1
AP CALCULUS AB STUDY GUIDE.
Matt Salek
Mrs. Edwards
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Table of Contents
Unit 1: Limits and Continuity
3
Unit 1.1: Rate and Change of Limits
3
Unit 1.2: Limits Involving Infinity
4
Unit 1.3: Continuity
5
Unit 2: Derivatives
6
Unit 2.1: Derivative of a Function
7
Unit 2.2: Differentiability
7
Unit 2.3: Rules for Differentiation
7
Unit 2.4: Velocity and Other Rates of Change
8
Unit 2.5: Derivatives of Trigonometric Functions
8
Unit 3: More Derivatives
9
Unit 3.1: Chain Rule
10
Unit 3.2: Implicit Differentiation
10
Unit 3.3: Derivatives of Inverse Trigonometric Functions
10
Unit 3.4: Derivatives of Exponential and Logarithmic Functions
10
Unit 4: Applications of Derivatives
12
Unit 4.1: Extreme Value of Functions
12
Unit 4.2: Mean Value Theorem
13
Unit 4.3: Connecting f’ and f” with the Graph of f
14
Unit 4.4: Modeling and Optimization
15
Unit 4.5: Linearization, Sensitivity, and Differentials
17
Unit 4.6: Related Rates
17
Unit 5: The Definite Integral
Unit 5.1: Estimating with Finite Sums
Unit 5.2: Definite Integrals
Unit 5.3: Definite Integrals and Antiderivatives
Unit 5.4: Fundamental Theorem of Calculus
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Unit 5.5: Trapezoidal Sums
Unit 6: Differential Equations and Mathematical Modeling
Unit 6.1: Slope Fields and Euler’s Method
Unit 6.2: Antidifferentiation by substitution
U-Substitution
Unit 6.3: Antidifferentiation by Parts (not on exam)
Unit 6.4: Exponential Growth and Decay
Separable Differential Equations
Unit 6.5: Logistic Growth
Unit 7: Applications of Definite Integrals
Unit 7.1: Accumulation and Net Change
Unit 7.2: Areas in the Plane
Unit 7.3: Volumes
Unit 7.5: Applications from Science and Statistics
Unit 8: Sequences, L’Hospital’s Rule, and Improper Integrals
Unit 8.2: L’Hopital’s Rule
Main Ideas and Things to Memorise
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Unit 1: Limits and Continuity
1.1 Rate of Change of Limits
Video: https://www.youtube.com/watch?v=4Up5gsDeluw
Average and Instantaneous Velocity
Average Velocity:
∆
∆
(
=
( )
∆
∆
Instantaneous Velocity
∆ )
∆
=
(
)
( )
, lim
(
)
( )
→
Ex. 𝑓(𝑎) = 16𝑎
lim
(
)
→
*get h out of the bottom
lim (32𝑎 + 16ℎ), lim (32𝑎 + 16(0)) = 32𝑎
→
→
Properties of Limits:
Video: https://www.youtube.com/watch?v=yaUpAIzYjrg
lim 𝑓(𝑥) = 𝐿 and lim 𝑔(𝑥) = 𝑀
→
→
1. Sum rule: lim 𝑓(𝑥) + 𝑔(𝑥) = 𝐿 + 𝑀
→
2. Difference Rule: lim 𝑓(𝑥) − 𝑔(𝑥) = 𝐿 − 𝑀
→
3. Product Rule: lim 𝑓(𝑥) ∗ 𝑔(𝑥) = 𝐿 ∗ 𝑀
→
4. Constant Multiple Rule: lim 𝑘 ∗ 𝑔(𝑥) = 𝑘 ∗ 𝑀
→
5. Quotient Rule: lim
→
( )
( )
=
6. Power Rule: lim(𝑓(𝑥)) = 𝐿
→
lim 𝑓(𝑥) = 10 and lim 𝑔(𝑥) = 5; k = 1; r = 2; s = 4
→
→
1. Sum rule: lim 𝑓(𝑥) + 𝑔(𝑥) = 10 + 5
→
2. Difference Rule: lim 𝑓(𝑥) − 𝑔(𝑥) = 10 − 5
→
3. Product Rule: lim 𝑓(𝑥) ∗ 𝑔(𝑥) = 10 ∗ 5
→
4. Constant Multiple Rule: lim 𝑘 ∗ 𝑔(𝑥) = 1 ∗ 5
→
5. Quotient Rule: lim
→
( )
( )
=
5
6. Power Rule: lim(𝑓(𝑥)) = 10
→
One-Sided and Two Sided Limits:
A function f(x) has a limit as x approaches c if and only if the right-hand and left-hand
limits at c exist and are equal.
