Harold’s Calculus Notes
Cheat Sheet
24 February 2016
AP Calculus
Limits
Definition of Limit
Let f be a function defined on an open
interval containing c and let L be a real
number. The statement:
lim 𝑓(𝑥) = 𝐿
𝑥→𝑎
means that for each 𝜖 > 0 there exists a
𝛿 > 0 such that
if 0 < |𝑥 − 𝑎| < 𝛿, then |𝑓(𝑥) − 𝐿| < 𝜖
Tip :
Direct substitution: Plug in 𝑓(𝑎) and see if
it provides a legal answer. If so then L =
𝑓(𝑎).
The Existence of a Limit
The limit of 𝑓(𝑥) as 𝑥 approaches a is L if
and only if:
lim 𝑓(𝑥) = 𝐿
𝑥→𝑎 −
lim 𝑓(𝑥) = 𝐿
𝑥→𝑎 +
𝟐
Prove that 𝒇(𝒙) = 𝒙 − 𝟏 is a continuous function.
Definition of Continuity
A function f is continuous at c if for
every 𝜀 > 0 there exists a 𝛿 > 0 such that
|𝑥 − 𝑐| < 𝛿 and |𝑓(𝑥) − 𝑓(𝑐)| < 𝜀.
Tip: Rearrange |𝑓(𝑥) − 𝑓(𝑐)| to have
|(𝑥 − 𝑐)| as a factor. Since |𝑥 − 𝑐| < 𝛿 we
can find an equation that relates both 𝛿
and 𝜀 together.
|𝑓(𝑥) − 𝑓(𝑐)|
= |(𝑥 2 − 1) − (𝑐 2 − 1)|
= |𝑥 2 − 1 − 𝑐 2 + 1|
= |𝑥 2 − 𝑐 2 |
= |(𝑥 + 𝑐)(𝑥 − 𝑐)|
= |(𝑥 + 𝑐)| |(𝑥 − 𝑐)|
Since |(𝑥 + 𝑐)| ≤ |2𝑐|
|𝑓(𝑥) − 𝑓(𝑐)| ≤ |2𝑐||(𝑥 − 𝑐)| < 𝜀
𝟏
So given 𝜀 > 0, we can choose 𝜹 = |𝟐𝒄| 𝜺 > 𝟎 in the
Definition of Continuity. So substituting the chosen 𝛿
for |(𝑥 − 𝑐)| we get:
1
|𝑓(𝑥) − 𝑓(𝑐)| ≤ |2𝑐| (| | 𝜀) = 𝜀
2𝑐
Since both conditions are met, 𝑓(𝑥) is continuous.
𝑠𝑖𝑛 𝑥
=1
𝑥→0 𝑥
𝑙𝑖𝑚
Two Special Trig Limits
1 − 𝑐𝑜𝑠 𝑥
=0
𝑥→0
𝑥
𝑙𝑖𝑚
Copyright © 2015-2016 by Harold Toomey, WyzAnt Tutor
1
Derivatives
Definition of a Derivative of a Function
Slope Function
(See Larson’s 1-pager of common derivatives)
𝑓(𝑥 + ℎ) − 𝑓(𝑥)
𝑓 ′ (𝑥) = lim
ℎ→0
ℎ
Notation for Derivatives
The Constant Rule
The Power Rule
The General Power Rule
The Constant Multiple Rule
The Sum and Difference Rule
Position Function
Velocity Function
Acceleration Function
Jerk Function
The Product Rule
The Quotient Rule
The Chain Rule
Exponentials (𝒆𝒙 , 𝒂𝒙 )
Logorithms (𝐥𝐧 𝒙 , 𝐥𝐨𝐠 𝒂 𝒙)
Sine
Cosine
Tangent
Secent
Cosecent
Cotangent
Copyright © 2015-2016 by Harold Toomey, WyzAnt Tutor
2
𝑓(𝑥) − 𝑓(𝑐)
𝑥→𝑐
𝑥−𝑐
𝑑𝑦 ′ 𝑑
′ (𝑥), (𝑛) (𝑥),
𝑓
𝑓
, 𝑦 , [𝑓(𝑥)], 𝐷𝑥 [𝑦]
𝑑𝑥
𝑑𝑥
𝑑
[𝑐] = 0
𝑑𝑥
𝑑 𝑛
[𝑥 ] = 𝑛𝑥 𝑛−1
𝑑𝑥
𝑑
[𝑥] = 1 (𝑡ℎ𝑖𝑛𝑘 𝑥 = 𝑥 1 𝑎𝑛𝑑 𝑥 0 = 1)
𝑑𝑥
𝑑 𝑛
[𝑢 ] = 𝑛𝑢𝑛−1 𝑢′ 𝑤ℎ𝑒𝑟𝑒 𝑢 = 𝑢(𝑥)
𝑑𝑥
𝑑
[𝑐𝑓(𝑥)] = 𝑐𝑓 ′ (𝑥)
𝑑𝑥
𝑑
[𝑓(𝑥) ± 𝑔(𝑥)] = 𝑓 ′ (𝑥) ± 𝑔′ (𝑥)
𝑑𝑥
1
𝑠(𝑡) = 𝑔𝑡 2 + 𝑣0 𝑡 + 𝑠0
2
𝑣(𝑡) = 𝑠 ′ (𝑡) = 𝑔𝑡 + 𝑣0
𝑎(𝑡) = 𝑣 ′ (𝑡) = 𝑠 ′′ (𝑡)
𝑗(𝑡) = 𝑎′ (𝑡) = 𝑣 ′′ (𝑡) = 𝑠 (3) (𝑡)
𝑑
[𝑓𝑔] = 𝑓𝑔′ + 𝑔 𝑓 ′
𝑑𝑥
𝑑 𝑓
𝑔𝑓 ′ − 𝑓𝑔′
[ ]=
𝑑𝑥 𝑔
𝑔2
𝑑
[𝑓(𝑔(𝑥))] = 𝑓 ′ (𝑔(𝑥))𝑔′ (𝑥)
𝑑𝑥
𝑑𝑦 𝑑𝑦 𝑑𝑢
=
·
𝑑𝑥 𝑑𝑢 𝑑𝑥
𝑑 𝑥
𝑑 𝑥
[𝑒 ] = 𝑒 𝑥 ,
[𝑎 ] = (ln 𝑎) 𝑎 𝑥
𝑑𝑥
𝑑𝑥
𝑑
1
𝑑
1
[ln 𝑥] = ,
[log 𝑎 𝑥] =
𝑑𝑥
𝑥
𝑑𝑥
(ln 𝑎) 𝑥
𝑑
[𝑠𝑖𝑛(𝑥)] = cos(𝑥)
𝑑𝑥
𝑑
[𝑐𝑜𝑠(𝑥)] = −𝑠𝑖𝑛(𝑥)
𝑑𝑥
𝑑
[𝑡𝑎𝑛(𝑥)] = 𝑠𝑒𝑐 2(𝑥)
𝑑𝑥
𝑑
