MTH240- Calculus II
Summer 2025
Section 3.1 – Integration by Parts
Instructor: Saeid Samiezadeh
Email: saeid.samiezadeh@torontomu.ca
Office: VIC 748
1
Review (Useful Identities)
𝑎𝑚 𝑏𝑚 = 𝑎𝑏 𝑚
𝑎𝑚 𝑛 = 𝑎 𝑚𝑛
Exponential identities
𝑎𝑚 𝑏𝑚 = 𝑎𝑏 𝑚
𝑎2 − 𝑏2 = ሺ𝑎 − 𝑏)ሺ𝑎 + 𝑏)
Difference of two squares
𝑎3 ± 𝑏3 = 𝑎 ± 𝑏 𝑎 2 ∓ 𝑎𝑏 + 𝑏2
Sum and difference of two cubes
𝑎 ± 𝑏 2 = 𝑎2 ± 2𝑎𝑏 + 𝑏2
Square of sum and difference
𝑎 ± 𝑏 3 = 𝑎3 ± 3𝑎2 𝑏 + 3𝑎𝑏2 ± 𝑏3
Cube of sum and difference
𝑐
𝑐 𝑐
𝑎+𝑏 𝑎 𝑏
but
≠ +
= +
𝑎+𝑏 𝑎 𝑏
𝑐
𝑐 𝑐
Splitting a fraction
ln𝑎𝑏 = ln𝑎 + ln𝑏
𝑎
ln = ln𝑎 − ln𝑏
𝑏
Natural Logarithmic Identities
2
Review on Integration & Integration by Parts
Section 3.1
3
Review of Integration
Antiderivative Formulas
U-substitution
4
Review of Integration
Antiderivative Formulas
U-substitution
x5 x3
 7x − x dx = 7 5 − 3 + C
4
2
7

1
z
1z 2
 5 2 1 − 12 
 2 1 
−
+C
z  z −  dz =  z − z  dz =
7 4 1
4
4z 



2
2
2 7 2 1 12
= z − z +C
7
2
2
2
1  7
x
x
 8
2
1  2 + x dx = 6 + 2 lnx  =  6 + 2 ln 2 − 6 − 0 = 6 + 2 ln 2
1
2
2
3
5
Review of Integration
Antiderivative Formulas
U-substitution
Substitution = Change of variable (e.g. from x to u)
 f ( x)dx   g (h( x))h( x)dx   g (u )du
u
du
1. Choose your substitution function
2. Differentiate it
3. Rewrite the integral in terms of the new variable
4. In case of indefinite integral, write the answer in terms of the original variable
* The idea is to make the integral solvable by changing the variables
Example:
2
x
1
−
x
dx