lim 𝑓(𝑥) = 𝐿, lim 𝑓(𝑥) = 𝐿 𝑎𝑛𝑑 lim 𝑓(𝑥) = 𝐿
→

→
→
Limit fails to exist if:
o lim 𝑓(𝑥) 𝑑𝑜𝑒𝑠 𝑛𝑜𝑡 𝑒𝑥𝑖𝑠𝑡
→
o
lim 𝑓(𝑥) 𝑑𝑜𝑒𝑠 𝑛𝑜𝑡 𝑒𝑥𝑖𝑠𝑡
→
o Both of these one-sided limits exist, but have different values
Ex.
lim 𝑓(𝑥) = − 1
→
lim 𝑓(𝑥) = 1
→
lim 𝑓(𝑥) = 𝐷𝑁𝐸
→
lim 𝑓(𝑥) = 4
→
lim 𝑓(𝑥) = 4
→
lim 𝑓(𝑥) = 4
→
Practice Problems:
1. Determine is the func on is con nuous: 𝑓(𝑥) =
2. 𝑓(𝑥) =
√
, lim 𝑓(𝑥) =
→
(
(
)
)(
)
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3.
1.2 Limits Involving Infinity
Limits Involving Infinity:
Ex. The func on above has ver cal asymptotes at x =−2 and x = 3, and a horizontal asymptote at y =1.
Looking at the graph, we can determine the following limits:
lim 𝑓(𝑥) = ∞
→
lim 𝑓(𝑥) = −∞
→
lim 𝑓(𝑥) = ∞
→
lim 𝑓(𝑥) = 1
→
The Squeeze Theorem:
If the graph of a function lies between the graphs of two other functions, and if the two other
functions share a limit at a certain point, then the function in between also shares that same limit.
If. 𝑔(𝑥) ≤ 𝑓(𝑥) ≤ ℎ(𝑥) and lim 𝑔(𝑥) = lim ℎ(𝑥) = 𝐿 Then. lim 𝑓(𝑥) = 𝐿
→
→
Video: https://www.youtube.com/watch?v=lVIb9iJ-rRI
→
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Ex. Using the Squeeze Theorem, prove lim 𝑥 𝑠𝑖𝑛
=0
→
−𝑥 ≤ 𝑥 𝑠𝑖𝑛
≤𝑥
lim (−𝑥 ) = 0, and lim (𝑥 ) = 0, Therefore lim 𝑥 𝑠𝑖𝑛
→
→
→
=0
Practice Problems:
Evaluate (a) lim 𝑓(𝑥) and (b) lim 𝑓(𝑥)
→
→
1. 𝑓(𝑥) = 2𝑒
2. 𝑓(𝑥) = √
−𝑒
− 10𝑒
1.3 Continuity
Continuity: Limit fails to exist if:

lim 𝑓(𝑥) 𝑑𝑜𝑒𝑠 𝑛𝑜𝑡 𝑒𝑥𝑖𝑠𝑡

lim 𝑓(𝑥) 𝑑𝑜𝑒𝑠 𝑛𝑜𝑡 𝑒𝑥𝑖𝑠𝑡

Both of these one-sided limits exist, but have different values
→
→
Intermediate Value Theorem:
Video: https://www.youtube.com/watch?v=9wEHwFrUyOU
The Intermediate Value Theorem applies to continuous functions on an interval a,b]. If d is any
value between f(a) and f(b), then there must be at least one number c between a and b such that
f(c) = d.
Ex. Guarente that there is a zero in the given interval. 𝑔(𝑥) = 𝑥 − 3 on [-1,2]
𝑔(−1) = (−1) − 3 = −2
𝑔(2) = (2) − 3 = 1
𝑔(𝑥) = 𝑥 − 3 = 0
Practice Problems:
Find if the fuctions are continuous at the given points.