[𝑠𝑒𝑐(𝑥)] = 𝑠𝑒𝑐(𝑥) 𝑡𝑎𝑛(𝑥)
𝑑𝑥
𝑑
[𝑐𝑠𝑐(𝑥)] = − 𝑐𝑠𝑐(𝑥) 𝑐𝑜𝑡(𝑥)
𝑑𝑥
𝑑
[𝑐𝑜𝑡(𝑥)] = −𝑐𝑠𝑐 2 (𝑥)
𝑑𝑥
𝑓 ′ (𝑐) = lim
Applications of Differentiation
Rolle’s Theorem
f is continuous on the closed interval [a,b],
and
f is differentiable on the open interval (a,b).
If f(a) = f(b), then there exists at least one number c in
(a,b) such that f’(c) = 0.
𝑓(𝑏) − 𝑓(𝑎)
𝑏−𝑎
𝑓(𝑏) = 𝑓(𝑎) + (𝑏 − 𝑎)𝑓′(𝑐)
Find ‘c’.
𝑃(𝑥)
𝐼𝑓 lim 𝑓(𝑥) = lim
=
𝑥→𝑐
𝑥→𝑐 𝑄(𝑥)
Mean Value Theorem
If f meets the conditions of Rolle’s
Theorem, then
L’Hôpital’s Rule
𝑓 ′ (𝑐) =
0 ∞
{ , , 0 • ∞, 1∞ , 00 , ∞0 , ∞ − ∞} , 𝑏𝑢𝑡 𝑛𝑜𝑡 {0∞ },
0 ∞
𝑃(𝑥)
𝑃′ (𝑥)
𝑃′′ (𝑥)
𝑡ℎ𝑒𝑛 lim
= lim ′
= lim ′′
=⋯
𝑥→𝑐 𝑄(𝑥)
𝑥→𝑐 𝑄 (𝑥)
𝑥→𝑐 𝑄 (𝑥)
Graphing with Derivatives
Test for Increasing and Decreasing
Functions
The First Derivative Test
The Second Deriviative Test
Let f ’(c)=0, and f ”(x) exists, then
Test for Concavity
Points of Inflection
Change in concavity
Analyzing the Graph of a
Function
x-Intercepts (Zeros or Roots)
y-Intercept
Domain
Range
Continuity
Vertical Asymptotes (VA)
Horizontal Asymptotes (HA)
Infinite Limits at Infinity
Differentiability
Relative Extrema
Concavity
Points of Inflection
1. If f ’(x) > 0, then f is increasing (slope up) ↗
2. If f ’(x) < 0, then f is decreasing (slope down) ↘
3. If f ’(x) = 0, then f is constant (zero slope) →
1. If f ’(x) changes from – to + at c, then f has a relative
minimum at (c, f(c))
2. If f ’(x) changes from + to - at c, then f has a relative
maximum at (c, f(c))
3. If f ’(x), is + c + or - c -, then f(c) is neither
1. If f ”(x) > 0, then f has a relative minimum at (c, f(c))
2. If f ”(x) < 0, then f has a relative maximum at (c, f(c))
3. If f ’(x) = 0, then the test fails (See 1𝑠𝑡 derivative test)
1. If f ”(x) > 0 for all x, then the graph is concave up ⋃
2. If f ”(x) < 0 for all x, then the graph is concave down ⋂
If (c, f(c)) is a point of inflection of f, then either
1. f ”(c) = 0 or
2. f ” does not exist at x = c.
(See Harold’s Illegals and Graphing Rationals Cheat
Sheet)
f(x) = 0
f(0) = y
Valid x values
Valid y values
No division by 0, no negative square roots or logs
x = division by 0 or undefined
lim− 𝑓(𝑥) → 𝑦 and lim+ 𝑓(𝑥) → 𝑦
𝑥→∞
𝑥→∞
lim− 𝑓(𝑥) → ∞ and lim+ 𝑓(𝑥) → ∞
𝑥→∞
𝑥→∞
Limit from both directions arrives at the same slope
Create a table with domains, f(x), f ’(x), and f ”(x)
If 𝑓 ”(𝑥) → +, then cup up ⋃
If 𝑓 ”(𝑥) → −, then cup down ⋂
f ”(x) = 0 (concavity changes)
Copyright © 2015-2016 by Harold Toomey, WyzAnt Tutor
3
Approximating with
Differentials
𝑓(𝑥𝑛 )
𝑓′(𝑥𝑛 )
𝑦 = 𝑚𝑥 + 𝑏
𝑦 = 𝑓 ′ (𝑐)(𝑥 − 𝑐) + 𝑓(𝑐)
Newton’s Method
Finds zeros of f, or finds c if f(c) = 0.