6
Review of Integration
Antiderivative Formulas
 x 1 − x dx
2
du
−
2
U-substitution
3ൗ
𝑑𝑢
1
2
1ൗ
1
𝑢
⇒ න 𝑢ሺ− ) = − න𝑢 2 𝑑𝑢 = −
+𝐶
2
2
2 3ൗ
2
1 − 𝑥2
=−
3
3ൗ
2
+𝐶
7
Review of Integration
Antiderivative Formulas
U-substitution
Examples
නsin4𝑥 𝑑𝑥
නsin𝑥 cos𝑥𝑑𝑥
8
Review of Integration
Antiderivative Formulas
U-substitution
5
ln𝑥
න
𝑑𝑥
𝑥
1
𝑒𝑥
඲ 2𝑥
𝑑𝑥
𝑒 +1
9
Integration by Parts
• Substitution (in integrals) corresponds to the Chain Rule (in derivatives)
• The rule that corresponds to the Product Rule is Integration by Parts.
d
 f ( x) g ( x) = f ( x) g ( x) + f ( x) g ( x)
dx
u = f ( x)  du = f ( x)dx
f ( x) g ( x) =  g ( x) f ( x)dx +  f ( x) g ( x)dx
v = g ( x)  dv = g ( x)dx
u
v
v
du
u
dv
𝑢𝑣 = න𝑣𝑑𝑢 + න𝑢𝑑𝑣
 udv = uv −  vdu
10
Integration by Parts
Example:
x
xe
 dx
11
Integration by Parts
Examples
• If two functions are present, choose u according to L I A T E !
Logarithmic, Inverse Trigonometric, Algebraic, Trigonometric, Exponential
e.g.
 (2 t − 4) sin(3t )dt
12
Integration by Parts
Examples
ln x
1 x 2 dx
2
13
Integration by Parts
• Integration by parts (IBP) could be employed to integrate a single function.
e.g. නlnሺ 𝑥 2 + 1) 𝑑𝑥
14
Integration by Parts
Example
2
x
 sinxdx
15
Integration by Parts
• After each application of IBP, watch for a constant multiple of the original integral.
e.g.
x
e
 cosxdx
16
Integration by Parts
• Substitution is sometimes required prior to using IBP.
e.g.
 cos(lnx)dx
17
Integration by Parts
• Substitution is sometimes required prior to using IBP.
 x e dx
5 x2
18
MTH240- Calculus II
Summer 2025
Section 3.2 – Trigonometric Integrals
Instructor: Saeid Samiezadeh
Email: saeid.samiezadeh@torontomu.ca
Office: VIC 748
1
Trigonometric Integrals
U-substitution and trigonometric identities can be combined to solve integrals in
the following forms
Type1: නsin𝑚 𝑥 cos𝑛 𝑥𝑑𝑥
Type2: නtan𝑚 𝑥 sec 𝑛 𝑥𝑑𝑥
m and n are integers and
𝑚 ≥ 0, 𝑛 ≥ 0
Type3: නsin𝑚𝑥 cos𝑛𝑥𝑑𝑥
2
Trigonometric Integrals
Some useful trigonometric identities
sin2 𝑥 + cos 2 𝑥 = 1
sec 2 𝑥 = 1 + tan2 x
1
sin2 𝑥 = (1 − cos2𝑥)
2
Half-angle identities
1
cos 2 𝑥 = (1 + cos2𝑥)
2
1
sin𝐴cos𝐵 = sin(𝐴 − 𝐵) + sin(𝐴 + 𝐵)
2
1
sin𝐴sin𝐵 = cos 𝐴 − 𝐵 − cos(𝐴 + 𝐵)
2
1
cos𝐴cos𝐵 = cos(𝐴 − 𝐵) + cos(𝐴 + 𝐵)
2
sin2𝑥 = 2sin𝑥cos𝑥 ,
Product-to-sum identities
cos2𝑥 = 2cos2 𝑥 − 1
3
Trigonometric Integrals
Type 1
නsin𝑚 𝑥 cos 𝑛 𝑥𝑑𝑥
Type 2
m and n are integers and
Type 3
𝑚 ≥ 0, 𝑛 ≥ 0
4
Trigonometric Integrals
Type 1
නsin𝑚 𝑥 cos 𝑛 𝑥𝑑𝑥
Type 2
m and n are integers and
Type 3
𝑚 ≥ 0, 𝑛 ≥ 0
If either m or n is odd → save one of the odd factors and use sin2 𝑥 = 1 − cos 2 𝑥
or cos 2 𝑥 = 1 − sin2 𝑥, then use u-substitution.
Example: නsin2 𝑥 cos 3 𝑥𝑑𝑥
5
Trigonometric Integrals
Type 1
නsin𝑚 𝑥 cos 𝑛 𝑥𝑑𝑥
Type 2
Type 3
m and n are integers and
𝑚 ≥ 0, 𝑛 ≥ 0
1
sin2 𝑥 = (1 − cos2𝑥)
2
1
cos 2 𝑥 = (1 + cos2𝑥)
2
If both m and n are even → use half-angle identities
Example: නsin2 𝑥 cos 2 𝑥𝑑𝑥
*Note that m and n can be zero, i.e. නsin𝑚 𝑥 𝑑𝑥 or නcos 𝑛 𝑥 𝑑𝑥
6
Trigonometric Integrals
Type 1
නtan𝑚 𝑥 sec 𝑛 𝑥𝑑𝑥
Type 2
Type 3
m and n are integers
If the power of sec𝑥 is even >> save a factor of sec 2 𝑥 and use sec 2 𝑥 = 1 + tan2 x
then use u-substitution (𝑢 = tan𝑥).
Example: නtan3 𝑥 sec 4 𝑥𝑑𝑥
*Note that m and n can be zero, i.e. නtan𝑚 𝑥 𝑑𝑥 or න𝑠𝑒𝑐 𝑛 𝑥 𝑑𝑥
7
Trigonometric Integrals
Type 1
නtan𝑚 𝑥 sec 𝑛 𝑥𝑑𝑥
Type 2
Type 3
m and n are integers
If the power of tan𝑥 is odd >> save a factor of sec𝑥tan𝑥 and use tan2 𝑥 = sec 2 𝑥 − 1
then express the remaining factors as sec𝑥 .
Example: නtan3 𝑥 sec 2 𝑥𝑑𝑥
• Integrals like නcot 𝑚 𝑥 csc 𝑛 𝑥𝑑𝑥 could also be solved in the same manner.
8
Trigonometric Integrals
Type 1
Type 2
Type 3
නsin𝑚𝑥sin𝑛𝑥𝑑𝑥
නcos𝑚𝑥cos𝑛𝑥𝑑𝑥
m and n are integers
නsin𝑚𝑥cos𝑛𝑥𝑑𝑥
Example: නsin 3𝑥 cos 2𝑥 𝑑𝑥
9
Trigonometric Integrals
Reminder
නtan 𝑥 𝑑𝑥 = ln|sec 𝑥| + 𝐶
නsec 𝑥 𝑑𝑥 = ln|sec 𝑥 + tan 𝑥| + 𝐶
(tan 𝑥)′ = sec 2 𝑥
(sec 𝑥)′ = sec 𝑥 tan 𝑥
10
Trigonometric Integrals
Additional Examples
𝜋
2
5
න sin 𝑥𝑑𝑥
0
න
1
𝑑𝑥
sin𝑥 − 1
නtan3 𝑥 𝑑𝑥
නcos 2 𝑥 − sin2𝑥 𝑑𝑥
නsin7 𝑥cos5 𝑥 𝑑𝑥
න𝑥sin3 𝑥 𝑑𝑥
නcsc 4 𝑥cot 6 𝑥 𝑑𝑥
න𝑥 sec 𝑥 tan 𝑥 𝑑𝑥
11
MTH240- Calculus II
Summer 2025
Section 3.3 – Trigonometric Substitution
Instructor: Saeid Samiezadeh
Email: saeid.samiezadeh@torontomu.ca
Office: VIC 748
1
Trigonometric Substitution
Some integrals containing radical functions could be solved by substituting
trigonometric functions.
Expression
Substitution
Identity
𝜋
𝜋
1 − sin2 𝜃 = cos 2 𝜃
𝜋
2
𝜋
2
1 + tan2 𝜃 = sec 2 𝜃
𝑎2 − 𝑥 2
𝑥 = 𝑎sin𝜃,
−2 ≤ 𝜃 ≤ 2
𝑎2 + 𝑥 2
𝑥 = 𝑎tan𝜃,
− <𝜃<
𝑥 2 − 𝑎2
𝜋
3𝜋
𝑥 = 𝑎sec𝜃, 0 ≤ 𝜃 < 2 or 𝜋 ≤ 𝜃 < 2
sec 2 𝜃 − 1 = tan2 𝜃
* 𝑎 is a constant
2
Trigonometric Substitution
3
Example: න
0
𝑥
36 − 𝑥 2
𝑑𝑥
3
Trigonometric Substitution
3
Example: න
0
𝑥
36 − 𝑥 2
𝑑𝑥
4
Trigonometric Substitution
Steps
1- Choose the suitable substitution according to the table.
2- Write the integral in terms of 𝜃 (change the boundaries in case of having a
definite integral).
3- Integrate the function and write the final answer in terms of the initial variable
(i.e. x). The final answer does not have any trigonometric functions.
1
𝑐
𝑎
csc𝜃 =
=
sin𝜃 =
sin𝜃 𝑎
𝑐
1
𝑐
𝑐
𝑏
sec𝜃
=
=
cos𝜃 =
𝑎
cos𝜃 𝑏
𝑐
1
𝑏
𝑎
cot𝜃
=
=
tan𝜃 =
tan𝜃 𝑎
𝑏
𝑏
5
Trigonometric Substitution
Example න
1
𝑥2
𝑥2 + 4
𝑑𝑥
6
Trigonometric Substitution
2
• Trigonometric substitution could also be used to integrate functions like 𝑥 ± 𝑎
or
𝑛
2
2
𝑎 ±𝑥 2
(where n is an integer) for instance, e.g.,඲
1
3
𝑥 2 + 3 ൗ2
𝑛
2 2
𝑑𝑥
7
Trigonometric Substitution
• Completing the square can sometimes be used to make the radical expression
1
න
𝑑𝑥 .
ready for trig. Substitution, e.g.,
2
𝑥 − 6𝑥 + 13
8
Trigonometric Substitution
• Some integrals containing standard radical expressions could be solved without
using trig. substitution, e.g., ඲
8𝑥 − 6
5
2𝑥 2 − 3𝑥 + 5 ൗ2
𝑑𝑥
9
Trigonometric Substitution
Additional Examples:
඲
1
3
3 + 4𝑥 − 4𝑥 2 ൗ2
𝑑𝑥
𝑥2 − 1
඲
𝑑𝑥
𝑥3
10
MTH240- Calculus II
Summer 2025
Section 3.4 – Partial Fractions
Instructor: Saeid Samiezadeh
Email: saeid.samiezadeh@torontomu.ca
Office: VIC 748
1
Integrating using Partial Fractions
A rational function is in the form of
𝑃(𝑥 ሻ
𝑄(𝑥 ሻ
in which P(x) and Q(x) are polynomials.
𝑥3 + 1
Examples: ඲
𝑑𝑥
𝑥−1
න
𝑥
𝑑𝑥
𝑥 2 − 3𝑥 + 2
* Some rational functions could be easily integrated using direct formula or usubstitution. Examples:
−1
1
+𝐶
න 2 𝑑𝑥 =
𝑥
𝑥
1
න
𝑑𝑥 = ln|𝑥 + 5| + 𝐶
𝑥+5
3𝑥 2 − 4𝑥
඲ 3
𝑑𝑥
𝑥 − 2𝑥 2
𝑑𝑢
𝑢 = 𝑥 3 − 2𝑥 2
⇒න
= ln|𝑢| + 𝐶 = ln|𝑥 3 − 2𝑥 2 | + 𝐶
𝑢
𝑑𝑢 = (3𝑥 2 − 4𝑥ሻ𝑑𝑥
2
Integrating using Partial Fractions
𝑃(𝑥 ሻ
To integrate඲
𝑑𝑥
𝑄(𝑥 ሻ
If deg(𝑃(𝑥ሻሻ < deg(𝑄(𝑥ሻሻ
⇒
If deg(𝑃(𝑥ሻሻ ≥ deg(𝑄(𝑥ሻሻ
⇒
𝑥4 + 1
Proper rational function, e.g.,
𝑥 5 + 4𝑥 3
𝑥6
Improper rational function, e.g.,
𝑥2 − 4
* An improper rational function should be converted to a proper one before integrating.
2𝑥 3 − 9𝑥 2 + 15
඲
𝑑𝑥
2𝑥 − 5
3
Integrating using Partial Fractions
Steps to do partial fraction decomposition
1
𝑥 3 + 4𝑥
1- Factor the denominator as much as you can.
1
𝑥(𝑥 2 + 4ሻ
* If you get quadratic functions, make sure that they are irreducible.
2- Write a partial fraction for each factor. The number of fractions is equal to the
number of factors in the denominator.
Bx+C
1
A
= +
𝑥(𝑥 2 + 4ሻ 𝑥 𝑥 2 + 4
3- Choose an expression for the numerators as follows:
Linear function (e.g. 4x) ⇒ A constant (e.g. A)
Quadratic function (e.g. 5𝑥 2 + 4) ⇒ Linear function (e.g. 𝐴𝑥 + 𝐵 )
4
Integrating using Partial Fractions
4- To find the constants, the method of method of strategic substitution should
be used. Multiply both sides of the equation by the initial denominator.
1
A Bx + C
=
+ 2
 1 = A( x 2 + 4) + ( Bx + C ) x
2
x( x + 4) x x + 4
5- Instead of expanding the equation, plug in some numbers in the equation and
solve for the constants.
5
Integrating using Partial Fractions
To sum up:
6
Integrating using Partial Fractions
To sum up:
7
Integrating using Partial Fractions
Reminder:
න
𝑐
𝑑𝑥
𝑎𝑥 + 𝑏
1
1
𝑥
−1
න 2
𝑑𝑥 = tan
+𝐶
𝑥 + 𝑎2
𝑎
𝑎
8
Integrating using Partial Fractions
Additional Examples: Evaluate the following integrals
න
6
𝑑𝑥
𝑥2 − 1
න
𝑑𝑥
1 + 𝑒𝑥
𝑥2
඲ 2
𝑑𝑥
𝑥 +𝑥+2
Write the partial fraction decomposition for the following fractions. DO NOT
evaluate the coefficients.
5𝑥 − 4
𝑥3 + 1
𝑥 3 + 2𝑥 + 1
𝑥2 − 𝑥 − 2
𝑥5 + 1
𝑥 2 − 𝑥ሻ(𝑥 4 + 2𝑥 2 + 1
4𝑥
𝑥3 + 𝑥2 + 𝑥 + 1
9
MTH240- Calculus II
Summer 2025
Section 3.7 – Improper Integrals
Instructor: Saeid Samiezadeh
Email: saeid.samiezadeh@torontomu.ca
Office: VIC 748
1
Review: L’Hospital’s Rule
L’Hospital’s Rule says that the limit of a quotient of functions is equal to the limit
𝑓(𝑥 ሻ
𝑓 ′ (𝑥 ሻ