1. 𝑓(𝑥) =
,
(a) x=-2
(b) x=0
(c) x=5
1 − 3𝑥, 𝑥 < −6
2. 𝑓(𝑥) = 𝑥 , −6 ≤ 𝑥 < 1
(a) x=-6
(b) x=1
2 − 𝑥, 𝑥 ≥ 1
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Unit 2: Derivatives
2.1 Deriative of a Function
Video: https://www.youtube.com/watch?v=-aTLjoDT1GQ
Definition of a derivative:
(
𝑓 (𝑥) = lim
)
( )
→
Ex. 𝑓(𝑥) = 𝑥 , lim
(
)
→
, lim (3𝑥 + 3𝑥ℎ + ℎ ) = 3𝑥 ,
→
Practice Problems:
Use the definition of the derivative to find the derivative of the following functions.
1. 𝑓(𝑥) =
2. 𝑓(𝑥) = √3𝑥 − 4
2.2 Differentiability
Video: https://www.youtube.com/watch?v=fml0-ELYLaE
Nondifferentable at
1. Corner
2. Cusp
3. Vertical tangent
4. Discontinuity
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2.3 Rules for Differentiation
Video: https://www.khanacademy.org/math/ap-calculus-ab/ab-differentiation-1-new/ab-26a/v/derivative-properties-and-polynomial-derivatives
Basic Rules
1. Constant Rule:
𝑐 = 0,
5=0
2. Power Rule:
(𝑥 ) = 𝑛𝑥
,
3. Constant Multiple Rule:
(𝑥 ) = 5𝑥
,
(𝑐𝑢) = 𝑐
4. The Sum and Difference Rule:
(5𝑥) = 5
(𝑢 ± 𝑣) =
=5
,
±
(𝑥 + 𝑥 ) =
+
𝑥 = 1 + 2𝑥
5. The Product Rule: (𝑢𝑣) = 𝑢 + 𝑣 ,
𝑑𝑦
[(2𝑥 + 1)(𝑥 + 2)] = (2𝑥 + 1)(1) + (𝑥 + 2)(2) = 3𝑥 + 3
𝑑𝑥
6. The Quotient Rule:
=
,
=
( )
(
( )
)
=−
Practice Problems:
Find the derivative of the given function.
1. 𝑓(𝑥) = 𝑥(3𝑥 − 9)
2. 𝑓(𝑥) =
2.4 Velocity and Other Rates of Change
Video: https://www.youtube.com/watch?v=6t3mYHIWF2A
Instantanious Rate of Change is the dirivative.
𝑓 (𝑎) = lim
(
)
( )
→
Instantanious Velocity is the derivative of the position function s=f(t) with respect to time.
𝑣(𝑡) =
= lim
∆ →
(
∆ )
( )
∆
Ex. What is the velocity of the possition function 𝑓(𝑡) = 2𝑡 + 5 when t=2
𝑓 (𝑥) = 4𝑥 = 4(2) = 8
Acceleration is the derivative of Velocity or the Second Derivative of the possition function
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𝑎(𝑡) =
=
Ex. The possition of a train is described as 𝑓(𝑡) = 3𝑡 + 5𝑡 − 4, what is the acceleration
of the train at t=5
𝑓”(𝑡) = 𝑓 (9𝑡 + 5) = 18𝑡,
𝑓"(5) = 18(5) = 90
Practice Problems:
1. Find the velocity of the following possition function
𝑝(𝑥) = 4𝑥 + 5𝑥
2. Find the acceleration of the following possition function
a. 𝑝(𝑥) = 5𝑥 −
𝑥
2.5 Derivatives of Trigonometric Functions
Video: https://www.youtube.com/watch?v=PEqCa0U77mc
Trig Function Chart
Practice Problems:
Find the derivative of the following function
1. 2 cos(𝑥) − 6 sec(𝑥) + 3
2. 𝑓(𝑥) =
( )
( )
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Unit 3: More Derivatives
3.1 Chain Rule
Video: https://www.khanacademy.org/math/ap-calculus-ab/ab-differentiation-2-new/ab-31a/v/chain-rule-introduction
Chain Rule: if f is differentiable at point u=g(x), and g is differentiable at x, then the composite
function (f(g(x)) =(g(f(x)) is differentiable at x, and
(𝑓°𝑔) (𝑥) = 𝑓 𝑔(𝑥) ∗ 𝑔′(𝑥)
Ex. 𝑓(𝑥) = (2𝑥 + 4) , then 𝑓 (𝑥) = 3(2𝑥 + 4) (4𝑥)
Ex. 𝑓(𝑥) =
(
)
, then 𝑓 (𝑥) =
( )(
) (
) ( )
3.2 Implicit Differentiation
Video: https://www.youtube.com/watch?v=xbviQHhU1rA
Implicit Differentiation: The processof finding
1.
