded region determined by three semi-circles.
Solution. Let r1 , r2 be the radii of the two small circles, and r be the radius
of the large c
Math 118 Honors Calculus II Winter, 2003-2004
Solutions to Homework #4
Dr. Michael Li
#8.6 If In = cosn x dx, show
In =
n1
1
cosn1 x sin x +
In2 ,
n
n
n = 1, 2, 3 . . . .
Proof.
In =
cosn x dx =
cosn1 x cos x dx =
cosn1 x d(sin x)
= cosn1 x sin x
sin x d
Math 118 Honors Calculus II Winter, 2003-2004
Solutions to Homework #1
Dr. Michael Li
#7.1 (20 point) Find f (x).
(a) f (x) = 3x . f (x) = 3x log x.
(b) f (x) = xx . f (x) =
d
(ex log x )
dx
= ex log x (x log x) = ex log x (log x + 1) =
xx (log x + 1).
(c
Math 118 Honors Calculus II Winter, 2003-2004
Solutions to Homework #3
Dr. Michael Li
#8.1. Evaluate integrals.
(a)
sin5 x cos x dx =
sin5 x d(sin x)
u=sin x
=
u5 du = u6 /6 + c
= sin6 x/6 + c.
2
(b)
ex x dx =
(c)
1
x
(d)
u=x2 1
2
2
1
2
sin
2
1
eu du = 2
Math 118 Honors Calculus II Winter, 2003-2004
Solutions to Homework #5
Dr. Michael Li
#9.1 Show that the area of a trapezoid is 1 (b1 + b2 )h, where b1 , b2 are the
2
lengths of the parallel sides and h is the distance between them.
Proof. See Figure 1. W
Math 118 Winter 2015 Lecture 16
Puzzle. 1
lim
x!0
(Jan. 30, 2015)
sin(tan x) tan(sin x)
:
arcsin(arctan x) arctan(arcsin x)
(1)
Note. This lecture is based on Inside Interesting Integrals by Paul J. Nahin, Springer
2015.
Symmetry.
Example 1. Calculate
Sol
Math 118 Winter 2015 Homework 2 Solutions
Due Thursday Jan. 22 3pm in Assignment Box
Question 1. (10 pts) Calculate the following indenite integrals through change of variables.
Please provide enough details.
a) (3 pts)
x2
1 x2 dx;
b) (3 pts)
x
dx;
1x
c)
Math 118 Winter 2015 Lecture 8
(Jan. 16, 2015)
Integration of rational functions.
Goal:
Know: a; b; A; B; C 2 R; k 2 N. How to do
Z
Z
A dx
Bx+C
;
dx:
(x a)k
[(x a)2 + b2]k
Z
P (x)
dx:
Q(x)
(1)
(2)
Exercise 1. How?
Idea: There always holds
X
X
P (x)
A
= P0
Math 118 Winter 2015 Lecture 9
(Jan. 19, 2015)
More examples.
Example 1. Calculate
Solution. We write
Z
x2 + 2
dx.
(x + 1)2 (x 2)
x2 + 2
A
B
C
=
+
+
:
2 (x 2)
x + 1 (x + 1)2
(x + 1)
(x 2)
(1)
2
Multiply both sides by x 2 and set x = 2, we see that A = 3 .
Math 118 Winter 2015 Lecture 2
Recall
the denition of indenite integral: The indenite integral for a function f : (a; b) 7! R
is the following set:
Denoted
(Jan. 7, 2015)
Z
fF (x)j F 0 = f on (a; b)g:
(1)
f (x) dx.
the structure of this set: If there is F
Math 118 Winter 2015 Lecture 4
(Jan. 9, 2015)
Recall
Z
1
x dx =
x1+ + C
1+
Z
dx
= ln jxj + C;
x
Z
2 R;
= 1;
/
(1)
(2)
ex dx = ex + C;
(3)
Z
cos x dx = sin x + C;
(4)
sin x dx = cos x + C;
(5)
Z
= tan x + C;
(6)
= cot x + C;
(7)
= arcsin x + C;
(8)
= arc
Math 118 Winter 2015 Lecture 5
Z
Recall
(Jan. 12, 2015)
d(cabin)
= log(cabin) + C = houseboat:
cabin
(1)
Type I substitution.