of the quotient of their derivatives i.e. lim
, provided that:
= lim ′
𝑥→𝑎 𝑔(𝑥 ሻ
𝑥→𝑎 𝑔 (𝑥 ሻ
1- Both 𝑓(𝑥 ሻ and 𝑔(𝑥 ሻ are differentiable and 𝑔(𝑥ሻ ≠ 0 on an interval I that
contains 𝑎.
2- We have an indeterminate form of type
∞
0
or .
∞
0
𝑓′ (𝑥൯
3- lim ′ ሻ exists or is +∞ or −∞.
𝑥→𝑎 𝑔 (𝑥
1
𝐿𝐻
0
ln𝑥
𝑥 = lim 1 = 1
indeterminate
=
lim
Example: lim
=
𝑥→1 1 𝑥→1 𝑥
0
𝑥→1 𝑥 − 1
2
Review: L’Hospital’s Rule
• Note that limits like
∞
0
or
are not indeterminate and do not require
0
∞
L’Hospital’s rule.
ln𝑥
𝑥→0 𝑥
Example: lim+
• L’Hospital’s rule cannot directly be used on limits resulting in 0 × ∞ and ∞ × 0.
0
∞
They have to be converted to or first.
0
Example: lim 𝑥𝑒 −𝑥
∞
2
𝑥→∞
3
Improper Integrals
A definite integral of a function on 𝑎, 𝑏 gives the area under a curve between 𝑎
and 𝑏.
𝑏
න 𝑓(𝑥ሻ𝑑𝑥 = 𝑆
𝑎
This area may not always be finite.
4
Improper Integrals
• The possibility of having an infinite area would make a definite integral
improper.
• An improper integral is a definite integral that has either or both limits infinite
(i.e. +∞ or −∞ ) OR the function in the integral approaches infinity at one or
more points in the range of integration.
• Based on the definitions above, there are two types of improper integrals.
• An improper integral is convergent if it results in a finite number.
5
Improper Integrals
Type 2
Discontinues Integrands
Type 1
Infinite Intervals
If either or both limits are infinite.
𝑡
𝑡
+∞
If න 𝑓(𝑥ሻ𝑑𝑥 exists for every number 𝑡 ≥ 𝑎
𝑓(𝑥ሻ𝑑𝑥 = lim න 𝑓(𝑥ሻ𝑑𝑥
න
𝑎
𝑡→+∞ 𝑎
𝑏
𝑏
𝑎
𝑏
If න 𝑓(𝑥ሻ𝑑𝑥 exists for every number 𝑡 ≤ 𝑏
න 𝑓(𝑥ሻ𝑑𝑥 = lim න 𝑓(𝑥ሻ𝑑𝑥
−∞
𝑡→−∞ 𝑡
+∞
𝑎
න
−∞
𝑡
+∞
𝑓(𝑥ሻ𝑑𝑥 = න 𝑓(𝑥ሻ𝑑𝑥 + න
−∞
𝑓(𝑥ሻ𝑑𝑥
𝑎 could be any real number
𝑎
6
Improper Integrals
Type 1
Infinite Intervals
+∞
න
𝑎
Type 2
Discontinues Integrands
𝑡
𝑓(𝑥ሻ𝑑𝑥 = lim න 𝑓(𝑥ሻ𝑑𝑥
𝑡→+∞ 𝑎
The integral is convergent if the limit exists and divergent if the limit does not exist.
7
Improper Integrals
Type 1
Infinite Intervals
Type 2
Discontinues Integrands
∞
1
𝑑𝑥
Example: න
𝑥
1
8
Improper Integrals
Type 1
Infinite Intervals
∞
Example: න
1
Type 2
Discontinues Integrands
1
𝑑𝑥
𝑥2
9
Improper Integrals
Type 1
Infinite Intervals
∞
Example: න
1
Type 2
Discontinues Integrands
1
𝑑𝑥
𝑥2
10
Improper Integrals
Type 1
Infinite Intervals
Type 2
Discontinues Integrands
∞
𝑒𝑥
Example: ඲
𝑑𝑥
𝑒 2𝑥 + 3
0
11
Improper Integrals
Type 1
Infinite Intervals
Type 2
Discontinues Integrands
An Important example:
∞
1
න 𝑝 𝑑𝑥 is convergent if 𝑝 > 1 and divergent if 𝑝 ≤ 1
1 𝑥
12
Improper Integrals
Type 2
Discontinues Integrands
Type 1
Infinite Intervals
Comparison Theorem
Suppose that 𝑓(𝑥ሻ and 𝑔(𝑥ሻ are continuous functions with 𝑓(𝑥ሻ ≥ 𝑔(𝑥ሻ ≥ 0
for 𝑥 ≥ 𝑎
∞
∞
(a) If න 𝑓(𝑥ሻ𝑑𝑥 is convergent, then න 𝑔(𝑥ሻ𝑑𝑥 is convergent.
a
a
∞
∞
(b) If න 𝑔(𝑥ሻ𝑑𝑥 is divergent, then න 𝑓(𝑥ሻ𝑑𝑥 is divergent.
a
a
13
Improper Integrals
Type 1
Infinite Intervals
Type 2
Discontinues Integrands
∞
3 cos 2 𝑥
𝑑𝑥 converges or diverges?
Example: Determine if ඲
1 + 𝑥2
0
14
Improper Integrals
Type 2
Discontinues Integrands
Type 1
Infinite Intervals
If the integrant approaches infinity at either limit or at some point between the
b
limits; න 𝑓(𝑥ሻ𝑑𝑥
𝑎
b
If 𝑓(𝑥ሻ is discontinuous at 𝑎
𝑏
න 𝑓(𝑥ሻ𝑑𝑥 = lim+ න 𝑓(𝑥ሻ𝑑𝑥
𝑎
𝑡→𝑎
b
If 𝑓(𝑥ሻ is discontinuous at 𝑏
If 𝑓(𝑥ሻ is discontinuous at 𝑐
where 𝑎 < 𝑐 < 𝑏
𝑡
𝑡
න 𝑓(𝑥ሻ𝑑𝑥 = lim− න 𝑓(𝑥ሻ𝑑𝑥
𝑎
𝑡→𝑏
b
𝑐
𝑎
𝑏
න 𝑓(𝑥ሻ𝑑𝑥 = න 𝑓(𝑥ሻ𝑑𝑥 + න 𝑓(𝑥ሻ𝑑𝑥
𝑎
𝑎
𝑐
15
Improper Integrals
Type 2
Discontinues Integrands
Type 1
Infinite Intervals
b
𝑏
න 𝑓(𝑥ሻ𝑑𝑥 = lim+ න 𝑓(𝑥ሻ𝑑𝑥
𝑎
𝑡→𝑎
𝑡
The integral is convergent if the limit exist
and divergent if the limit does not exist.
16
Improper Integrals
Type 2
Discontinues Integrands
Type 1
Infinite Intervals
Reminder
1
Undefined
0
1
= lim+ = +∞
𝑥→0 𝑥
1
lim
𝑥→0 𝑥
1
= lim− = −∞
𝑥→0 𝑥
ln0
Undefined
= lim+ ln𝑥 = −∞
𝑥→0
lim ln𝑥
𝑥→0
= lim− ln𝑥 Undefined
𝑥→0
17
Improper Integrals
Type 1
Infinite Intervals
Type 2
Discontinues Integrands
4
1
𝑑𝑥
2
𝑥
0
Example: න
18
Improper Integrals
Type 1
Infinite Intervals
Type 2
Discontinues Integrands
4
1
𝑑𝑥
1
−
𝑥
1
Example: න
19
Improper Integrals
Type 1
Infinite Intervals
Type 2
Discontinues Integrands
3
1
𝑑𝑥
Example: න 2
𝑥
−
6𝑥
+
5
0
20
Improper Integrals
Type 1
Infinite Intervals
Type 2
Discontinues Integrands
Cont’d:
21
Improper Integrals
Type 1
Infinite Intervals
Type 2
Discontinues Integrands
What can go wrong?
3
3
1
𝑥−5
1
1
1
1
| = − ln5
න 2
𝑑𝑥 = − ln|𝑥 − 1| + ln|𝑥 − 5| = ln|
4
𝑥−1 0
4
4
4
0 𝑥 − 6𝑥 + 5
3
0
But we know that this integral is divergent.
22
Improper Integrals
Additional Examples:
∞
1
න
3 𝑑𝑥
𝑥
ln𝑥
𝑒
∞
න (𝑦 3 − 3𝑦 2 ሻ 𝑑𝑦
−∞
∞
3
𝑑𝑥
4
−2 𝑥
න
𝑥2
඲
𝑑𝑥
9 + 𝑥6
−∞
2
න 𝑥 2 ln𝑥 𝑑𝑥
0
23
MTH240- Calculus II
Summer 2025
Sections 4.1, 4.3, 4.5 - Differential Equations
Instructor: Saeid Samiezadeh
Email: saeid.samiezadeh@torontomu.ca
Office: VIC 748
1
Differential Equations
Some familiar types of equations:
• Algebraic equations, e.g., 3𝑥 + 5 = 0 , 𝑥 2 − 4𝑥 + 4 = 0
• Exponential equations, e.g., 22𝑥−4 = 8
• Trigonometric equations, e.g., cos2𝑥 = 1 + 4sin𝑥
Solving an equation ⇒
Finding the unknown which is a number or numbers
in the above equations
Differential Equations
A differential equation is a mathematical equation that relates a function with its
derivatives. The equation can contain the function, its derivatives, and the variable.
The unknown is a function and not a number in these equations.
Example: 𝑦 ′ = 2𝑥 ⇒
𝑑𝑦
= 2𝑥 ⇒ 𝑦 = 𝑥 2 + 𝐶
𝑑𝑥
The order of the differential equation ⇒ The number of highest derivative in a
differential equation
Example:
𝑦 ′ − 3𝑥𝑦 = 0
𝑦 ″ − 4𝑦 = 0
⇒ Only first order DEs will be covered in this
course
Second-order
First-order
Differential Equations
Example: What is the order of the following differential equations?
Differential Equations
Verifying Solutions of Differential Equations
Example: Verify that the function 𝑦 = 𝑒 −3𝑥 + 2𝑥 + 3 is a solution to the differential
equation 𝑦 ′ + 3𝑦 = 6𝑥 + 11.
Differential Equations
General and Particular Solutions
Differential Equations
Example: Find the particular solution to the differential equation y′ = 2x passing through
the point (2, 7).
Differential Equations
Initial-Value Problems
• Usually, a given differential equation has an infinite number of solutions.
• To choose one solution, more information is needed. Some specific information
that can be useful is an initial value, which is an ordered pair that is used to find
a particular solution.
• A differential equation together with one or more initial values is called an
initial-value problem.
• The general rule is that the number of initial values needed for an initial-value
problem is equal to the order of the differential equation.
Differential Equations
Example:
Differential Equations
Two types of first-order differential equations will be studied:
1- Separable differential equations
2- Linear differential equations
Differential Equations
Separable DE
Linear DE
A separable equation is a first-order differential equation in which the expression
𝑑𝑦
for
can be factored as a function of x times a function of y.
𝑑𝑥
In other words, it can be written in the forms of
𝑑𝑦
𝑑𝑦 𝑓(𝑥 )
or
= 𝑓(𝑥). 𝑔(𝑦)
=
𝑑𝑥
𝑑𝑥 ℎ(𝑦)
To solve it ℎ(𝑦)𝑑𝑦 = 𝑓(𝑥)𝑑𝑥 ⇒ නℎ(𝑦)𝑑𝑦 = න𝑓(𝑥)𝑑𝑥
The solution may be found implicitly or explicitly as a function of x.
Differential Equations
Separable DE
Example: Solve 𝑦 ′ = 2𝑥
It is not a single curve but a family of curves.
Example: Solve
𝑑𝑦
= 𝑥 2𝑦2
𝑑𝑥
Linear DE
Differential Equations
Separable DE
Linear DE
* An initial condition is required to define the constant after integration.
Example: Solve 𝑦 ′ cos𝑦 = sin2 𝑡 cos𝑡 , 𝑦(0) =
𝜋
6
Differential Equations
Additional Examples:
Solve the following differential equations
(𝑦 2 + 𝑥𝑦 2 )𝑦 ′ = 1
𝑑𝑦 ln𝑥
=
𝑑𝑥 𝑥𝑦
𝑦(1) = 2
𝑦 ′ tan𝑥 = 𝑎 + 𝑦
𝜋
𝑦
=𝑎
3
𝜋
0<𝑥<
2
Differential Equations
Separable DE
Linear DE
A first order linear differential equation is one that can be written in the form
𝑑𝑦
+ 𝑃(𝑥)𝑦 = 𝑄(𝑥)
𝑑𝑥
Example 𝑦 ′ + 𝑥𝑦 = 𝑥 2
𝑑𝑦
Note that this differential equation is not separable as it is not possible to write 𝑑𝑥
as a function of x times a function of y.
𝑑𝑦
= 𝑄 𝑥 − 𝑃(𝑥)𝑦
𝑑𝑥
Differential Equations
Separable DE
Linear DE
Example: Put each of the following first-order linear differential equations into
standard form. Identify p(x) and q(x) for each equation.
Differential Equations
Separable DE
Linear DE
Differential Equations
Separable DE
Linear DE
How to solve a Linear DE?
𝟏
Note: You may use the formula 𝒚 = 𝝁(𝒙) ‫ 𝒙𝒅 𝒙 𝒒 𝒙 𝝁 ׬‬+ 𝑪 to get the solution in Step 4.
Differential Equations
Separable DE
Example: Solve