𝑥𝑛+1 = 𝑥𝑛 −
Tangent Line Approximations
Function Approximations with
Differentials
Related Rates
𝑓(𝑥 + ∆𝑥) ≈ 𝑓(𝑥) + 𝑑𝑦 = 𝑓(𝑥) + 𝑓 ′ (𝑥) 𝑑𝑥
Steps to solve:
1. Identify the known variables and rates of change.
𝑚
(𝑥 = 2 𝑚; 𝑦 = −3 𝑚; 𝑥 ′ = 4 ; 𝑦 ′ = ? )
𝑠
2. Construct an equation relating these quantities.
(𝑥 2 + 𝑦 2 = 𝑟 2 )
3. Differentiate both sides of the equation.
(2𝑥𝑥 ′ + 2𝑦𝑦 ′ = 0)
4. Solve for the desired rate of change.
𝑥
(𝑦 ′ = − 𝑥 ′ )
𝑦
5. Substitute the known rates of change and quantities
into the equation.
2
8𝑚
(𝑦 ′ = −
⦁4=
)
−3
3 𝑠
Copyright © 2015-2016 by Harold Toomey, WyzAnt Tutor
4
Summation Formulas
𝑛
∑ 𝑐 = 𝑐𝑛
𝑖=1
𝑛
∑𝑖 =
𝑖=1
𝑛
𝑛(𝑛 + 1) 𝑛2 𝑛
=
+
2
2 2
∑ 𝑖2 =
𝑖=1
𝑛
𝑛(𝑛 + 1)(2𝑛 + 1) 𝑛3 𝑛2 𝑛
=
+ +
6
3
2 6
2
𝑛
3
∑ 𝑖 = (∑ 𝑖 ) =
𝑖=1
𝑛
Sum of Powers
𝑖=1
𝑛2 (𝑛 + 1)2 𝑛4 𝑛3 𝑛2
=
+ +
4
4
2
4
𝑛(𝑛 + 1)(2𝑛 + 1)(3𝑛2 + 3𝑛 − 1) 𝑛5 𝑛4 𝑛3 𝑛
∑ 𝑖4 =
=
+ + −
30
5
2
3 30
𝑖=1
𝑛
∑ 𝑖5 =
𝑖=1
𝑛
∑ 𝑖6 =
𝑖=1
𝑛
∑ 𝑖7 =
𝑖=1
𝑛2 (𝑛 + 1)2 (2𝑛2 + 2𝑛 − 1) 𝑛6 𝑛5 5𝑛4 𝑛2
=
+ +
−
12
6
2
12 12
𝑛(𝑛 + 1)(2𝑛 + 1)(3𝑛4 + 6𝑛3 − 3𝑛 + 1)
42
𝑛2 (𝑛 + 1)2 (3𝑛4 + 6𝑛3 − 𝑛2 − 4𝑛 + 2)
24
𝑛
𝑘−1
(𝑛 + 1)𝑘+1
1
𝑘+1
𝑆𝑘 (𝑛) = ∑ 𝑖 =
−
∑(
) 𝑆𝑟 (𝑛)
𝑘+1
𝑘+1
𝑟
𝑘
𝑖=1
𝑛
𝑟=0
𝑛
𝑛
2
∑ 𝑖(𝑖 + 1) = ∑ 𝑖 + ∑ 𝑖 =
𝑖=1
𝑛
Misc. Summation Formulas
𝑖=1
𝑖=1
𝑛(𝑛 + 1)(𝑛 + 2)
3
1
𝑛
∑
=
𝑖(𝑖 + 1) 𝑛 + 1
𝑖=1
𝑛
∑
𝑖=1
1
𝑛(𝑛 + 3)
=
𝑖(𝑖 + 1)(𝑖 + 2) 4(𝑛 + 1)(𝑛 + 2)
Copyright © 2015-2016 by Harold Toomey, WyzAnt Tutor
5
Numerical Methods
𝑛
𝑏
𝑃0 (𝑥) = ∫ 𝑓(𝑥) 𝑑𝑥 = lim ∑ 𝑓(𝑥𝑖∗ ) ∆𝑥𝑖
‖𝑃‖→0
𝑎
𝑖=1
where 𝑎 = 𝑥0 < 𝑥1 < 𝑥2 < ⋯ < 𝑥𝑛 = 𝑏
and ∆𝑥𝑖 = 𝑥𝑖 − 𝑥𝑖−1
and ‖𝑃‖ = 𝑚𝑎𝑥{∆𝑥𝑖 }
Types:
 Left Sum (LHS)
 Middle Sum (MHS)
 Right Sum (RHS)
Riemann Sum
𝑛
𝑏
𝑃0 (𝑥) = ∫ 𝑓(𝑥) 𝑑𝑥 ≈ ∑ 𝑓(𝑥̅𝑖 ) ∆𝑥 =
𝑎
𝑖=1
∆𝑥[𝑓(𝑥̅1 ) + 𝑓(𝑥̅2 ) + 𝑓(𝑥̅3 ) + ⋯ + 𝑓(𝑥̅𝑛 )]
𝑏−𝑎
where ∆𝑥 =
Midpoint Rule
𝑛
1
and 𝑥̅𝑖 = 2 (𝑥𝑖−1 + 𝑥𝑖 ) = 𝑚𝑖𝑑𝑝𝑜𝑖𝑛𝑡 𝑜𝑓 [𝑥𝑖−1 , 𝑥𝑖 ]
Error Bounds: |𝐸𝑀 | ≤
𝑏
𝐾(𝑏−𝑎)3
24𝑛2