Z
Z
Z
0(x) dx = f (u) du = F (u) + C = F (u(x) + C;
f(x) dx = f1(u(x) u
1
1
1
(2)
Here f (x) = f1(u(x) u 0(x).
Type II substituti
Math 118 Winter 2015 Lecture 3
Recall
Z
1
x dx =
x1+ + C
1+
Z
dx
= ln jxj + C;
x
Z
2 R;
(Jan. 8, 2015)
= 1;
/
(1)
(2)
ex dx = ex + C;
(3)
Z
cos x dx = sin x + C;
(4)
sin x dx = cos x + C;
(5)
Z
= tan x + C;
(6)
= cot x + C;
(7)
= arcsin x + C;
(8)
= arc
Math 118 Winter 2015 Lecture 6
(Jan. 14, 2015)
Integration by parts.
Observation: u; v dierentiable, then
Taking indenite integral:
(u v) 0 = u 0 v + u v 0 =) u v 0 = (u v) 0 u 0 v:
Z
Use dierential symbol:
u v 0 dx = u v
Z
u dv = u v
Z
Z
(1)
v u 0 dx:
Math 118 Winter 2015 Lecture 7
Integration by parts.
Z
Z
f (x) dx =
Z
=
(Jan. 15, 2015)
u(x) v 0(x) dx
u dv
= u(x) v(x)
= u(x) v(x)
Z
Z
v du
(1)
v(x) u 0(x)dx:
Hope: v(x) u 0(x) is easier to integrate than f (x).
Z
Example 1. Calculate
(3 x2 7 x + 1) ex
Math 118 Winter 2015 Lecture 11
P (x)
Q(x)
We have seen that rational functions
(Jan. 22, 2015)
can in theory1 always be integrated. Now we show
Birational functions of cos x and sin x.
that another large class of functions also enjoys this property.
A po
Math 118 Winter 2015 Lecture 15
(Jan. 29, 2015)
Recall:
Integration by parts:
holds if
Z
b
a
u(x) v 0(x) dx = u(b) v(b) u(a) v(a)
Z
b
u 0(x) v(x) dx:
(1)
a
i. u; v are continuous on [a; b];
ii. u; v are dierentiable on (a; b);
iii. u 0; v 0 are integrabl
Math 118 Winter 2015 Lecture 13
(Jan. 26, 2015)
Note. All the functions below should be understood as complex functions of complex variables.
However most of the main ideas could be roughly understood with this fact ignored.
Note. This lecture is based ma
Math 118 Honors Calculus II Winter, 2003-2004
Solutions to Homework #7
Dr. Michael Li
#10.15 Let h =
ba
n ,
i
fi = f (a + n (b a) and
Ln = h[f0 + f1 + + fn1 ],
Rn = h[f1 + f2 + + fn ],
Sn = h[f1/2 + f3/2 + + fn1/2 ].
Ln , Rn , Sn are called the left endpo
Math 118 Honors Calculus II Winter, 2003-2004
Solutions to Homework #9
Dr. Michael Li
#11.7 Prove that nn e1n < n! < nn+1 e1n and that
(n!)1/n
1
= .
n
n
e
lim
Proof. The limit follows from the inequality and the Squeeze Principle. The inequality is
equiva
Math 118: Honours Calculus II
Dr. J. Bowman 09:00-12:00 April 25, 2006
Final Exam
First Name:
Last Name:
Student ID:
Question
Score
Maximum
1 2 3 4 5 6 7 8 9 Total
4 4 3 5 6 3 2 4 2
33
No calculators or formula sheets. Check that you have 7 pages.
1. Eval
Math 118: Honours Calculus I
Dr. J. Bowman 09:00-12:00 April 23, 2014
Final Exam
Last Name:
First Name:
Question
1 2 3 4 5 6 Total
Score
Maximum 14 4 2 4 6 5 35
Student ID:
No calculators or formula sheets. Check that you have 6 pages.
1. Evaluate
(a)
log
Math 118: Honours Calculus II
Winter, 2014
Assignment 5
March 11 due March 24
1. Suppose f and g are both integrable on [a, b]. Let
b
(xf (t) + g(t)2 dt
Q(x) =
a
for x R and dene
b
.
A=
f 2 (t) dt,
a
b
.
B=2
f (t)g(t) dt,
a
b
.
C=
g 2 (t) dt.
a
(a) Use th