Integral Calculus With Applications to Commerce and Social Sciences
MATHEMATIC 105

Spring 2015
LECTURE 11: APPROXIMATING AREAS UNDER CURVES
MINGFENG ZHAO
January 28, 2015
Approximating areas by Riemann sums
Figure 1. Approximation of the area
Denition 1. Suppose [a, b] is a closed interval, for any positive integer n, a partition with n subinterval
Integral Calculus With Applications to Commerce and Social Sciences
MATHEMATIC 105

Spring 2015
Quiz 3 for MATH 105 SECTION 205
February 04, 2015
Given Name
Family Name
Student Number
2
1. (a) (1 point) Write down the midpoint Riemann sum with n = 50 to approximate
1
notations.
50
(a)
k=1
Answer.
Since n = 5, then x =
So the midpoint Riemann sum is:
Integral Calculus with Applications to Commerce and Social Sciences
MATHEMATIC 105

Fall 2014
70. Use the partial fraction decomposition
A
Bx + C
1
= + 2
2+x
+x
x
x +x +1
2 + x + 1) + B x 2 + C x
A(x
=
x3 + x2 + x
A+B =0
A + C = 0 A = 1, B = 1, C = 1.
A=1
x3
Therefore,
x3
=
dx
+ x2 + x
dx
x +1
dx
x
x2 + x + 1
= ln x
= ln x
1
2
u2 + 3
4
u+
L
Integral Calculus with Applications to Commerce and Social Sciences
MATHEMATIC 105

Fall 2014
12.
n=10
sin(n + 1 )
2
=
ln ln n
n=10
(1)n
converges by the alternating series test but only condiln ln n
1
1
diverges to innity by comparison with
.
tionally since
ln ln n
n
n=10
n=10
(ln ln n < n for n 10.)
25. Let u = x 1. Then x = 1 + u, and
x ln x =
Integral Calculus with Applications to Commerce and Social Sciences
MATHEMATIC 105

Fall 2014
30. Since
x2
1
3 for all x 0, therefore
x
x5 + 1
x2
dx
x5 + 1
0
1
x2
=
dx +
5
0 x +1
1
x2
dx +
5
0 x +1
= I1 + I2 .
I =
1
1
x2
dx
x5 + 1
dx
x3
Since I1 is a proper integral (nite) and I2 is a convergent improper integral, (see Theorem
2), therefore I con
Integral Calculus with Applications to Commerce and Social Sciences
MATHEMATIC 105

Fall 2014
10.
x = cos t + sin t, y = cos t sin t, (0 t 2 )
The circle x 2 + y 2 = 2, traversed clockwise, starting and ending at (1, 1).
17.
x = et t, y = 4et/2 , (0 t 2). Length is
2
(et 1)2 + 4et dt
L=
0
2
2
(et + 1)2 dt =
=
0
= (et + t)
(et + 1) dt
0
2
0
= e2 +
Integral Calculus with Applications to Commerce and Social Sciences
MATHEMATIC 105

Fall 2014
75.
dx
tan x + sin x
cos x d x
=
sin x(1 + cos x)
Let z = tan(x/2),
cos x =
1 z2
,
1 + z2
2 dz
1 + z2
2z
sin x =
1 + z2
dx =
1 z 2 2 dz
1 + z2 1 + z2
=
2z
1 z2
1+
1 + z2
1 + z2
1
1 z2
(1 z 2 ) dz
=
dz
=
z(1 + z 2 + 1 z 2 )
2
z
1
z2
+C
= ln z
2
4
1
x
1
Integral Calculus With Applications to Commerce and Social Sciences
MATHEMATIC 105

Spring 2015
Quiz 1 for MATH 105 SECTION 205
January 14, 2015
Given Name
Family Name
Student Number
1. Let u = 3, 4 and v = 1, 2 , then
(a) (1 point) Compute u + 2v.
5, 0
(a)
Answer.
3, 4 + 2 1, 2
u + 2v =
=
3, 4 + 2, 4
=
3 + 2, 4 + 4
=
5, 0
(b) (1 point) Find the ang
Integral Calculus With Applications to Commerce and Social Sciences
MATHEMATIC 105

Spring 2015
Quiz 2B for MATH 105 SECTION 205
January 23, 2015
Given Name
Family Name
Student Number
1. (a) (1 point) Find the value of a such that the gradient of f (x, y, z) = aex + xy + sin(z) at point (0, 0, 0) is
orthogonal to a normal vector of the plane 2x z =
Integral Calculus With Applications to Commerce and Social Sciences
MATHEMATIC 105

Spring 2015
LECTURE 10: LAGRANGE MULTIPLIERS
MINGFENG ZHAO
January 26, 2015
Lagrange multipliers
Consider the optimization problem:
f (x, y)
max / min
objective function
g(x, y) = 0
constraint.
Method of Lagrange Multipliers:
If
g(x, y) = 0 for all (x, y) such that
Integral Calculus With Applications to Commerce and Social Sciences
MATHEMATIC 105