𝑃1 (𝑥) = ∫ 𝑓(𝑥) 𝑑𝑥 ≈
𝑎
∆𝑥
[𝑓(𝑥0 ) + 2𝑓(𝑥1 ) + 2𝑓(𝑥3 ) + ⋯ + 2𝑓(𝑥𝑛−1 )
2
+ 𝑓(𝑥𝑛 )]
𝑏−𝑎
where ∆𝑥 =
𝑛
and 𝑥𝑖 = 𝑎 + 𝑖∆𝑥
Trapezoidal Rule
Error Bounds: |𝐸𝑇 | ≤
𝑏
𝐾(𝑏−𝑎)3
12𝑛2
𝑃2 (𝑥) = ∫ 𝑓(𝑥)𝑑𝑥 ≈
𝑎
Simpson’s Rule
∆𝑥
[𝑓(𝑥0 ) + 4𝑓(𝑥1 ) + 2𝑓(𝑥2 ) + 4𝑓(𝑥3 ) + ⋯
3
+ 2𝑓(𝑥𝑛−2 ) + 4𝑓(𝑥𝑛−1 ) + 𝑓(𝑥𝑛 )]
Where n is even
𝑏−𝑎
and ∆𝑥 = 𝑛
and 𝑥𝑖 = 𝑎 + 𝑖∆𝑥
𝐾(𝑏−𝑎)5
Error Bounds: |𝐸𝑆 | ≤ 180𝑛4
[MATH] fnInt(f(x),x,a,b), [MATH] [1] [ENTER]
TI-84 Plus
TI-Nspire CAS
Copyright © 2015-2016 by Harold Toomey, WyzAnt Tutor
6
Example: [MATH] fnInt(x^2,x,0,1)
1
1
∫ 𝑥 2 𝑑𝑥 =
3
0
[MENU] [4] Calculus [3] Integral
[TAB] [TAB]
[X] [^] [2] [TAB]
[TAB] [X] [ENTER]
Integration
∫ 𝑓 ′ (𝑥) 𝑑𝑥 = 𝑓(𝑥) + 𝐶
Basic Integration Rules
Integration is the “inverse” of
differentiation, and vice versa.
𝑑
∫ 𝑓(𝑥) 𝑑𝑥 = 𝑓(𝑥)
𝑑𝑥
𝑓(𝑥) = 0
∫ 0 𝑑𝑥 = 𝐶
𝑓(𝑥) = 𝑘 = 𝑘𝑥 0
∫ 𝑘 𝑑𝑥 = 𝑘𝑥 + 𝐶
The Constant Multiple Rule
The Sum and Difference Rule
The Power Rule
𝑓(𝑥) = 𝑘𝑥 𝑛
∫ 𝑘 𝑓(𝑥) 𝑑𝑥 = 𝑘 ∫ 𝑓(𝑥) 𝑑𝑥
∫[𝑓(𝑥) ± 𝑔(𝑥)] 𝑑𝑥 = ∫ 𝑓(𝑥) 𝑑𝑥 ± ∫ 𝑔(𝑥) 𝑑𝑥
∫ 𝑥 𝑛 𝑑𝑥 =
𝑥 𝑛+1
+ 𝐶, 𝑤ℎ𝑒𝑟𝑒 𝑛 ≠ −1
𝑛+1
𝐼𝑓 𝑛 = −1, 𝑡ℎ𝑒𝑛 ∫ 𝑥 −1 𝑑𝑥 = ln|𝑥| + 𝐶
𝑑
The General Power Rule
If 𝑢 = 𝑔(𝑥), 𝑎𝑛𝑑 𝑢′ = 𝑔(𝑥) then
𝑑𝑥
𝑢𝑛+1
𝑛 ′
∫ 𝑢 𝑢 𝑑𝑥 =
+ 𝐶, 𝑤ℎ𝑒𝑟𝑒 𝑛 ≠ −1
𝑛+1
𝑛
Reimann Sum
∑ 𝑓(𝑐𝑖 ) ∆𝑥𝑖 ,
𝑤ℎ𝑒𝑟𝑒 𝑥𝑖−1 ≤ 𝑐𝑖 ≤ 𝑥𝑖
𝑖=1
‖∆‖ = ∆𝑥 =
𝑛
Definition of a Definite Integral
Area under curve
𝑏
lim ∑ 𝑓(𝑐𝑖 ) ∆𝑥𝑖 = ∫ 𝑓(𝑥) 𝑑𝑥
‖∆‖→0
𝑏
Swap Bounds
Additive Interval Property
𝑏−𝑎
𝑛
𝑎
𝑖=1
𝑎
∫ 𝑓(𝑥) 𝑑𝑥 = − ∫ 𝑓(𝑥) 𝑑𝑥
𝑎
𝑏
𝑐
𝑏
𝑏
∫ 𝑓(𝑥) 𝑑𝑥 = ∫ 𝑓(𝑥) 𝑑𝑥 + ∫ 𝑓(𝑥) 𝑑𝑥
𝑎
𝑎
𝑏
The Fundamental Theorem of
Calculus
𝑐
∫ 𝑓(𝑥) 𝑑𝑥 = 𝐹(𝑏) − 𝐹(𝑎)
𝑎
𝑥
𝑑
∫ 𝑓(𝑡) 𝑑𝑡 = 𝑓(𝑥)
𝑑𝑥
𝑔(𝑥)
The Second Fundamental
Theorem of Calculus
𝑑
∫ 𝑓(𝑡) 𝑑𝑡 = 𝑓(𝑔(𝑥))𝑔′ (𝑥)
𝑑𝑥
𝑎
ℎ(𝑥)
(See Harold’s Fundamental
Theorem of Calculus Cheat Sheet)
Mean Value Theorem for
Integrals
𝑎
𝑑
∫ 𝑓(𝑡) 𝑑𝑡 = 𝑓(ℎ(𝑥))ℎ′ (𝑥) − 𝑓(𝑔(𝑥))𝑔′(𝑥)
𝑑𝑥
𝑔(𝑥)
𝑏
∫ 𝑓(𝑥) 𝑑𝑥 = 𝑓(𝑐)(𝑏 − 𝑎) Find ‘𝑐’.