𝑦 ′ − 𝑦 = 𝑥𝑒 𝑥
Linear DE
Differential Equations
Separable DE
Linear DE
Sometimes the DE needs to be manipulated to obtain the standard form.
Example: Solve the initial value problem 𝑥 2 𝑦 ′ − x𝑦 = 1
𝑥>0
𝑦 1 =2
Differential Equations
Additional Examples
Solve the following differential equations
𝑥𝑦 ′ + 𝑦 = 𝑥
𝑥𝑦 ′ = 𝑦 + 𝑥 2 sin𝑥
𝑦(𝜋) = 0
𝑑𝑟
𝑡 ln𝑡 + 𝑟 = 𝑡𝑒 𝑡
𝑑𝑡
21
MTH240- Calculus II
Summer 2025
Section 5.1 - Sequences
Instructor: Saeid Samiezadeh
Email: saeid.samiezadeh@torontomu.ca
Office: VIC 748
1
Sequences
A sequence is an infinite set of numbers written in a defined order.
e.g. 1, 2, 3, 4, ...
𝑛
𝑎1 , 𝑎2 , 𝑎3 , 𝑎4 , ….
𝑎𝑛
First term
General term
It is also denoted by 𝑎1 , 𝑎2 , 𝑎3 , 𝑎4 , . . . or 𝑎𝑛 ∞
𝑛=1
Some examples
𝑛 ∞
𝑛 + 1 𝑛=1
1 2 3 4
𝑛
, , , ,...,
,...
2 3 4 5
𝑛+1
−1 𝑛 (𝑛 + 1ሻ
2𝑛
3 4 5
−1 𝑛 (𝑛 + 1ሻ
−1, , − , , . . . ,
,...
4 8 16
2𝑛
∞
𝑛 − 4 𝑛=4
0,1, 2, 3, . . . , 𝑛 − 4, . . .
2
Sequences
Example: Find an explicit formula for the general term of the sequence
1 3 5 7
− , ,− , ...
2 4 6 8
𝑎𝑛 = −1 𝑛
2𝑛 − 1
2𝑛
Limit of Sequence: A sequence has the limit L if for every 𝜀 > 0 , there is an
integer N such that
|𝑎𝑛 − 𝐿| < 𝜀
n>𝑁
⇒
and we write lim 𝑎𝑛 = 𝐿
𝑛→∞
3
Sequences
Theorem: If lim 𝑓(𝑥ሻ = 𝐿 and 𝑓(𝑛ሻ = 𝑎𝑛 , where n is an integer, then lim 𝑓(𝑛ሻ = 𝐿
𝑥→∞
𝑛→∞
Convergence/Divergence Test for Sequences
A sequence is convergent if lim 𝑓(𝑛ሻ exists.
𝑛→∞
and is divergent if lim 𝑓(𝑛ሻ does not exist (e.g. lim 𝑓(𝑛ሻ = ∞)
𝑛→∞
𝑛→∞
4
Sequences
Example: Determine whether the following sequences are convergent or divergent.
1
𝑛
1
lim = 0
𝑛→∞ 𝑛
Convergent
1
𝑛3
1
=0
𝑛→∞ 𝑛3
Convergent
𝑒 −𝑛
lim 𝑒 −𝑛= lim
lim
𝑛→∞
1
=0
𝑛→∞ 𝑒 𝑛
Convergent
Remark:
1
lim 𝑟 = 0 for r > 0
𝑛→∞ 𝑛
5
Sequences
Example: Determine whether the following sequences are convergent or divergent.
2𝑒 𝑛 + 1
𝑒𝑛
𝑛
ln 𝑛
tan−1 (𝑛
6
Sequences
Squeeze Theorem for Sequences
If 𝑎𝑛 ≤ 𝑏𝑛 ≤ 𝑐𝑛 for 𝑛 ≥ 𝑛0 , and lim 𝑎𝑛 = lim 𝑐𝑛 = 𝐿 , then lim 𝑏𝑛 = 𝐿 .
𝑛→∞
𝑛→∞
𝑛→∞
Sequence 𝑏𝑛 is squeezed between
𝑎𝑛 and 𝑐𝑛 .
7
Sequences
Squeeze Theorem for Sequences
sin(𝑛2 ሻ
Example: Find the limit of the sequence 𝑎𝑛 =
𝑛2
8
Sequences
Alternating sequence
An alternating sequence is one whose terms are alternately positive and negative.
−1 𝑛
−1 1 −1 1 −1
Example:
=
, ,
, ,
,...
3𝑛
3 6 9 12 15
Theorem: If lim |𝑎𝑛 | = 0, then lim 𝑎𝑛 = 0 .
𝑛→∞
𝑛→∞
9
Sequences
−1 𝑛
Example: Find the limit of the sequence 𝑎𝑛 =
.
𝑛+1
10
Sequences
Reminder from Calculus I
11
Sequences
Theorem: If lim 𝑎𝑛 = 𝐿 , and the function f is continuous at L, then
𝑛→∞
lim 𝑓(𝑛ሻ = 𝑓(𝐿ሻ
𝑛→∞
Example: Find the limit of the sequence 𝑎𝑛 = 𝑒
Example: Find the limit of the sequence lim sin
𝑛→∞
𝑛2
3𝑛2 +4
.
𝜋
.
𝑛
12
Sequences
An important example: The sequence 𝑟 𝑛 is convergent if −1 < 𝑟 ≤ 1, and is
divergent elsewhere.
0 −1 < 𝑟 < 1
This is because lim 𝑟 𝑛 = ቐ
𝑛→∞
1
𝑟=1
Monotone sequence: A sequence is monotone/monotonic if it is either increasing or
decreasing.
An increasing sequence is one in which 𝑎𝑛 < 𝑎𝑛+1 for 𝑛 ≥ 1 , i.e., 𝑎1 < 𝑎2 < 𝑎3 <. . .
A decreasing sequence is one in which 𝑎𝑛 > 𝑎𝑛+1 for 𝑛 ≥ 1 , i.e., 𝑎1 > 𝑎2 > 𝑎3 >. . .
13
Sequences
Example: The sequence
Because
1
is decreasing.
𝑛+3
1 1 1
> > >. . .
4 5 6
1
Also
𝑛+3
′
=
−1
< 0 for 𝑛 ≥ 1
𝑛+3 2
Bounded Sequence:
A sequence 𝑎𝑛 is bounded above if there is a number M such that
𝑎𝑛 ≤ 𝑀
for all 𝑛 ≥ 1
And it is bounded below if there is if there is a number m such that
𝑚 ≤ 𝑎𝑛
for all 𝑛 ≥ 1
If it is bounded above and below, then is a bounded sequence.
14
Sequences
Example: The sequence 𝑛 is bounded below since 1 ≤ 𝑛
and the sequence
𝑛
𝑛
is
bounded
above
since
<1
𝑛2 + 1
𝑛2 + 1
Monotone Convergence
Theorem:
Every bounded, monotonic
sequence is convergent.
Example: The sequence
so it is convergent.
1
𝑛
𝑒
1
is bounded ( 1 < 𝑒 𝑛 < 𝑒) and decreasing
1 ′
𝑒𝑛 =
1
1
𝑛
− 2 𝑒
𝑛
15
Sequences
Additional Examples
List the first five terms of the sequence
𝑎𝑛
3𝑛
𝑎
=
,
𝑎
=
6
𝑛+1
1
𝑛
1 + 2𝑛
Find a formula for the general term of the sequence
4 8 16
−3,2, − , , − , . . .
3 9 27
Determine if the sequence converges or diverges. If it converges, find the limit.
𝑎𝑛 =
𝑛2
𝑛3 + 4𝑛
−1 𝑛+1 𝑛
𝑎𝑛 =
𝑛+ 𝑛
𝑎𝑛 = 𝑛 sin
1
𝑛
2−𝑛 cos (𝑛𝜋ሻ
16
MTH240- Calculus II
Spring 2025
Section 5.2 – Infinite Series
Instructor: Saeid Samiezadeh
Email: saeid.samiezadeh@torontomu.ca
Office: VIC 748
1
Series
A series is the sum of the terms of a sequence.
5
1+2+3+4+5= ෍𝑛
𝑛=1
In this course, we are particularly interested in infinite series:
∞
෍ 𝑎𝑛
෍ 𝑎𝑛
𝑛=1
∞
෍ 𝑎𝑛 = 𝑎1 + 𝑎2 + 𝑎3 + 𝑎4 +. . . +𝑎𝑛 +. . .
𝑛=1
2
Series
Can the sum of infinitely many numbers be finite?
How can we find the sum (value) of a series?
Partial sum in series
∞
1 1 1 1
1
1
=
+
+
+
+.
.
.
+
Consider the series ෎ 𝑛
𝑛 +. . .
2
4
8
16
2
2
3
𝑠3 = ෎
𝑛=1
1
1 1 1
= + +
2𝑛 2 4 8
𝑛=1
5
1
1 1 1 1
1
𝑠5 = ෎ 𝑛 = + + +
+
2
2 4 8 16 32
𝑛
𝑠𝑛 = ෎
1 1 1 1 1
1
=
+
+
+
+.
.
.
+
2𝑛
2𝑖 2 4 8 16
𝑖=1
𝑛=1
3
Series
∞
Definition: Given a series ෍ 𝑎𝑛 = 𝑎1 + 𝑎2 + 𝑎3 + 𝑎4 +. . . , let 𝑠𝑛 denotes its
nth partial sum:
𝑛=1
𝑛
𝑠𝑛 = ෍ 𝑎𝑖 = 𝑎1 + 𝑎2 + 𝑎3 + 𝑎4 +. . . +𝑎𝑛
𝑖=1
If the sequence 𝑠𝑛 is convergent and lim 𝑠𝑛 = 𝑠 exists as a real number, then
𝑛→∞
the series ෍ 𝑎𝑛 is called convergent, and we write
∞
𝑎1 + 𝑎2 + 𝑎3 + 𝑎4 +. . . +𝑎𝑛 +. . . = 𝑠
or
෍ 𝑎𝑛 = 𝑠
𝑛=1
The number is called the sum of the series. If the sequence 𝑠𝑛 is divergent, then
the series is called divergent.
* This definition can only be used if we have/can get the sum of the nth term.
4
Series
∞
Example: Suppose we know that the sum of the first n terms of the series ෍ 𝑎𝑛
𝑛=1
is:
3𝑛2
𝑠𝑛 = 𝑎1 + 𝑎2 + 𝑎3 + 𝑎4 +. . . +𝑎𝑛 = 2
5𝑛 + 1
Find the sum of the series.
5
Series
Geometric series
Geometric series is a special series in which each successive term is produced by
multiplying the previous term by a constant number.
* This constant number is called common ratio and is shown by r.
∞
𝑎 + 𝑎𝑟 + 𝑎𝑟 2 + 𝑎𝑟 3 +. . . +𝑎𝑟 𝑛−1 +. . . = ෍ 𝑎𝑟 𝑛−1
∞
𝑟≠0
𝑛=1
Example: ෍ 2𝑛 = 2 + 4 + 8 + 16 + 32+. . .
𝑛=1
Sum of geometric series
Assuming 𝑟 ≠ 1
𝑠𝑛 = 𝑎 + 𝑎𝑟 + 𝑎𝑟 2 + 𝑎𝑟 3 +. . . +𝑎𝑟 𝑛−1
𝑟𝑠𝑛 =
Subtracting
𝑎𝑟 + 𝑎𝑟 2 + 𝑎𝑟 3 +. . . +𝑎𝑟 𝑛−1 + 𝑎𝑟 𝑛
𝑠𝑛 − 𝑟𝑠𝑛 = 𝑎 − 𝑎𝑟 𝑛
6
Series
Geometric series (cont’d)
𝑎(1 − 𝑟 𝑛 ሻ
𝑠𝑛 =
1−𝑟
To find the sum of the series
𝑎
𝑎
𝑎
𝑎𝑟 𝑛
𝑎(1 − 𝑟 𝑛 ሻ
=
−
lim𝑟 𝑛
= lim
−
lim 𝑠𝑛 = lim
𝑛→∞ 1 − 𝑟
𝑛→∞
𝑛→∞
1 − 𝑟 1 − 𝑟 1 − 𝑟 𝑛→∞
1−𝑟
When 𝑟 < 1 , lim𝑟 𝑛 = 0 and the sum is
𝑛→∞
𝑎
1−𝑟
Convergence of Geometric Series
∞
The geometric series ෍ 𝑎𝑟 𝑛−1 is convergent if |𝑟| < 1 , and its sum is
𝑛=1
The series is divergent if |𝑟| ≥ 1 .
𝑎
1−𝑟
first term of the series
7
Series
∞
Example: Is the series ෍ 22𝑛 31−𝑛 convergent or divergent?
𝑛=1
Theorem: If the series ෍ 𝑎𝑛 and ෍ 𝑏𝑛 are convergent, then so are the series ෍ 𝑐𝑎𝑛
(where c is a constant), ෍ 𝑎𝑛 + 𝑏𝑛 , and ෍ 𝑎𝑛 − 𝑏𝑛 , and
෍ 𝑐𝑎𝑛 = 𝑐 ෍ 𝑎𝑛
෍ 𝑎𝑛 ± 𝑏𝑛 = ෍ 𝑎𝑛 ± ෍ 𝑏𝑛
8
Series
Theorem: If the series ෍ 𝑎𝑛 is convergent, then lim 𝑎𝑛 = 0 .
𝑛→∞
Test for Divergence: If lim 𝑎𝑛 does not exist or lim 𝑎𝑛 ≠ 0 , then the series
𝑛→∞
𝑛→∞
෍ 𝑎𝑛 is divergent.
Important Remark: The test for divergence cannot be used to conclude that a series
is convergent. It simply means that if lim 𝑎𝑛 ≠ 0 or if the limit does not exist, the series
𝑛→∞
is divergent and if lim 𝑎𝑛 = 0 , this test is inconclusive and other methods must be used
𝑛→∞
to conclude that the series is convergent or divergent.
9
Series
∞
𝑛2 + 1
Example: Show that the series෎ 2
is divergent.
2𝑛 + 3
𝑛=1
10
Series
Telescoping series: A telescoping series is a series in which most of the terms
cancel in each of the partial sums, leaving only some of the last terms.
11
Series
Example: Determine if the following series is convergent or divergent. If it
converges, find its sum.
12
Series
Additional Examples
Determine if the geometric series is convergent or divergent.
∞
∞
5
෎ 𝑛
2
1
෎
4
𝑛=2
𝑛=0
𝑛
56−𝑛
10 − 2 + 0.4 − 0.08+. . .
Determine if the series is divergent, or state that the divergence test is inconclusive.
∞
∞
෍ arctan (𝑛ሻ
෍
𝑛=1
𝑛=1
𝑛
𝑛2 + 1
Determine if the series is convergent or divergent. If it is convergent, find its sum.
∞
෎
𝑛=2
2
𝑛2 − 1
∞
෍ ln
𝑛=1
𝑛
𝑛+1
13
MTH240- Calculus II
Summer 2025
Section 5.3 – The Integral Test for Series
Instructor: Saeid Samiezadeh
Email: saeid.samiezadeh@torontomu.ca
Office: VIC 748
1
The Integral Test
Consider the following series
There is no simple formula for sn. But the sum
can be compared with an integral.
2
The Integral Test
Now consider the following series
3
The Integral Test
∞
In the series ෍ 𝑎𝑛 where 𝑎𝑛 = 𝑓(𝑛ሻ and f is a continuous, positive, and
𝑛=1