Spring 2015
LECTURE 9: LAGRANGE MULTIPLIERS
MINGFENG ZHAO
January 23, 2015
Polar coordinates
For the point (x, y) = (0, 0) in the xyplane, let r = (x, y) and be the angle between xaxis and the vector from
(0, 0) to (x, y), then we have
x = r cos(),
and y = r sin(
Integral Calculus With Applications to Commerce and Social Sciences
MATHEMATIC 105

Spring 2015
LECTURE 13: DEFINITE INTEGRALS AND ANTIDERIVATIVES
MINGFENG ZHAO
February 02, 2015
Evaluating denite integrals
If f is integrable on [a, b], then
n
b
f (x )x,
k
f (x) = lim
a
n
k=1
where x0 = a < x1 < < xn = b is a regular partition of [a, b]. For the cho
Integral Calculus With Applications to Commerce and Social Sciences
MATHEMATIC 105

Spring 2015
LECTURE 14: FUNDAMENTAL THEOREM OF CALCULUS
MINGFENG ZHAO
February 04, 2015
Denition 1. Let F (x) be an antiderivative of f (x), that is, F (x) = f (x). The indenite integral,
f (x) dx :=
F (x) + C (where C is arbitrary constant), means nd the antiderivat
Integral Calculus With Applications to Commerce and Social Sciences
MATHEMATIC 105

Spring 2015
LECTURE 12: APPROXIMATING AREAS UNDER CURVES AND DEFINITE INTEGRALS
MINGFENG ZHAO
January 30, 2015
Example 1. For a regular partition x0 = a < x1 < < xn = b of [a, b], then x =
ba
and the Riemann sum can
n
be written:
n
f (x )x + f (x )x + + f (x )x =
1
2
Integral Calculus With Applications to Commerce and Social Sciences
MATHEMATIC 105

Spring 2015
LECTURE 15B: BASIC APPROACHES TO INTEGRATIONS
MINGFENG ZHAO
February 06, 2015
Basic approaches
Example 1. Compute
1
dx.
ex + ex
In fact, we have
ex
1
dx
+ ex
ex
dx
+1
1
du Let u = ex , then du = ex dx
u2 + 1
=
e2x
=
=
=
Example 2. Compute
tan1 (u) + C
tan
Integral Calculus With Applications to Commerce and Social Sciences
MATHEMATIC 105

Spring 2015
LECTURE 8: MAXIMUM/MINIMUM PROBLEMS
MINGFENG ZHAO
January 21, 2015
Absolute maximum and minimum values
Let R be a bounded closed subset in R2 , to nd the absolute maximum and minimum values of f on R:
Step 1: Find the maximum and minimum values of f on th
Integral Calculus With Applications to Commerce and Social Sciences
MATHEMATIC 105

Spring 2015
Quiz 2 for MATH 105 SECTION 205
January 21, 2015
Family Name
Given Name
Student Number
1. (a) (1 point) Let f (x, y) = y sin(xy), compute fyx (x, y).
(a)
Answer.
2y cos(xy) xy 2 sin(xy)
First Approach:
fy (x, y) = sin(xy) + y cos(xy) x
By the product rule
Integral Calculus with Applications to Commerce and Social Sciences
MATHEMATIC 105

Fall 2014
1. If an > 0 and
n
an+1
for all n, then
>
an
n+1
a2
1
>
a1
2
2
a3
>
a2
3
a1
2
2a2
a1
a3 >
>
3
3
a2 >
.
.
.
an
n1
>
an1
n
an >
a1
.
n
(This can be veried by induction.)
Therefore
n=1 an
diverges by comparison with the harmonic series
n=1
1
.
n
x
45.
S(x) =
Integral Calculus with Applications to Commerce and Social Sciences
MATHEMATIC 105

Fall 2012
12. The ellipses 3x 2 + 4y 2 = C all satisfy the differential equation
6x + 8y
dy
= 0,
dx
or
dy
3x
= .
dx
4y
A family of curves that intersect these ellipses at right angles must therefore have slopes
dy
4y
given by
=
. Thus
dx
3x
dx
dy
=4
3
y
x
3 ln y
Integral Calculus with Applications to Commerce and Social Sciences
MATHEMATIC 105

Fall 2012
Math 105  Practice Midterm 2
1 Compute the following:
a
d
dx
2 1
dt
x 1+t3
x
d
1
Solution: We rewrite this as dx 2 1+t3 dt and apply the Funda1
mental Theorem of Calculus to yield 1+x3 .
b
xdx
Solution: This needs to be split into two improper integrals
Integral Calculus with Applications to Commerce and Social Sciences
MATHEMATIC 105

Fall 2012
Math 105  Practice Midterm 1 for Midterm 2
Solutions
This practice midterm may be harder and/or longer than the real midterm.
Not all question will be worth the same number of points.
e x
dx, or show that it doesnt exist.
1. Evaluate
x
0
Lets rst evalua
Integral Calculus with Applications to Commerce and Social Sciences
MATHEMATIC 105

Fall 2012
Math 105 Assignment 9
Due the week of March 28
1. Recall problem 3 from Homework 8, the monopolist selling two products in
Canada and the UK. Let x be the number of units to be sold in Canada
and y the number of units to be sold in the UK. Due to the laws