The Average Value for a
Function
Copyright © 2015-2016 by Harold Toomey, WyzAnt Tutor
7
𝑎
𝑏
1
∫ 𝑓(𝑥) 𝑑𝑥
𝑏−𝑎 𝑎
Integration Methods
1. Memorized
See Larson’s 1-pager of common integrals
∫ 𝑓(𝑔(𝑥))𝑔′ (𝑥)𝑑𝑥 = 𝐹(𝑔(𝑥)) + 𝐶
Set 𝑢 = 𝑔(𝑥), then 𝑑𝑢 = 𝑔′ (𝑥) 𝑑𝑥
2. U-Substitution
∫ 𝑓(𝑢) 𝑑𝑢 = 𝐹(𝑢) + 𝐶
𝑢 = _____
𝑑𝑢 = _____ 𝑑𝑥
∫ 𝑢 𝑑𝑣 = 𝑢𝑣 − ∫ 𝑣 𝑑𝑢
𝑢 = _____
𝑑𝑢 = _____
3. Integration by Parts
𝑣 = _____
𝑑𝑣 = _____
Pick ‘𝑢’ using the LIATED Rule:
L – Logarithmic : ln 𝑥 , log 𝑏 𝑥 , 𝑒𝑡𝑐.
I – Inverse Trig.:
tan−1 𝑥 , sec −1 𝑥 , 𝑒𝑡𝑐.
A – Algebraic:
𝑥 2 , 3𝑥 60 , 𝑒𝑡𝑐.
T – Trigonometric: sin 𝑥 , tan 𝑥 , 𝑒𝑡𝑐.
E – Exponential:
𝑒 𝑥 , 19𝑥 , 𝑒𝑡𝑐.
D – Derivative of: 𝑑𝑦⁄𝑑𝑥
𝑃(𝑥)
∫
𝑑𝑥
𝑄(𝑥)
where 𝑃(𝑥) 𝑎𝑛𝑑 𝑄(𝑥) are polynomials
4. Partial Fractions
Case 1: If degree of 𝑃(𝑥) ≥ 𝑄(𝑥)
then do long division first
Case 2: If degree of 𝑃(𝑥) < 𝑄(𝑥)
then do partial fraction expansion
∫ √𝑎2 − 𝑥 2 𝑑𝑥
Substutution: 𝑥 = 𝑎 sin 𝜃
Identity: 1 − 𝑠𝑖𝑛2 𝜃 = 𝑐𝑜𝑠 2 𝜃
5a. Trig Substitution for √𝒂𝟐 − 𝒙𝟐
∫ √𝑥 2 − 𝑎2 𝑑𝑥
Substutution: 𝑥 = 𝑎 sec 𝜃
Identity: 𝑠𝑒𝑐 2 𝜃 − 1 = 𝑡𝑎𝑛2 𝜃
5b. Trig Substitution for √𝒙𝟐 − 𝒂𝟐
∫ √𝑥 2 + 𝑎2 𝑑𝑥
5c. Trig Substitution for √𝒙𝟐 + 𝒂𝟐
6. Table of Integrals
7. Computer Algebra Systems (CAS)
8. Numerical Methods
9. WolframAlpha
Substutution: 𝑥 = 𝑎 tan 𝜃
Identity: 𝑡𝑎𝑛2 𝜃 + 1 = 𝑠𝑒𝑐 2 𝜃
CRC Standard Mathematical Tables book
TI-Nspire CX CAS Graphing Calculator
TI –Nspire CAS iPad app
Riemann Sum, Midpoint Rule, Trapezoidal Rule,
Simpson’s Rule, TI-84
Google of mathematics. Shows steps. Free.