decreasing function on 1, ∞ , the series is convergent if and only if the improper
∞
integralන 𝑓(𝑥ሻ𝑑𝑥 is convergent.
1
In other words:
∞
∞
(i) If න 𝑓(𝑥ሻ𝑑𝑥 is convergent, then ෍ 𝑎𝑛 is convergent.
1
∞
𝑛=1
∞
(ii) If න 𝑓(𝑥ሻ𝑑𝑥 is divergent, then ෍ 𝑎𝑛 is divergent.
1
𝑛=1
4
The Integral Test
∞
Example: Test ෎
1
for convergence or divergence.
2
𝑛 +3
𝑛=1
Solution: The function 𝑓(𝑥ሻ =
1
Since 2
𝑥 +3
′
1
is continuous, positive, and decreasing
𝑥2 + 3
−2𝑥
= 2
<0.
2
𝑥 +3
𝑡
1
1
𝑑𝑥
න 2
𝑑𝑥 = lim න 2
𝑡→∞ 1 𝑥 + 3
𝑥
+
3
1
∞
5
The Integral Test
Remarks
In using Integral Test:
1. The series does not have to start at n=1. It can start from any positive integer.
However, this positive number would be the lower limit for the improper
integral.
∞
1
Example: ෎
𝑛−1 2
∞
න
4
1
𝑑𝑥
𝑥−1 2
𝑛=4
2. The series does not have to be always decreasing. What matters is that f be
eventually decreasing. This means that f could be decreasing for x larger than N.
6
The Integral Test
Remarks (cont’d)
In using integral test:
3. The value (sum) of the series is not necessarily equal to the value of the
corresponding integral.
∞
∞
෍ 𝑎𝑛 ≠ න 𝑓(𝑥ሻ𝑑𝑥
𝑛=1
1
7
The Integral Test
∞
Example: For what values of p is the series ෎
1
convergent?
𝑝
𝑛
𝑛=1
1
1
Solution: If p < 0, then the limit lim 𝑝 = ∞. If p = 0, then lim 𝑝 = 1
𝑛→∞ 𝑛
𝑛→∞ 𝑛
In either case, the limit is not zero and the series is divergent by divergence test.
If p > 0, then the function 𝑓(𝑥ሻ =
1
is continuous, positive, and decreasing.
𝑥𝑝
We showed in chapter 7 that (see Lecture Slide 5).
∞
1
න 𝑝 𝑑𝑥 is convergent if 𝑝 > 1 and divergent if 𝑝 ≤ 1
1 𝑥
So the series ෍
∞
1
𝑝
𝑛=1 𝑛
is convergent for 𝑝 > 1 by the integral test.
8
The Integral Test
P-series
∞
The series ෎
1
is called p-series and is convergent for 𝑝 > 1 . The series
𝑛𝑝
𝑛=1
is divergent for 𝑝 ≤ 1 .
Example: Is the series convergent or divergent?
∞
෎
𝑛=1
1
3
𝑛 ൗ5
∞
1
෎ 1.2
𝑛
𝑛=1
9
The Integral Test
Harmonic Series
∞
The series ෎
1
1 1 1 1
= 1 + + + + +. . . is called harmonic series and is
𝑛
2 3 4 5
𝑛=1
divergent since it is a p-series with 𝑝 ≤ 1 .
10
The Integral Test
Additional Examples
Determine if the series is convergent or divergent.
∞
෍ 𝑛2 𝑒 −𝑛
3
𝑛=1
∞
∞
1 1
1
1
1+ +
+
+
+. . .
8 27 64 125
෍ 3−ln n
𝑛=1
∞
1
෎
𝑛 ln𝑛 2
෎
𝑛=2
𝑛=1
1
𝑛2 + 6𝑛 + 13
∞
cos 2 𝑛
Explain why the Integral Test cannot be used on ෎
1 + 𝑛2
𝑛=1
Find the values of p for which the series is convergent.
∞
෎
𝑛=1
ln𝑛
𝑛𝑝
11
MTH240- Calculus II
Summer 2025
Section 5.4 – The Comparison Tests for Series
Instructor: Saeid Samiezadeh
Email: saeid.samiezadeh@torontomu.ca
Office: VIC 748
1
The Comparison Tests
The Direct Comparison Test
Suppose that ෍ 𝑎𝑛 and ෍ 𝑏𝑛 are series with positive terms.
(i) If ෍ 𝑏𝑛 is convergent and 𝑎𝑛 ≤ 𝑏𝑛 for n > N, then ෍ 𝑎𝑛 is also convergent.
(ii) If ෍ 𝑎𝑛 is divergent and 𝑎𝑛 ≤ 𝑏𝑛 for n > N, then ෍ 𝑏𝑛 is also divergent.
where N is a positive integer.
2
The Comparison Tests
∞
1
Example: Determine if the series ෎ 2
converges or diverges.
𝑛 + 10
𝑛=1
3
The Comparison Tests
Solution:
𝑛2 + 10 > 𝑛2
∞
1
2
𝑛=1 𝑛
෍
1
1
⇒ 2
<
𝑛 + 10 𝑛2
∞
⇒ ෍
1
2
𝑛=1 𝑛 +10
Converges by the direct
comparison test
converges (p-series series with p=2 > 1)
4
The Comparison Tests
∞
Example: Determine if the series ෎
ln𝑛
converges or diverges.
𝑛3
𝑛=1
5
The Comparison Tests
∞
Example: Determine if the series ෎
1
𝑛−3
converges or diverges.
𝑛=4
6
The Comparison Tests
The Limit Comparison Test
𝑎𝑛
Suppose that ෍ 𝑎𝑛 and ෍ 𝑏𝑛 are series with positive terms. If lim
=𝑐
𝑛→∞ 𝑏𝑛
where c is a finite and positive number, then either both series converge or both diverge.
7
The Comparison Tests
Example: Determine if the series ෍
∞
1
converges or diverges.
𝑛+3
𝑛=1
8
The Comparison Tests
∞
𝑛2 −2𝑛+1
Example: Determine if the series ෍
3
𝑛=1 3𝑛 +𝑛−7
converges or diverges.
9
The Comparison Tests
Example: Determine if the series ෍
∞
𝑛=1
1
2𝑛 +1
converges or diverges.
10
The Comparison Tests
∞
𝑛4 − 2𝑛2 + 3
Example: Determine if the series ෎
converges or diverges.
2𝑛6 − 𝑛 + 5
𝑛=1
11
The Comparison Tests
12
The Comparison Tests
Example: Determine if the series following converges or diverges.
∞
෎
ln𝑛
𝑛2
𝑛=1
13
The Comparison Tests
Solution (cont’d):
14
The Comparison Tests
Additional Examples
Determine if the series converges or diverges.
∞
∞
෎
𝑘sin2 𝑘
෎
1 + 𝑘3
1
𝑘 3Τ2 + 1
𝑘=1
𝑘=1
∞
∞
sin(1Τ𝑛ሻ
෎
𝑛3
𝑛 + 4𝑛
෎
𝑛 + 6𝑛
𝑛=1
𝑛=1
∞
1
෎ 1+
𝑛
𝑛=1
∞
2
3
𝑒 −𝑛
෎
𝑛
𝑛3 + 4𝑛 + 3
𝑛=1
15
MTH240- Calculus II
summer 2025
Section 5.5 – Alternating Series
Instructor: Saeid Samiezadeh
Email: saeid.samiezadeh@torontomu.ca
Office: VIC 748
1
Alternating Series
Definition
An alternating series is a series whose terms are alternately positive and negative.
∞
Example:
1−
1 1 1 1 1
+ − + − +. . .
2 3 4 5 6
෎ −1 𝑛−1
1
𝑛
𝑛=1
Convergence Test for Alternating Series
∞
The alternating series ෍ −1 𝑛−1 𝑏𝑛 = 𝑏1 − 𝑏2 + 𝑏3 − 𝑏4 + 𝑏5 −. . . is convergent
𝑛=1
if
(i) 𝑏𝑛+1 ≤ 𝑏𝑛
(i.e. is non-increasing) for
𝑛≥𝑁
(ii) lim 𝑏𝑛 = 0
𝑛→∞
2
Alternating Series
Convergence Test for Alternating Series (cont’d)
3
Alternating Series
∞
Example: Test the series ෎ −1 𝑛−1
1
for convergence or divergence.
𝑛
𝑛=1
4
Alternating Series
Solution (cont’d):
5
Alternating Series
∞
Example: Test the series ෎ −1 𝑛+1
3n
for convergence or divergence.
3𝑛 + 2
𝑛=1
6
Alternating Series
∞
Example: Test the series ෍ −1 𝑛 tan−1 (𝑛) for convergence or divergence.
𝑛=1
7
Alternating Series
Alternating Series Estimation Theorem
∞
If 𝑠 = ෍ −1 𝑛−1 𝑏𝑛 is the sum of an alternating series which is convergent, and
𝑛=1
𝑠𝑛 represents its partial sum then
|𝑅𝑛 | = |𝑠 − 𝑠𝑛 | ≤ 𝑏𝑛+1
where 𝑅𝑛 is the remainder.
8
Alternating Series
∞
−1 𝑛
Example: How many terms of the series෎
do we need to add in order to
𝑛6
𝑛=1
find the sum correct to 4 decimal places?
9
Alternating Series
Solution (cont’d):
Thus, by Alternating Series Estimation theorem, we should have 0.0001 ≤ 𝑏𝑛+1
𝑏𝑛 =
1
𝑛6
𝑏3 = 0.00137 > 0.0001
𝑏4 = 0.00024 > 0.0001
𝑏5 = 0.000064 < 0.0001
𝑛+1= 5 ⇒𝑛 =4
That is, since the 5th term is less than the desired error (0.0001), we need to add
the first 4 terms to get the sum to the desired accuracy.
10
Alternating Series
Additional Examples: Test the series for convergence or divergence.
∞
෎ −1 𝑛
∞
𝑛
𝑛3 + 2
𝑛=1
𝑛=1
cos 𝑛𝜋
𝑛2
𝑛=1
∞
෎ −1 𝑛 𝑛 sin
෎
1
𝑛
∞
෍ −1 𝑛 ( 𝑛 + 1 − 𝑛൯
𝑛=1
11
MTH240- Calculus II
Summer 2025
Section 5.6 – Absolute Convergence and Ratio and
Root Tests
Instructor: Saeid Samiezadeh
Email: saeid.samiezadeh@torontomu.ca
Office: VIC 748
1
Absolute Convergence
Definition
A series ෍ 𝑎𝑛 is called absolutely convergent if the series of absolute values ෍ |𝑎𝑛 |
is convergent.
∞
Example: The series ෎ −1 𝑛
1
is absolutely convergent, because
2
𝑛
𝑛=1
∞
෎ | −1 𝑛
𝑛=1
∞
1
1
| = ෎ 2 is convergent (p-series with p = 2 > 1)
𝑛2
𝑛
𝑛=1
2
Absolute Convergence
∞
Example: Show that ෎ −1 𝑛
1
is convergent but not absolutely convergent,
𝑛
𝑛=1
3
Absolute Convergence
Definition
If a series is convergent but not absolutely convergent, it is conditionally convergent.
∞
Example: So based on the above-mentioned definition෎ −1 𝑛
convergent.
1
is conditionally
𝑛
𝑛=1
4
Absolute Convergence
Theorem: If a series ෍ 𝑎𝑛 is absolutely convergent, then it is convergent.
∞
Example: Determine if the series ෎
cos (𝑛)
is convergent or divergent.
𝑛2
𝑛=1
5
Absolute Convergence
6
The Ratio Test
7
The Ratio Test
∞
2𝑛
Example: Use the Ratio Test to determine if the series෎
converges.
𝑛!
𝑛=1
𝑎𝑛+1
|
𝑛→∞ 𝑎𝑛
lim |
8
The Ratio Test
Example: Use the Ratio Test to determine when the p-series converges.
9
The Root Test
10
The Root Test
∞
4𝑛2 − 3
Example: Use the Root Test to determine if the series෎
7𝑛2 + 6
𝑛
converges.
𝑛=1
lim
𝑛→∞
𝑛
|𝑎𝑛 |
11
The Root Test
Remarks on the Ratio and Root Tests
• The Ratio and Root Tests come with an absolute value, so they can be used on
alternating series and would determine if these series are absolutely convergent
or divergent.
• If the Ratio test is inconclusive, then so is the Root test and vice versa.
12
Absolute Convergence and
Ratio and Root Tests
Additional examples: Determine if the series is absolutely convergent,
conditionally convergent or divergent.
∞
∞
∞
−1 𝑛
−2 𝑛
෎
𝑛𝑛
−3 𝑛
෎
(2𝑛 + 1)!
෎
𝑛=0
𝑛=0
𝑛=1
∞
∞
∞
෎
𝑛=1
𝑛𝑛
𝑛!
5𝑛 + 1
𝑛!
෎
100𝑛
−2𝑛
෎
𝑛+1
𝑛=1
𝑛=2
5𝑛
13
This document is intended to summarize various tests used to determine convergence/divergence of series
in Calculus II (MTH 240). It CANNOT be used in anyway as an aid in closed-book tests/quizzes.
Series or Test
Form of Series
Condition for
Divergence
 ar a  0
n
r 1
r 1
Does not apply
lim an  0
n 0