2.
3.
4.
.
Differentiate both sides of the equasion with respect to x
Collet the terms with dy/dx on one side of the equasion
Factor out dy/dx
Solve for dy/dx
Ex. 𝑓(𝑥) = 𝑥 + 𝑦 − 9𝑥𝑦 = 0, then 𝑓 (𝑥) = 3𝑥
0 = 3𝑥 +
(3𝑦 − 9)
Ex. 4𝑥 + 3𝑦 = 9𝑥
−(
𝑥 + 6𝑦
)
+ 3𝑦
−9
=
= 18𝑥
=
Practice Problems:
Differentiate the following equasion.
1. 𝑓(𝑥) =
2. 𝑓(𝑥) = 𝑥 √5𝑥 − 𝑥
3.3 Derivatives of Inverse Trigonometric Functions
Video: https://www.youtube.com/watch?v=KbYW9FDm-Zk
=0
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Ex. 𝑓(𝑥) = 𝑠𝑖𝑛 (𝑥)
𝑓 (2𝜋) =
1
1 − (2𝜋)
Practice Problems:
Find the derivatives of the given function.
1. 𝑓(𝑥) = 5𝑥 − sec
(𝑥)
( )
2. 𝑓(𝑥) =
3.4 Derivatives of Exponential and Logarithmic Functions
Video: https://www.youtube.com/watch?v=zmnh448y_ZU
Derivative of a natural log: (ln = logbase(e) = 1)
Ex.
ln(𝑢) =
ln(3𝑥) =
(3)
Ex.
) (𝑥
Derivative of a logorthmic function
log 𝑢 = ( )
( )
log (
+ 1) =
(
)(
)
Derivative of an Exponential Function
(𝑎 ) = 𝑎 𝑙𝑛𝑎
Ex.
Practice Problems:
Find the derivative of the given function.
1. 𝑓(𝑥) = 4 log (𝑥) − ln 𝑥
2. 𝑓(𝑥) =
( )
(𝑥 + 1)( ) = (𝑥 + 1)( ) (ln(𝑥 + 1))(1)
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Unit 4: Applications of Derivatives
4.1 Extreme Values of Functions
Absolute Extreme Values
Let f be a function with domain D. Then f(c) is the
(a) Absolute maximum value on D if and only if 𝑓(𝑥) ≤ 𝑓(𝑐) for all x in D
(b) Absolute minimum value on D if and only if 𝑓(𝑥) ≥ 𝑓(𝑐) for all x in D
(Absolute/ global maximum and minimum values are also called absolute extrema)
Local extreme Values
If a function f has a local maximum or minumum value at an interior point c of it domain,
and if f’ exists at c, then 𝑓 (𝑐) = 0.
Critical Point
A Point in the interior of the domain of a function f at which f’=0 or f’ does not exist is a
critical point of f.
Candidates Test for Absolute Extrema
The absolute extrema of a fuction f on a closed interval can be determined by evaluating
the function at the critical number and at each endpoint of the interval. The Maximum of these
values is the absolute maximum, while the minimum of these values is the absolute minimum.
Video: https://www.youtube.com/watch?v=gB1F9IvHNqg
Ex. Determine the Absolute minumum and maximum values of 𝑓(𝑥) = (2𝑥 + 1) on the
interval [-2,2]
𝑓 (𝑥) = 2(2𝑥 + 1)(2)
𝑓(−2) = 9, 𝑓 −
(8𝑥 + 4) = 0
𝑥 = − (Critical point)
1
= 0, 𝑓(2) = 25
2
Absolute minimum at x=-1/2, Absolute Maximum at x=2
Practice Problems:
Determine the absolute extema of the given function on the specified interval
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1. 𝑓(𝑥) = 8𝑥 + 81𝑥 − 42𝑥 − 8 on [-8, 2]
2. 𝑓(𝑥) = 𝑥 (10 − 𝑥) on [-10, 15]
4.2 Mean Value Theorem
The Mean Value Theorem states that if f is continuous on [a,b] and differentiable on (a,b), then
there is at least one point between a and b at which the instantaneous rate of change of f is equal
to its average range of change over the entire interval. In other words, there is at least one value c
in the interval (a,b) for which 𝑓 (𝑐) =
( )
( )
.