www.wolframalpha.com
Copyright © 2015-2016 by Harold Toomey, WyzAnt Tutor
8
Partial Fractions
Condition
Case I: Simple linear (𝟏𝒔𝒕 degree)
Case II: Multiple degree linear (𝟏𝒔𝒕 degree)
Case III: Simple quadratic (𝟐𝒏𝒅 degree)
Case IV: Multiple degree quadratic (𝟐𝒏𝒅
degree)
Example Expansion
Typical Solution for Cases I & II
Typical Solution for Cases III & IV
Series
(http://en.wikipedia.org/wiki/Partial_fraction_deco
mposition)
𝑃(𝑥)
𝑓(𝑥) =
𝑄(𝑥)
where 𝑃(𝑥) 𝑎𝑛𝑑 𝑄(𝑥) are polynomials
and degree of 𝑃(𝑥) < 𝑄(𝑥)
𝐴
(𝑎𝑥 + 𝑏)
𝐴
𝐵
𝐶
+
+
2
(𝑎𝑥 + 𝑏) (𝑎𝑥 + 𝑏)
(𝑎𝑥 + 𝑏)3
𝐴𝑥 + 𝐵
2
(𝑎𝑥 + 𝑏𝑥 + 𝑐)
𝐴𝑥 + 𝐵
𝐶𝑥 + 𝐷
𝐸𝑥 + 𝐹
+
+
(𝑎𝑥 2 + 𝑏𝑥 + 𝑐) (𝑎𝑥 2 + 𝑏𝑥 + 𝑐)2 (𝑎𝑥 2 + 𝑏𝑥 + 𝑐)3
𝑃(𝑥)
(𝑎𝑥 + 𝑏)(𝑐𝑥 + 𝑑)2 (𝑒𝑥 2 + 𝑓𝑥 + 𝑔)
𝐴
𝐵
𝐶
𝐷𝑥 + 𝐸
=
+
+
+
(𝑎𝑥 + 𝑏) (𝑐𝑥 + 𝑑) (𝑐𝑥 + 𝑑)2 (𝑒𝑥 2 + 𝑓𝑥 + 𝑔)
𝑎
∫
𝑑𝑥 = 𝑎 𝑙𝑛|𝑥 + 𝑏| + 𝐶
𝑥+𝑏
𝑎
𝑎
𝑥
∫ 2
𝑑𝑥 = 𝑡𝑎𝑛−1 ( ) + 𝐶
2
𝑥 +𝑏
𝑏
𝑏
Arithmetic
Geometric
lim 𝑎𝑛 = 𝐿 (Limit)
𝑛→∞
Sequence
Example: (𝑎𝑛 , 𝑎𝑛+1 , 𝑎𝑛+2 , …)
𝑛
𝑆𝑛 = ∑ 𝑎𝑘
Summation Notation
𝑘=1
Summation Expanded
Sum of n Terms (finite series)
𝑆𝑛 = 𝑎1 + 𝑎2 + ⋯ + 𝑎𝑛−1 + 𝑎𝑛
(Partial Sum)
𝑛−1
𝑛
𝑘
𝑆𝑛 = ∑ 𝑎0 𝑟 = ∑ 𝑎0 𝑟 𝑘−1
𝑘=0
𝑆𝑛 = 𝑎0 + 𝑎0 𝑟 + 𝑎0 𝑟 2 + ⋯ + 𝑎0 𝑟 𝑛−1
𝑎1 + 𝑎𝑛
𝑆𝑛 = 𝑛 (
)
2
𝑆𝑛 =
𝑛
(2𝑎1 + (𝑛 − 1)𝑑)
2
𝑘=1
𝑆𝑛 = 𝑎0
1 − 𝑟𝑛
1−𝑟
𝑎(1 − 𝑟 𝑛 )
𝑎
=
𝑛→∞
1−𝑟
1−𝑟
𝑆 = lim
Sum of ∞ Terms (infinite
series)
Recursive nth Term
Explicit nth Term
𝑆→∞
𝑎𝑛 = 𝑎𝑛−1 + 𝑑
𝑎𝑛 = 𝑎1 + 𝑑(𝑛 − 1)
Copyright © 2015-2016 by Harold Toomey, WyzAnt Tutor
9
only if |𝑟| < 1
where 𝑟 is the radius of convergence
and (−𝑟, 𝑟) is the interval of
convergence
𝑎𝑛 = 𝑎𝑛−1 𝑟
𝑎𝑛 = 𝑎0 𝑟 𝑛−1
Convergence Tests
(See Harold’s Series Convergence Tests Cheat Sheet)
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
Convergence Tests
𝑛𝑡ℎ Term
Geometric Series
p-Series
Alternating Series
Integral
Ratio
Root
Direct Comparison
Limit Comparison
Telescoping
∞
∑(𝑏𝑛 − 𝑏𝑛+1 )
𝑛=1
Telescoping Series
Converges if lim 𝑏𝑛 = 𝐿
𝑛→ ∞
Diverges if N/A
Sum: 𝑆 = 𝑏1 − 𝐿
Taylor Series
+∞
Power Series
∑ 𝑎𝑛 (𝑥 − 𝑐)𝑛 = 𝑎0 + 𝑎1 (𝑥 − 𝑐) + 𝑎2 (𝑥 − 𝑐)2 + ⋯
𝑛=0
+∞
Power Series About Zero
∑ 𝑎𝑛 𝑥 𝑛 = 𝑎0 + 𝑎1 𝑥 + 𝑎2 𝑥 2 + ⋯
𝑛=0
+∞
Maclaurin Series
Taylor series about zero
𝒇(𝒙) ≈ 𝑷𝒏 (𝒙) = ∑
𝒏=𝟎
𝒇(𝒏) (𝟎) 𝒏
𝒙
𝒏!
𝑓(𝑥) = 𝑃𝑛 (𝑥) + 𝑅𝑛 (𝑥)
+∞
𝑓 (𝑛) (0) 𝑛 𝑓 (𝑛+1) (𝑥 ∗ ) 𝑛+1
=∑
𝑥 +
𝑥
𝑛!
(𝑛 + 1)!
Maclaurin Series with Remainder
𝑛=0
where 𝑥 ≤ 𝑥 ∗ ≤ 𝑚𝑎𝑥
and lim 𝑅𝑛 (𝑥) = 0
𝑥→+∞
+∞
Taylor Series
𝑓(𝑥) ≈ 𝑃𝑛 (𝑥) = ∑
𝑛=0
𝑓 (𝑛) (𝑐)
(𝑥 − 𝑐)𝑛
𝑛!