Divergence Test
Comments


Geometric Series
Condition for
Convergence
a
n 1
n
n 
a
1 r
n0
Useful for comparison tests
If convergent,  ar n 
Cannot be used to prove
convergence

 a where
n
n 1
Integral Test
1
an  f (n) and f is
continuous, positive, converges
and decreasing

P-series

1
n
n 1
p
f ( x)dx

 f ( x)dx
1
diverges
The value of the integral is not
the value of the series.
p 1
p 1
Useful for comparison tests
lim sn  
lim sn  
The sum of the series can be
found by lim sn .

 a  b whose
n 1
Telescoping
Series
n
n
partial sums ( sn )
only have a fixed
number of terms
after cancellation

Direct
Comparison Test
 a where a  0
n 1
n
n
and bn  0
Alternating
Series Test
 a where a  0
n 1
n
n
and bn  0

 (1) b
n
n
n 1
an  bn and

b
n 1

Limit
Comparison Test
n 
n 
n 
bn  an and

n
converges
a
0  lim n  
n  b
n

 bn diverges
n 1
an

n  b
n
0  lim

and  bn
and  bn
converges
diverges
bn is nonincreasing and
lim bn  0
Does not
apply
n 1
n 1
n 

Ratio Test
a
n 1
a
n 1
Absolute
Convergence
By Saeid Samiezadeh
n 1
n 1
n 1
The test determines absolute
convergence.


 a is given; you supply  b
n 1
n
n 1
The test determines absolute
convergence.
If lim bn  0 , divergence test
n 
shall be used.
lim
an 1
1
n  a
n
lim
an 1
1
n  a
n
Inconclusive if lim
n
lim n an  1
lim n an  1
Inconclusive if lim n an  1
Does not
apply
Applies to arbitrary series
n 
n 
an 1
1
n  a
n
n 