Video: https://www.youtube.com/watch?v=SL2RobwU_M4
Ex. 𝑓(𝑥) = 𝑥 , [3,7]
Average Rate of Change
( )
( )
=
= 10
If f is
1. Differentiable everywhere and
2. Continuous
Then, the Mean Value Theorem guarantees that there is at least one c between 3 and 7 for which
𝑓 (𝑐) = 10
𝑓 (𝑥) = 𝑓′(𝑐)
𝑓 (𝑥) = 2𝑥 = 10
x=5
Increasing Function, Decreasing Function
1. F increases on I if 𝑥 < 𝑥 → 𝑓(𝑥 ) < 𝑓(𝑥 )
2. F decreases on I if 𝑥 < 𝑥 → 𝑓(𝑥 ) > 𝑓(𝑥 )
Antiderivative
A function is an antiderivative of a function f(x) if f’(x) = f(x) for all x in the domain of f.
The process of finding an antiderivative is antidifferentiation.
Practice Problems:
15
Determine all the number(s) c which satisfy the conclusion of the Mean Value Theorem for the
given function and interval.
1. 𝑓(𝑥) = 4𝑥 − 8𝑥 + 7𝑥 − 2 on [2, 5]
2. 𝑓(𝑥) = 8𝑥 + 𝑒
on [-2, 3]
4.3 Connecting f’ and f” with the Graph of f
First Derivative Test for Extrema
Video: https://www.youtube.com/watch?v=G5wlKltW7pM
On a continuous function f(x), at critical points c:
1. If f’ changes sign from possitive to negative at c (f’ > 0 for x < c and f’ < 0 and x > c),
then f has a local maximum value at c.
2. If f’ changes sign from negative to positive at c (f’ < 0 for x < c and f’ > 0 for x > c), the
nf has a minimum value at c
3. If f’ does not change sign at c (f’ has the same sign on both sides of c), then f has no local
etreme value at c.
4. At a left endpoint, if f’ < 0 (f’ > 0) for x > a, then f has a local maximum (minimum)
value at a.
5. At a right endpoint, if f’ < 0 (f’ > 0) for x > a, then f has a local minimum (maximum)
value at a.
Ex. 𝑓(𝑥) = (3𝑥 + 9𝑥 − 30)
𝑓 (𝑥) = 6𝑥 + 9
Crit numbers at -1.5
𝑓 (−2) = −3; 𝑓 (0) = 9
Local minimum at x=-1.5
Concavity: The graph of a function y = f(x) is
a. Concave up on an interval I if y’ is increasing on I.
b. Concave down on an interval I if y’ is decreasing on I.
Second Derivative Test for Concavity
The graph of a twice-differentiable function y = f(x) is
a. Concave up on any interval y” > 0
b. Concave down on any interval where y” < 0
Video: https://www.youtube.com/watch?v=G8GAsYkZlpE
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Point of Inflection
A point where the graph of a function has a tangent line and where the cancavity changes
is a point of inflection
Video: https://www.youtube.com/watch?v=-JcjvkBXdKU
Ex. 𝑓(𝑥) = 9𝑥 − 4𝑥 − 2𝑥 + 3
108𝑥 − 8 = 0
𝑓 (𝑥) = 36𝑥 − 8𝑥 − 2
Critical Number at 𝑥 = ±
𝑓"(𝑥) = 108𝑥 − 8
𝑓"(-1)=100, f"(0) = −8, 𝑓"(1) = 100
Concave up from −∞, −
Concave down from −
Point of inflection at −
,
,∞
,
and
Summery of Curve Sketching
Practice Problems:
Determine the inflection points of the function
1. 𝑓(𝑥) = 12 + 6𝑥 − 𝑥
2. 𝑓(𝑥) = 𝑥(𝑥 + 4)
4.4 Modeling and Optimization
Video: https://www.youtube.com/watch?v=lx8RcYcYVuU
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A Strategy for Optomization
1. Understand the Problem: Red the problem carefully. Identify the information you need to
solve the problem.