𝑓(𝑥) = 𝑃𝑛 (𝑥) + 𝑅𝑛 (𝑥)
+∞
Taylor Series with Remainder
𝑓 (𝑛) (𝑐)
𝑓 (𝑛+1) (𝑥 ∗ )
𝑛
=∑
(𝑥 − 𝑐) +
(𝑥 − 𝑐)𝑛+1
𝑛!
(𝑛 + 1)!
𝑛=0
where 𝑥 ≤ 𝑥 ∗ ≤ 𝑐
and lim 𝑅𝑛 (𝑥) = 0
𝑥→+∞
Copyright © 2015-2016 by Harold Toomey, WyzAnt Tutor
10
Common Series
Exponential Functions
∞
𝑥𝑛
𝑓𝑜𝑟 𝑎𝑙𝑙 𝑥
𝑛!
𝑥
𝑒 =∑
𝑛=0
∞
𝑎 𝑥 = 𝑒 𝑥 ln(𝑎) = ∑
𝑛=0
Natural Logarithms
(𝑥 ln(𝑎))𝑛
𝑓𝑜𝑟 𝑎𝑙𝑙 𝑥
𝑛!
∞
ln (1 − 𝑥) = ∑
𝑛=1
∞
∞
1 + 𝑥 ln(𝑎) +
𝑥𝑛
𝑓𝑜𝑟 |𝑥| < 1
𝑛
(−1)𝑛 (𝑥 − 1)𝑛
ln (𝑥) = ∑
𝑓𝑜𝑟 |𝑥| < 1
𝑛
𝑛=1
1+𝑥+
𝑥+
(𝑥 − 1) +
(−1)𝑛−1 𝑛
𝑥 𝑓𝑜𝑟 |𝑥| < 1
𝑛
ln (1 + 𝑥) = ∑
𝑛=1
𝑥2 𝑥3 𝑥4
+ + +⋯
2! 3! 4!
(𝑥 ln(𝑎))2 (𝑥 ln(𝑎))3
+
+⋯
2!
3!
𝑥2 𝑥3 𝑥4 𝑥5
+ + + +⋯
2
3
4
5
(𝑥 − 1)2 (𝑥 − 1)3 (𝑥 − 1)4
+
+
+⋯
2
3
4
𝑥−
𝑥2 𝑥3 𝑥4 𝑥5
+ − + −⋯
2
3
4
5
Geometric Series
∞
1
= ∑(−1)𝑛 (𝑥 − 1)𝑛 𝑓𝑜𝑟 0 < 𝑥 < 2
𝑥
𝑛=0
1 − (𝑥 − 1) + (𝑥 − 1)2 − (𝑥 − 1)3 + (𝑥 − 1)4 + ⋯
∞
1
= ∑(−1)𝑛 𝑥 𝑛 𝑓𝑜𝑟 |𝑥| < 1
1+𝑥
𝑛=0
1 − 𝑥 + 𝑥2 − 𝑥3 + 𝑥4 − ⋯
∞
1
= ∑ 𝑥 𝑛 𝑓𝑜𝑟 |𝑥| < 1
1−𝑥
1 + 𝑥 + 𝑥2 + 𝑥3 + 𝑥4 + ⋯
1
= ∑ 𝑛𝑥 𝑛−1 𝑓𝑜𝑟 |𝑥| < 1
(1 − 𝑥)2
1 + 2𝑥 + 3𝑥 2 + 4𝑥 3 + 5𝑥 4 + ⋯
𝑛=0
∞
𝑛=1
∞
1
(𝑛 − 1)𝑛 𝑛−2
=∑
𝑥
𝑓𝑜𝑟 |𝑥| < 1
3
(1 − 𝑥)
2
1 + 3𝑥 + 6𝑥 2 + 10𝑥 3 + 15𝑥 4 + ⋯
𝑛=2
Binomial Series
+∞
𝑟
(1 + 𝑥) = ∑ ( ) 𝑥 𝑛
𝑛
𝑟
𝑛=0
𝑓𝑜𝑟 |𝑥| < 1 and all complex r where
𝑛
𝑟
𝑟−𝑘+1
( )=∏
𝑛
𝑘
𝑘=1
𝑟(𝑟 − 1)(𝑟 − 2) … (𝑟 − 𝑛 + 1)
=
𝑛!
Trigonometric Functions
∞
sin (𝑥) = ∑
𝑛=0
1 + 𝑟𝑥 +
(−1)𝑛 2𝑛+1
𝑥
𝑓𝑜𝑟 𝑎𝑙𝑙 𝑥
(2𝑛 + 1)!
∞
cos (𝑥) = ∑
𝑛=0
(−1)𝑛 2𝑛
𝑥
𝑓𝑜𝑟 𝑎𝑙𝑙 𝑥
(2𝑛)!
Copyright © 2015-2016 by Harold Toomey, WyzAnt Tutor
11
𝑟(𝑟 − 1) 2 𝑟(𝑟 − 1)(𝑟 − 2) 3
𝑥 +
𝑥 +⋯
2!
3!
𝑥−
𝑥3 𝑥5 𝑥7 𝑥9
+ − + −⋯
3! 5! 7! 9!
1−
𝑥2 𝑥4 𝑥6 𝑥8
+ − + −⋯
2! 4! 6! 8!
∞
𝐵2𝑛 (−4)𝑛 (1 − 4𝑛 ) 2𝑛−1
𝑥
(2𝑛)!
