Integral Calculus with Applications to Commerce and Social Sciences
MATH 105

Winter 2013
LECTURE 25: EXPECTED VALUE, VARIANCE, AND STANDARD DEVIATION
MINGFENG ZHAO
March 11, 2015
Cumulative distribution function and probability density function
Theorem 1. Let X be a continuous random variable, then
I. For the cumulative distribution function
Integral Calculus with Applications to Commerce and Social Sciences
MATH 105

Winter 2013
Friday, 8 January
Math 105, section 207
Quiz 1
Name:
Student number:
Time: 10 minutes
Note: Each wrong answer has 13 negative score and each right answer has 1 positive
score. If you are not sure about a question, leave it blank.
1. LetP = (5, 11, 12) and
Integral Calculus with Applications to Commerce and Social Sciences
MATH 105

Winter 2013
Quiz 6
Math 105, Section 204/208
DUE Thursday, April 6, 4:00pm
Please submit your quiz on or before Thursday, April 6, 4:00pm. Your answers may be typed or handwritten.
Simplify all answers, and show your work. You DO NOT need to submit this document wit
Integral Calculus with Applications to Commerce and Social Sciences
MATH 105

Winter 2013
Quiz 6
Math 105, Section 204/208
DUE Thursday, April 6, 4:00pm
Please submit your quiz on or before Thursday, April 6, 4:00pm. Your answers may be typed or handwritten.
Simplify all answers, and show your work. You DO NOT need to submit this document wit
Integral Calculus with Applications to Commerce and Social Sciences
MATH 105

Winter 2013
Integration Methods
Elyse Yeager
Math 105
We have learned several fancy methods of integration. Although occasionally one method might be the
obvious choice, in general it is quite difficult to decide how to proceed when faced with an antiderivative.
The
Integral Calculus with Applications to Commerce and Social Sciences
MATH 105

Winter 2013
Optimization Practice
Elyse Yeager
Math 105
Problems
Question 1. Find all local extrema of the function f (x, y) = x2 + y 3 + xy + 5x.
Question 2. Let f (x, y) = x2 + y 2 2x + 2. Find the largest and smallest values f (x, y) attains over the
region R = cf
Integral Calculus with Applications to Commerce and Social Sciences
MATH 105

Winter 2015
1. Find the maximum of
2
(b) f (x, y) = ex
Math 105: Problems 2
(a) f (x, y) = x2 6x cos y + 9
+y 2 4
with
xy
g(x, y) = x2 + y 2 4 = 0
For part (b), rst establish the result using a Lagrange multiplier, then redo the calculation
by avoiding such a multip
Integral Calculus with Applications to Commerce and Social Sciences
MATH 105

Winter 2015
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Integral Calculus with Applications to Commerce and Social Sciences
MATH 105

Winter 2013
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Integral Calculus with Applications to Commerce and Social Sciences
MATH 105

Winter 2013
LECTURE 32: POWER SERIES
MINGFENG ZHAO
March 27, 2015
X
Theorem 1 (Divergence test). If a series
series
X
ak converges, then lim ak = 0. That means, if lim ak 6= 0, then the
k
k=1
k
ak diverges.
k=1
Theorem 2 (Integral test). If f (x) is a continuous, pos
Integral Calculus with Applications to Commerce and Social Sciences
MATH 105

Winter 2013
Math 105  Section 201  Homework Problems
Set 1: due January 26, 8 am, inclass
1. Let r = ha, b, ci be a vector and denote x , y , z to be the smallest angle
between r and the positive x, y, and z axes respectively. Note that
this means all three angles
Integral Calculus with Applications to Commerce and Social Sciences
MATH 105

Winter 2013
LECTURE 33: POWER SERIES AND TAYLOR SERIES
MINGFENG ZHAO
March 30, 2015
X
ck+1 
, then the radius of convergence of the power
ck 
k=0
X
X
1
series
ck (x a)k is R = . In particular, the power series
ck (x a)k absolutely converges for all x a < R,
r
k
Integral Calculus with Applications to Commerce and Social Sciences
MATH 105

Winter 2013
LECTURE 31: THE COMPARISON TESTS AND POWER SERIES
MINGFENG ZHAO
March 25, 2015
Theorem 1 (Divergence test). If a series
X
series
ak diverges.
X
ak converges, then lim ak = 0. That means, if lim ak 6= 0, then the
k
k
Theorem 2 (Integral test). If f (x) is
Integral Calculus with Applications to Commerce and Social Sciences
MATH 105

Winter 2013
LECTURE 26: REVIEW FOR MIDTERM 2
MINGFENG ZHAO
March 13, 2015
Example 1. Evaluate lim
n
X
6(k 1)2
r
1+2
(k 1)3
.
n3
n3
1
First, you should split off a x = in the sum, then
n
r
r
n
n
X
1 X 6(k 1)2
6(k 1)2
(k 1)3
(k 1)3
=
.
1+2
1+2
3
3
2
n
n
n
n
n3
n
k=1
k=
Integral Calculus with Applications to Commerce and Social Sciences
MATH 105

Winter 2013
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Integral Calculus with Applications to Commerce and Social Sciences
MATH 105

Winter 2013
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Integral Calculus with Applications to Commerce and Social Sciences
MATH 105

Winter 2013
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Integral Calculus with Applications to Commerce and Social Sciences
MATH 105

Winter 2015
Math 105, Assignment 4
Due 20150316, 4:00 pm
In the following problems you are expected to justify your answers unless stated otherwise.
Answers without any explanation will be given a mark of zero. The assignment needs to be
in my hand before I leave t
Integral Calculus with Applications to Commerce and Social Sciences
MATH 105

Winter 2015
Math 105, Assignment 1
Due 20150116, 4:00 pm
In the following problems you are expected to justify your answers unless stated otherwise.
Answers without any explanation will be given a mark of zero. The assignment needs to be
in my hand before I leave t
Integral Calculus with Applications to Commerce and Social Sciences
MATH 105

Winter 2015
Math 105, Assignment 3
Due 20150225, 4:00 pm
In the following problems you are expected to justify your answers unless stated otherwise.
Answers without any explanation will be given a mark of zero. The assignment needs to be
in my hand before I leave t
Integral Calculus with Applications to Commerce and Social Sciences
MATH 105

Winter 2015
Math 105, Assignment 2
Due 20150128, 4:00 pm
In the following problems you are expected to justify your answers unless stated otherwise.
Answers without any explanation will be given a mark of zero. The assignment needs to be
in my hand before I leave t