a

 an is given; you supply  bn
n

Root Test

n
a
n 1
n
converges
n
MTH240- Calculus II
Summer 2025
Section 6.1 – Power Series
Instructor: Saeid Samiezadeh
Email: saeid.samiezadeh@torontomu.ca
Office: VIC 748
1
Power Series
Definition: A power series is of the form
∞
෍ 𝑐𝑛 𝑥 𝑛 = 𝑐0 + 𝑐1 𝑥 + 𝑐2 𝑥 2 + 𝑐3 𝑥 3 +. . .
𝑛=0
where x is a variable and the 𝒄𝒏 ’s are constants called the coefficients of the series.
* The domain of the series is the set of all x for which the series converges.
Example: If 𝑐𝑛 = 1 for all n, the power series becomes
∞
෍ 𝑥 𝑛 = 1 + 𝑥 + 𝑥 2 + 𝑥 3 +. . . +𝑥 𝑛 +. . .
𝑛=0
* Comparing the series with a geometric series, it is convergent when |𝑥| < 1.
2
Power Series
Definition: A more general case is the series of the form
∞
෍ 𝑐𝑛 𝑥 − 𝑎 𝑛 = 𝑐0 + 𝑐1 (𝑥 − 𝑎) + 𝑐2 𝑥 − 𝑎 2 + 𝑐3 𝑥 − 𝑎 3 +. . . +𝑐𝑛 𝑥 − 𝑎 𝑛 +. . .
𝑛=0
which is called a power series in 𝑥 − 𝑎 or a power series centered at 𝑎 or a
power series about 𝑎.
Definition: The set of values of x for which the series converges is its interval of
convergence. The radius of convergence of the power series, denoted by R, is the
distance from the center of the series to the boundary of the interval of convergence.
Remark: Different tests for series could be used to determine interval and radius of
convergence in a power series. The common tests are the Ratio and Root tests.
3
Power Series
Example: Find the interval and radius of convergence for ෍
∞
𝑥𝑛
.
𝑛
𝑛=0
4
Power Series
Solution (cont’d):
5
Power Series
∞
𝑛 𝑥+2 𝑛
Example: Find the interval and radius of convergence for ෎
3𝑛+1
𝑛=0
Solution: Using the Ratio test
6
Power Series
7
Power Series
∞
Theorem: For a given power series ෍ 𝑐𝑛 𝑥 − 𝑎 𝑛 there are only three possibilities:
𝑛=0
(i) The series converges only when 𝑥 = 𝑎.
(ii) The series converges for all 𝑥.
(iii) There is a positive number such that the series converges if |𝑥 − 𝑎| < 𝑅
and diverges if 𝑥 − 𝑎 > 𝑅 .
8
Power Series
∞
Example: Find the interval and radius of convergence for ෍ 𝑛! 𝑥 𝑛
𝑛=1
Solution: Using the Ratio test
The only way to satisfy r < 1 is to take 𝑥 = 0, in which case the power series has a
value of 0. Thus, 𝑅 = 0 and 𝐼 = 0 .
9
Power Series
Additional Examples: Find the interval and radius of convergence for the series.
∞
∞
𝑛2 𝑥 2𝑛
෎
𝑛!
෎ −1 𝑛
𝑛=1
𝑛=2
∞
∞
𝑛
෍ 𝑛 𝑥+1 𝑛
4
𝑥𝑛
4𝑛 ln𝑛
෍ 𝑛! 2𝑥 − 1 𝑛
𝑛=1
𝑛=1
∞
∞
1
෎ sin𝑛 ( )𝑥 𝑛
𝑛
2𝑥 − 1 𝑛
෎
5𝑛 𝑛
𝑛=1
𝑛=1
10
Representations of
Functions as Power Series
• It is possible to represent certain types of functions as sums of power series by
manipulating geometric series or by differentiating or integrating such series.
∞
Example: The series ෍ 𝑥 𝑛 = 1 + 𝑥 + 𝑥 2 + 𝑥 3 +. . . +𝑥 𝑛 +. . . has a sum of
𝑛=0
1
1−𝑥
when |𝑥| < 1. Consequently,
∞
1
= ෍ 𝑥 𝑛 = 1 + 𝑥 + 𝑥 2 + 𝑥 3 +. . . +𝑥 𝑛 +. . .
1−𝑥
𝑛=0
11
Representations of
Functions as Power Series
• There are many important applications for such representation, such as
integrating functions that do not have elementary antiderivatives, solving
differential equations, and approximating functions.
Manipulating a Function: We can manipulate a function to be able to write it as
a power series.
12
Representations of
Functions as Power Series
Example: Express
1
as the sum of a power series and find the interval of
2
1+𝑥
convergent.
13
Representations of
Functions as Power Series
14
Representations of
Functions as Power Series
Example: Find a power series representation for
convergence.
1
and find the interval of
𝑥+3
15
Representations of
Functions as Power Series
Differentiation and Integration of Power Series
∞
Theorem: If the power series ෍ 𝑐𝑛 𝑥 − 𝑎 𝑛 has the radius of convergence R > 0,
then the function
𝑛=0
∞
𝑓(𝑥) = 𝑐0 + 𝑐1 (𝑥 − 𝑎) + 𝑐2 𝑥 − 𝑎 2 +. . . = ෍ 𝑐𝑛 𝑥 − 𝑎 𝑛
𝑛=0
is differentiable (and therefore continuous) on the interval 𝑎 − 𝑅, 𝑎 + 𝑅 and
∞
(i) 𝑓 ′ (𝑥) = 𝑐1 + 2𝑐2 (𝑥 − 𝑎) + 3𝑐3 𝑥 − 𝑎 2 +. . . = ෍ 𝑛𝑐𝑛 𝑥 − 𝑎 𝑛−1
𝑛=1
∞
𝑥−𝑎 2
𝑥−𝑎 3
𝑥 − 𝑎 𝑛+1
(ii) න𝑓(𝑥 ) 𝑑𝑥 = 𝑐0 𝑥 + 𝑐1
+ 𝑐2
+. . . = 𝐶 + ෎ 𝑐𝑛
2
3
𝑛+1
𝑛=0
The radii of convergence of the power series in (i) and (ii) are both R.
17
Representations of
Functions as Power Series
16
Representations of
Functions as Power Series
Example: Find a power series representation for
convergence.
1
and find the radius of
2
𝑥−2
18
Representations of
Functions as Power Series
Example: Find a power series representation for ln(1 − 𝑥 ) and find the radius of
convergence.
19
Representations of
Functions as Power Series
Example: Use power series representation to evaluate න
1
𝑑𝑥 .
5
1+𝑥
20
Representations of
Functions as Power Series
Additional Examples: Find a power series representation for the function and
determine the radius of convergence.
𝑥5
7 − 9𝑥 3
1+𝑥
1−𝑥
tan−1 𝑥
𝑥+2
2𝑥 2 − 𝑥 − 1
𝑥+2
1−𝑥 3
21
MTH240- Calculus II
Summer 2025
Section 6.2 – Properties of Power Series
Instructor: Saeid Samiezadeh
Email: saeid.samiezadeh@torontomu.ca
Office: VIC 748
1
Combining Power Series
Some combining techniques such as partial fraction decomposition, differentiation
and integration of power series were covered in the previous lesson.
2
Combining Power Series
Example:
3
Combining Power Series
Example:
4
Multiplication of Power Series
5
Multiplication of Power Series
6
Multiplication of Power Series
Example:
7
Multiplication of Power Series
8
Multiplication of Power Series
Solution:
9
MTH240- Calculus II
Summer 2025
Section 6.3 – Taylor and Maclaurin Series
Section 6.4 – Working with Taylor Series
Instructor: Saeid Samiezadeh
Email: saeid.samiezadeh@torontomu.ca
Office: VIC 748
1
Taylor & Maclaurin Series
• We have learned to write certain functions as a power series. For instance,
∞
1
= 1 + 𝑥 + 𝑥 2 + 𝑥 3 + 𝑥 4 +. . . = ෍ 𝑥 𝑛
1−𝑥
|𝑥| < 1
𝑛=0
• We now try to answer the following questions:
1. Does every function have a power series representation?
2. When is a function equal to its power series representation?
3. What is the error in approximating a function with a power series?
2
Taylor & Maclaurin Series
For special case a = 0, the Taylor series becomes
𝑓 ′ (0)
𝑓 ″ (0) 2 𝑓 ‴ (0) 3 𝑓 4 (0) 4
𝑓 𝑛 (0) 𝑛
𝑓(𝑥) = ෍
𝑥 = 𝑓(0) +
𝑥+
𝑥 +
𝑥 +
𝑥 +. . .
1!
2!
3!
4!
𝑛=0 𝑛!
∞
This case series is called Maclaurin series.
Example: Find the Maclaurin series of the function 𝑓(𝑥) = 𝑒 𝑥 and its radius of
convergence.
3
Taylor & Maclaurin Series
Solution: If 𝑓(𝑥) = 𝑒 𝑥 , then 𝑓 𝑛 (𝑥) = 𝑒 𝑥 , so 𝑓 𝑛 0 = 𝑒 0 = 1 for all n. Thus,
∞
∞
𝑓 𝑛 (0) 𝑛
𝑥𝑛
𝑥 𝑥2 𝑥3
෎
𝑥 =෎
= 1 + + + +. . .
𝑛!
𝑛!
1! 2! 3!
𝑛=0
𝑛=0
4
Taylor & Maclaurin Series
Solution (cont’d): To find the radius of convergence, we use the Ratio test as follows
|𝑥|
𝑥 𝑛+1 𝑛!
𝑎𝑛+1
→ 0< 1
| = lim
lim |
| = lim |
𝑛
𝑛→∞ 𝑛 + 1
𝑛→∞ (𝑛 + 1)! 𝑥
𝑛→∞ 𝑎𝑛
so, by the Ratio Test, the series converges for all values of x.
Definition: The nth-degree Taylor polynomial of f at a is defined as
𝑛
𝑓 𝑖 (𝑎)
𝑃𝑛 (𝑥) = ෎
𝑥−𝑎 𝑖
𝑖!
𝑖=0
5
Taylor & Maclaurin Series
Example: Write the first three Taylor polynomials of the function 𝑓(𝑥) = 𝑒 𝑥 at 0.
Solution:
Definition: The remainder is defined as 𝑅𝑛 (𝑥) = 𝑓(𝑥) − 𝑃𝑛 (𝑥 ) .
6
Taylor & Maclaurin Series
Taylor’s Inequality: If |𝑓 𝑛+1 (𝑥 ൯| ≤ 𝑀 for |𝑥 − 𝑎| < 𝑑 , then the remainder 𝑅𝑛 (𝑥 )
of the Taylor series satisfies the inequality
|𝑅𝑛 (𝑥 )| ≤
𝑀
|𝑥 − 𝑎|𝑛+1
(𝑛 + 1)!
for all x in the interval of convergence
7
Taylor & Maclaurin Series
Example: Find the Taylor series for 𝑓(𝑥) = 𝑒 𝑥 at a = 2.
8
Taylor & Maclaurin Series
Solution (cont’d):
9
Taylor & Maclaurin Series
Example (you need a calculator for this example!)
10
Taylor & Maclaurin Series
11
Taylor & Maclaurin Series
Example: Find the Maclaurin series for sin x.
12
Taylor & Maclaurin Series
Example: Find the Maclaurin series for sin x and prove that it represents sin x for
all x.
Solutions:
𝑓(𝑥) = sin𝑥
𝑓(0) = 0
𝑓 ′ (𝑥) = cos𝑥
𝑓 ′ (0) = 1
𝑓 ″ (𝑥) = −sin𝑥
𝑓 ″ (0) = 0
𝑓 ‴ (𝑥) = −cos𝑥
𝑓 ‴ (0) = −1
𝑓 4 (𝑥) = sin𝑥
𝑓 4 (0) = 0
Since the derivatives repeat in a cycle of four, we can write the Maclaurin series as
follow:
𝑥3 𝑥5 𝑥7
𝑓 ′ (0)
𝑓 ″ (0) 2 𝑓 ‴ (0) 3
− +. . .
𝑓(0) +
𝑥+
𝑥 +
𝑥 +. . . = 𝑥 − +
3! 5! 7!
1!
2!
3!
∞
𝑥 2𝑛+1
𝑛
=෍
−1 (
2𝑛+1)!
𝑛=0
for all x
13
Taylor & Maclaurin Series
Some important Maclaurin series
14
Taylor & Maclaurin Series
The traditional notation for the coefficients in the binomial series is
𝑘(𝑘 − 1)(𝑘 − 2)(𝑘 − 3). . . (𝑘 − 𝑛 + 1)
𝑘
=
𝑛
𝑛!
,
𝑘
=1
0
and these numbers are called the binomial coefficients.
Example:
5)(4)(3
5
= 10
=
3
3!
1ൗ
2
5
15
Taylor & Maclaurin Series
Convergence of Binomial Series
∞
The binomial series 1 + 𝑥 𝑘 = ෍
𝑛=0
𝑘 𝑛
𝑥 converges when |𝑥| < 1 .
𝑛
16
Taylor & Maclaurin Series
Example: Find the Maclaurin series for the function 𝑓(𝑥) =
of convergence.
1
4−𝑥
and its radius
∞
Solution: Comparing the function with the binomial series 1 + 𝑥 𝑘 = ෍ 𝑘 𝑥 𝑛
𝑛
𝑛=0
∞
1
1
1
𝑥 − ൗ2 1
= ෎ − ൗ2
= 1 + (− )
=
2
4
𝑛
4−𝑥
𝑥 2
2 1−4
𝑛=0
1
1
𝑥 𝑛
−
4
17
Taylor & Maclaurin Series
Solution (cont’d):
∞
1
= 2෎
𝑛=0
1
+
=
3
5
−2 −2 −2
3!
− 1Τ2
𝑛
𝑥 𝑛
1
− 4 = 2 [1 +
1
3
1
−2
5
𝑥
−4
1
−2 −2 −2 ... −2−𝑛+1
𝑥 3
−
+. . . +
4
𝑛!
+
1
2
3
2
−
−
2!
𝑥 2
−4
𝑥 𝑛
−
+ ⋯]
4
1
1
1⋅3 2 1⋅3⋅5 3
1 ⋅ 3 ⋅ 5 ⋅. . .⋅ (2𝑛 − 1) 𝑛
1+ 𝑥+
𝑥
+
𝑥
+.
.
.
+
𝑥 +. . .
2
8
2! ⋅ 82
3! ⋅ 83
𝑛! ⋅ 8𝑛
𝑥
The series converges when | − | < 1 , that is, |𝑥| < 4 . So, the radius of
4
convergence is R = 4.
18
Taylor & Maclaurin Series
Example: Find the sum of the series
1
1
1
1
−
+
−
+. . .
1 × 2 2 × 22 3 × 23 4 × 24
19
Taylor & Maclaurin Series
2
Example: Evaluate න𝑒 −𝑥 𝑑𝑥 as an infinite series.
20
Taylor & Maclaurin Series
Additional Example:
Q1. Find the Maclaurin series for the given function.
𝑥cos
1 2
𝑥
2
3
8+𝑥
sin−1 𝑥
𝑒𝑥 − 𝑥 − 1
Q2. Find the limit lim
using Maclaurin series representation.
𝑥→0
𝑥2
Q3. Find the sum of the series ෍
∞
3𝑛
.
𝑛 𝑛!
5
𝑛=0
Q4. Approximate 𝑓(𝑥) = 𝑥ln𝑥 by a Taylor polynomial 𝑃3 (𝑥 ) with degree 3 at a = 1.
Use Taylor's Inequality to estimate the accuracy of the approximation 𝑓(𝑥) = 𝑃3 (𝑥 )
when x lies in the interval 0.5 , 1.5 .
21
Taylor & Maclaurin Series
Some important Maclaurin series
14