2. Develop a mathamatical model of the Problem: Draw pictures and label the paes that are
important to the problem. Introduce a variable to represent the quantity that can be
controlled. Using that variable, write a function whose extreme value gives the
information saught.
3. Graph the Function: Find the domain of the function. Determine what values of the
variable make sense in the problem.
4. Identify the Critical Points and the Endpoints: Find where the derivative is zero or fails to
exist.
5. Solve the Mathamatical Model: If unsure of the result, support or confirm your solution
with another method.
6. Interpret the Solution: Translate your mathematical result into the problem setting and
decide whether the result makes sense.
Alternate strategy:
1. Draw a picture.
2. Write a function for the quantity to be optimized (maximized or minimized).
3. Rewrite the function from the previous step to be in terms of a single independent variable.
This often involves using a secondary equation, called a constraint.
4. Determine the domain of interest.
5. Differentiate the function and find the relevant critical points.
6. Use the first derivative test, second derivative test, or candidates test to determine which of
the critical points or endpoints represent the optimal solution.
Ex. A manufacturer wants to construct a cylindrical container with a volume of 5 ft^3. Find the
dimensions of the container that will minimize the amount of material used.
1. The quantity to be optimized is the surface area of the
container. In terms of r and h, the surface area is given by the
function 𝑆 = 2𝜋𝑟 + 2𝜋𝑟ℎ
2. 𝑉 = 𝜋𝑟 ℎ = 5, therefor, ℎ =
and 𝑆 = 2𝜋𝑟 +
3. Domain is r > 0
4. 𝑆 (𝑥) = 4𝜋𝑟 −
, 4𝜋𝑟 −
= 0, 𝑟 = ±
5. Second derivative test 𝑆"(𝑥) = 4𝜋 +
= 12𝜋, sinse it it possitive, there is a
𝑆"(𝑥) = 4𝜋 +
minimum at x=
,
.
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Practice Problems:
1. We want to build a box whose base length is 6 times the base width and the box will
enclose 20 in3. The cost of the material of the sides is $3/in2 and the cost of the top and
bottom is $15/in2. Determine the dimensions of the box that will minimize the cost.
2. We want to construct a cylindrical can with a bottom but no top that will have a volume
of 30 cm3. Determine the dimensions of the can that will minimize the amount of
material needed to construct the can.
4.5 Linearization, Sensitivity, and Differentials
Video: https://www.youtube.com/watch?v=XQaCbFMnDo0
Differentials: Let y=f(x) be a differentiable function. The differential dx is an independent
variable. The differential dy is dy=f’(x)dx
Linearization: If f is differentiable at x=a, then the equasion of the tangent line,
𝐿(𝑥) = 𝑓(𝑎) + 𝑓′(𝑎)(𝑥 − 𝑎) defines the lineraization of f at a. The approximation
f(x)=L(x) is the standard linear approximaton of f at a. The point x=a is the center of the
approxomation.
Ex. Estimate the value of 𝑓(𝑥) = (𝑥 − 4) at x=6.003 without a calculator
𝑓 (𝑥 )(𝑥 − 𝑥 ) + 𝑓(𝑥 ),
𝑓 (𝑥) = 3(𝑥 − 4) (2𝑥),
𝑓(𝑥) = 36,864(𝑥 − 6) + 32,768
𝑓 (6) = 36,864
𝑓(6.003) = 36,864(0.003) + 32,768
Practice Problems:
Find the linear approxomation to the function at the given point.
1. 𝑓(𝑥) = 3𝑥𝑒
at x=5
2. 𝑓(𝑥) = 𝑥 − 6𝑥 + 3𝑥 − 7 at x=-3
4.6 Related Rates
Video: https://www.youtube.com/watch?v=ps-r4nti5Go
Ex. A tank of water in the shape of a cone is being filled with water at a rate of 12 m3/sec. The
base radius of the tank is 26 meters and the height of the tank is 8 meters. At what rate is the
depth of the water in the tank changing when the radius of the top of the water is 10 meters?
19
We want to determine h′ when r = 10 and we know that V′ = 12.