𝑛=1
𝜋
𝑓𝑜𝑟 |𝑥| <
2
Bernoulli Numbers:
1
1
1
1
𝐵0 = 1, 𝐵1 = , 𝐵2 = , 𝐵4 =
, 𝐵6 = ,
2
6
30
42
1
5
691
7
𝐵8 =
, 𝐵10 =
, 𝐵12 =
, 𝐵14 =
30
66
2730
6
∞
(−1)𝑛 𝐸2𝑛 2𝑛
sec (𝑥) = ∑
𝑥
(2𝑛)!
𝑛=0
𝜋
𝑓𝑜𝑟 |𝑥| <
2
Euler Numbers:
𝐸0 = 1, 𝐸2 = −1, 𝐸4 = 5, 𝐸6 = −61,
𝐸8 = 1,385, 𝐸10 = −50,521, 𝐸12 = 2,702,765
∞
(2𝑛)!
arcsin (𝑥) = ∑ 𝑛 2
𝑥 2𝑛+1
(2 𝑛!) (2𝑛 + 1)
tan (𝑥) = ∑
𝑥+2
𝑥3
𝑥5
𝑥7
𝑥9
+ 16 + 272 + 7936 − ⋯
3!
5!
7!
9!
1
2
17 7
2 9
= 𝑥 + 𝑥3 + 𝑥5 +
𝑥 +
𝑥 −⋯
3
15
315
945
1+
𝑥2
𝑥4
𝑥6
𝑥8
𝑥 10
+ 5 + 61 + 1385 + 50,521
+⋯
2!
4!
6!
8!
10!
𝑥+
𝑛=0
𝑓𝑜𝑟 |𝑥| ≤ 1
𝜋
arccos (𝑥) = − arcsin (𝑥)
2
𝑓𝑜𝑟 |𝑥| ≤ 1
∞
(−1)𝑛 2𝑛+1
arctan (𝑥) = ∑
𝑥
(2𝑛 + 1)
𝑥3
1 ∙ 3𝑥 5 1 ∙ 3 ∙ 5𝑥 7
+
+
+⋯
2∙3 2∙4∙5 2∙4∙6∙7
𝜋
𝑥3
1 ∙ 3𝑥 5 1 ∙ 3 ∙ 5𝑥 7
−𝑥−
−
−
−⋯
2
2∙3 2∙4∙5 2∙4∙6∙7
𝑥3 𝑥5 𝑥7 𝑥9
𝑥− + − + −⋯
3
5
7
9
𝑛=0
𝑓𝑜𝑟 |𝑥| < 1, 𝑥 ≠ ±𝑖
Hyperbolic Functions
∞
𝑒 𝑥 − 𝑒 −𝑥
𝑥 2𝑛+1
sinh (𝑥) =
=∑
𝑓𝑜𝑟 𝑎𝑙𝑙 𝑥
(2𝑛 + 1)!
2
𝑥+
𝑥3 𝑥5 𝑥7 𝑥9
+ + + +⋯
3! 5! 7! 9!
𝑒 𝑥 + 𝑒 −𝑥
𝑥 2𝑛
cosh (𝑥) =
=∑
𝑓𝑜𝑟 𝑎𝑙𝑙 𝑥
(2𝑛)!
2
1+
𝑥2 𝑥4 𝑥6 𝑥8
+ + + +⋯
2! 4! 6! 8!
𝑛=0
∞
∞
𝑛=0
𝐵2𝑛 4𝑛 (4𝑛 − 1) 2𝑛−1
tanh (𝑥) = ∑
𝑥
(2𝑛)!
𝑛=1
𝜋
𝑓𝑜𝑟 |𝑥| <
2
∞
(−1)𝑛 (2𝑛)!
arcsinh (𝑥) = ∑ 𝑛 2
𝑥 2𝑛+1
(2 𝑛!) (2𝑛 + 1)
𝑥3
𝑥5
𝑥7
𝑥9
𝑥 − 2 + 16 − 272 + 7936 − ⋯
3!
5!
7!
9!
1 3 2 5 17 7
2 9
𝑥− 𝑥 + 𝑥 −
𝑥 +
𝑥 −⋯
3
15
315
945
𝑥−
𝑛=0
𝑓𝑜𝑟 |𝑥| ≤ 1
∞
𝜋
2−𝑛
arccosh (𝑥) = 𝑖 − 𝑖 ∑
𝑥 2𝑛+1
2
𝑛! (2𝑛 + 1)
𝑛=0
𝑓𝑜𝑟 |𝑥| ≤ 1
∞
arctanh (𝑥) = ∑
𝑛=0
𝑥3
1 ∙ 3𝑥 5 1 ∙ 3 ∙ 5𝑥 7
+
−
+⋯
2∙3 2∙4∙5 2∙4∙6∙7
𝜋𝑖
𝑖 𝑥 3 𝑖 ∙ 1 ∙ 3𝑥 5 𝑖 ∙ 1 ∙ 3 ∙ 5𝑥 7
−𝑖𝑥−
−
−
−⋯
2
2∙3
2∙4∙5
2∙4∙6∙7
𝑥 2𝑛+1
(2𝑛 + 1)
𝑓𝑜𝑟 |𝑥| < 1, 𝑥 ≠ ±1
Copyright © 2015-2016 by Harold Toomey, WyzAnt Tutor
12
𝑥+
𝑥3 𝑥5 𝑥7 𝑥9
+ + + +⋯
3
5
7
9
Copyright © 2015-2016 by Harold Toomey, WyzAnt Tutor
13