Toronto Metropolitan University
MTH 240
Functions of several variables.
Limits andContinuity.
The content for this lecture is in sections 4.1 & 4.2.
TMU
MTH 240
Functions of several variables.
Definition: A function of two variables is a rule that assigns to
each ordered pair of real numbers (x, y ) in a set D a unique real
number denoted by f (x, y ). The set D is the domain of f and its
range is the set of values that f takes on, that is,
{f (x, y )|(x, y ) 2 D}.
Sometimes we write z = f (x, y ). The variables x and y are called
independent variables and z is the dependent variable.
If a function f is given by a formula and no domain is specified ,
then the domain of f is considered to be the set of all pairs (x, y )
for which the given expression is a well defined real number.
Example 1: Find the domain and range of the function
f (x, y ) = xy 2
Example p
2: Find the domain and range of the function
f (x, y) = x + y
Find and sketch the domain of the
Problem 1
function
f (x, y ) =
p
y+
p
25
x2
y2
If f is a function of two variables with domain D then, the graph
of f is the set of all points (x , y , z) 2 R3 such that z = f (x, y )
with (x, y ) 2 D.
Example 3: Graph g(x, y) =
p
9
x2
y2
Example 4: Sketch the graph of f (x, y ) =
2x
y +4
Level Curves
Example:
Limit
Definition: Let f be a function of two variables whose domain D
includes points arbitrarily closed to (a, b) then, we say that the
limit f (x, y ) as (x, y ) approaches (a, b) is L and we write
lim
(x,y )!(a,b)
f (x, y ) = L
if for every number ✏ >
p0 there is a corresponding > 0 such that
if (x, y ) 2 D and 0 < (x a)2 + (y b)2 < then
|f (x, y ) L| < ✏.
For functions of a single variable, the limit exists if both
lateral limits exist and are equal
If the lateral limits are di↵erent then the limit does not exist.
In the two dimensional space, there exists an infinite number
of ways of approaching a point
Theorem: If f (x, y ) ! L1 , as (x, y ) ! (a, b) along path C1 , and
f (x, y ) ! L2 as (x, y ) ! (a, b) along path C2 and L1 6= L2 then
lim
(x,y )!(a,b)
does not exist.
f (x, y )
Example 5: Prove that the limit
2x y
does not exist.
(x,y)!(0,0) 3x + y
lim
Example 6: Does the limit
3yx 2
exist?
(x,y )!(0,0) x 4 + 2y 2
lim
So far we have seen how to prove that the limit does not exist for
some examples.
How to prove that the limit does exist?
If it does exist, how to find it?
Try direct substitution
Example 7: Find the limit
lim
(x,y )!(1, 1)
e xy cos(x + y)
Example 8: Find the limit
x4 y4
(x,y )!(0,0) x 2 + y 2
lim
Continuity
Definition: A function f of two variables is continuous at (a, b) if
lim
(x,y )!(a,b)
f (x, y ) = f (a, b)
we say that f is continuous on D if f is continuous at every point
(a, b) in D.
Sums, di↵erences, products, etc. of continuous functions are
continuous on their domains.
Example 10: f (x, y) = x2 ln y
Example 11: Is function
8 4
y4
< x
f (x, y ) =
x2 + y2
:
0
continuous?
if (x, y ) 6= (0, 0)
if (x, y ) = (0, 0)
Remark: These ideas can be naturally extended to the case of
more variables
Additional examples:
Find the limit if it exists or show that the limit does not exist.
5y 4 cos2 x
Problem 2:
lim
(x,y )!(0,0) x 4 + y 4
Problem 3:
Problem 4:
x 2 ye y
lim
(x,y )!(0,0) x 4 + 4y 2
lim
(x,y ,z)!(⇡,0,1/3)
2
e y tan(xz)
Determine the set of points at which the function is continuous:
p
Problem 5:
F (x, y ) = 1 + x y
ex + ey
Problem 6:
H(x, y ) = xy 1
e
Problem 7:
(
xy
if (x, y ) 6= (0, 0)
2
x + xy + y 2
f (x, y ) =
0
if (x, y ) = (0, 0)
Toronto Metropolitan University
MTH 240
Sec 4.3:
Partial derivatives
TMU
MTH 240
Partial derivatives
If f is a function of two variables, its Partial Derivatives are the
functions fx and fy defined by
f (x + h, y )
h!0
h
fx (x, y ) = lim
f (x, y )
(x, y + h) f (x, y )
h!0
h
fy (x, y ) = lim
There are di↵erent notations fx (x, y ) = fx =
@f
@
=
f (x, y )
@x
@x
The notation f 0 is NEVER USED for partial derivatives
The previous definition has limited practical value.
To find partial derivatives with respect to one variable we regard
the other variables as constants.
Example 1: f (x, y ) = x 2 sin y .
Example 2: f (x, y , z) = 3yx 2 ln z + 2y 2 z 3
Partial derivatives of higher order
Second derivatives are derivatives of the first partial derivatives:
(fx )x = fxx =
@2f
@x 2
(fx )y = fxy =
@2f
@y @x
(fy )x = fyx =
@2f
@x@y
(fy )y = fyy =
@2f
@y 2
The property that mixed partial derivatives are equal is true for
most functions that one meet in practice:
Clairaut’s Theorem:If f is defined on a disk D that contains the
point (a, b). If the functions fxy and fyx are both continuous on D,
then fxy (a, b) = fyx (a, b).
Additional examples:
Find the first partial derivatives of the function:
Problem 1,
u(r .✓) = sin(r cos ✓)
y /z
Problem 2,
u =x
Problem 3,
Find the indicated partial derivative:
f (x, y ) = arctan(y /x): fx (2, 3)
Problem 4,
Use implicit di↵erentiation to find
@z/@x and @z/@y : x 2 y 2 + z 2 2z = 4
Problem 5,
z = f (x/y ).
Problem 6,
y
v =e xe
Find @z/@x and @z/@y :
Find all the second partial derivatives
Toronto Metropolitan University
MTH 240
Sec 4.5: Chain rule
TMU
MTH 240
Chain Rule for functions of a single variable
[f (g (x))]0 = f 0 (g (x)) · g 0 (x) =
df dg
·
dg dx
For functions of several variables we have analogous results. We
will present first some particular cases and finally the general
formula.
Case 1
z = f (x, y ) is di↵erentiable function of x and y where
x = g (t), y = h(t) are both di↵erentiable functions of t, then
dz
@z dx
@z dy
=
·
+
·
dt
@x dt
@y dt
Example 1: Find
dz
if z = cos(x + 4y ), x = 5t 4 , y = 1/t
dt
Case 2
Suppose z = f (x, y ) is di↵erentiable function of x and y where
x = g (s, t) and y = h(s, t) are di↵erentiable functions of s and t.
Then,
@z
@z @x
@z @y
=
·
+
·
@s
@x @s
@y @s
@z
@z @x
@z @y
=
·
+
·
@t
@x @t
@y @t
@z
@z
Example 2 Problem 10, Section 14.2. Find
and
if
@s
@t
s
t
z = e x+2y with x = and y = .
t
s
Chain Rule (General Version)
Suppose that u is a di↵erentiable function of the variables
x1 , x2 , ...xn and each xj is a di↵erentiable function of the m
variables t1 , t2 , ...tm then u is a function of t1 , t2 , ...tm and
@u
@u @x1
@u @x2
@u @xn
=
·
+
·
+ ...
·
@ti
@x1 @ti
@x2 @ti
@xn @ti
for each i = 1, ...m
Example 3: Problem 5, Section 14.5. Find
w = xe y /z and x = t 2 , y = 1
dw
if
dt
t, z = 1 + 2t
Higher derivatives
Example 4: For the function z = e x+2y , with x =
find
@2z
@s 2
s
t
and y = ,
t
s
Implicit Function Theorem
Suppose z is defined implicitly as a function of x and y by the
expression F (x, y , z) = 0. Then we have
@z
=
@x
@F
@x
@F
@z
@z
=
@y
@F
@y
@F
@z
and
provided
@F
6= 0.
@z
Additional examples:
Problem 1. Use the chain rule to find @z/@s
and@z/@t
p
s = e r cos ✓, r = st, ✓ = s 2 + t 2
Problem 2. Use a tree diagram to write out the chainrule for the
given case. Assume all functions are di↵erentiable.
u = f (x, y ), where x = x(r , s, t), y = y (r , s, t)
Problem 3. Use the chain rule to find the indicated partial
derivatives:
@u/@↵, @u/@
u = xe ty , x = ↵2 , y =
2
and @u/@ when ↵ =
1,
,t=
= 2,
2↵
=1
Additional examples(cont):
Problem 4. Use implicit differentiation to find @z/@x and
@z/@y :
e z = xyz
Problem 5.
y = e s sin t, show that
✓
@u
@x
◆2
+
✓
@u
@y
If u = f (x, y ) where x = e s cos t and
◆2
= e 2s
"✓
@u
@s
◆2
+
✓
@u
@t
◆2 #
Toronto Metropolitan University
MTH 240
Sec 4.7
: Maximum and minimum values
TMU
MTH 240
Definition: A function of two variables has a local maximum at
(a, b) if f (x, y )  f (a, b) when (x, y ) is near (a, b). The number
f (a, b) is called a local maximum value.
If f (x, y )  f (a, b) for all points in the domain of f , then f has an
absolute maximum at (a, b).
The definition for the minimum is analogous:
Definition: A function of two variables has a local minimum at
(a, b) if f (x, y ) f (a, b) when (x, y ) is near (a, b). The number
f (a, b) is called a local minimum value.
If f (x, y ) f (a, b) for all points in the domain of f , then f has an
absolute minimum at (a, b).
Theorem: If f (x, y ) has a local maximum or minimum at (a, b)
and the first-order partial derivatives of f exist there then:
fx (a, b) = 0 and fy (a, b) = 0
Example 1: f (x, y ) = x 2 + 2y 2 + 6x
4y + 3
Remark: A point (a, b) is called a critical point of f if both partial
derivatives are equal to 0, or if at least one of these partial
derivatives does not exist.
The second derivative test
Suppose the second partial derivatives of f (x, y ) are continuous on
a disk with center (a, b) and suppose that fx (a, b) = 0 and
fy (a, b) = 0 [(a, b) is a critical point].
Consider D = D(a, b) = fxx (a, b)fyy (a, b)
[fxy (a, b)]2
a) If D > 0 and fxx (a, b) > 0 then f (a, b) is a local minimum.
b) If D > 0 and fxx (a, b) < 0 then f (a, b) is a local maximum.
c) If D < 0 then f (a, b) is not a local maximum or minimum.
Example 1 (cont.)
fxx = 2
fyy = 4
fxy = 0
Some remarks:
i) Case (c): Saddle point.
ii) If D = 0 the test gives no information.
iii) D can be written as D =
fxx
fyx
fxy
fyy
2
Example 2: f (x, y) = e4y x
y2
Example 2 (cont.) Second derivative test:
Absolute Maximum and Minimum values
Theorem: If f is continuous on a closed bounded set D in R2 ,
then f attains an absolute maximum value f (x1 , y1 ) and an
absolute minimum value f (x2 , y2 ) at some points (x1 , y1 ) and
(x2 , y2 ) in D.
To find the extreme values of a continuous function f on a closed
bounded set D.
a) Find critical points of f in D.
b) Find the extremes of f on the boundary of D.
- The largest (smallest) of the values of f on steps a) and b) is the
absolute maximum (minimum) of f on D.
Additional Examples:
Problem 1. Find the local maximum and minimum values and
saddle point(s) of the function:
f (x, y ) = xy (1
x
y)
Find the absolute maximum and minimum values of f on the set D
Problem 2:
f (x, y ) = 4x + 6y x 2 y 2 ,
D = {(x, y )|0  x  4, 0  y  5}
Problem 3.
2
f (x, y ) = xy ,
D = (x, y )|x
0, y
0x 2 + y 2  3