=
→𝑟=
ℎ
𝑉 = 𝜋𝑟 ℎ → 𝑉 =
𝜋ℎ
𝑉 =
𝜋ℎ ℎ′
To finish off this problem all we need to do is determine the value of h for the time we are
interested in. This can easily be done from the similar triangle equation and the fact that we
know r = 10.
ℎ=
12 =
4
4
40
(10) =
𝑟=
13
13
13
169 40
3
𝜋( ) ℎ = 100𝜋ℎ → ℎ =
16 13
25𝜋
Practice Problems:
1. The angle of elevation is the angle formed by a horizontal line and a line joining the
observer’s eye to an object above the horizontal line. A person is 500 feet way from the
launch point of a hot air balloon. The hot air balloon is starting to come back down at a
rate of 15 ft/sec. At what rate is the angle of elevation, θ, changing when the hot air
balloon is 200 feet above the ground. See the (probably bad) sketch below to help
visualize the angle of elevation if you are having trouble seeing it.
2. Two people on bikes are at the same place. One of the bikers starts riding directly north at
a rate of 8 m/sec. Five seconds after the first biker started riding north the second starts to
ride directly east at a rate of 5 m/sec. At what rate is the distance between the two riders
increasing 20 seconds after the second person started riding?
20
Unit 5: The Definite Integral
5.1 Estimating With Finite Sums
Rectangular Approxomation Method (RAM)
o Area is base*height, adding up smaller sections of area.
o RRAM
o Height (right) pt1*base + Height pt2*base + …
o LRAM
o Height (left) pt1*base + y2*Δx + …
o MRAM
o Height (center)*base + …
5.2 Definate Sums
The Riemann Sum
𝑎 = 𝑎 + 𝑎 + ⋯+ 𝑎
o
o
o
o
+𝑎
n = number of segments (upper limit)
ak = function
k = Index of summation
1 = lower limit
The Definate Intergral as a Limit:
lim
→
𝑓(𝑐 )∆𝑥 =
𝑓(𝑥)𝑑𝑥
Ex:
𝑓(𝑥) = 3𝑥 − 2𝑥 + 5 𝑜𝑣𝑒𝑟 𝑡ℎ𝑒 𝑖𝑛𝑡𝑒𝑔𝑟𝑎𝑙 [−1,3]
21
lim
→
(3(𝑚 ) − 2𝑚 + 5)∆𝑥 =
(3𝑥 − 2𝑥 + 5)𝑑𝑥
Definate Intergral as an Accumulator Function:
𝐴(𝑥) =
𝑓(𝑡)𝑑𝑡
5.3 Definite Integrals and Antiderivatives
Rules for definate intergrals
1. Zero: ∫ 𝑓(𝑥)𝑑𝑥 = 0
a. Ex: ∫ 2𝑥 + 5 = 0
2. Additivity: ∫ 𝑓(𝑥)𝑑𝑥 + ∫ 𝑓(𝑥)𝑑𝑥 = ∫ 𝑓(𝑥)𝑑𝑥
a. Ex: ∫ 2𝑥 + 𝑥 + 1 = ∫ 2𝑥 𝑑𝑥 + ∫ 𝑥 𝑑𝑥 + ∫ 1 𝑑𝑥
3. Order of Integration: ∫ 𝑓(𝑥)𝑑𝑥 = − ∫ 𝑓(𝑥)𝑑𝑥
a. Ex: ∫ 2𝑥 = − ∫ 2𝑥
4. Constant Multiple: ∫ 𝑘𝑓(𝑥)𝑑𝑥 = 𝑘 ∫ 𝑓(𝑥)𝑑𝑥, ∫ −𝑓(𝑥)𝑑𝑥 = − ∫ 𝑓(𝑥)𝑑𝑥
a. Ex: ∫ −3𝑥 = −3 ∫ 𝑥
5. Sum and Difference: ∫ 𝑓(𝑥) ± 𝑔(𝑥) = ∫ 𝑓(𝑥) ± ∫ 𝑔(𝑥)
a. Ex: ∫ 2𝑥 + 𝑥 + 1 = ∫ 2𝑥 𝑑𝑥 + ∫ 𝑥 𝑑𝑥 + ∫ 1 𝑑𝑥
6. Min-Max Inequality: If max f and min f are the maximum and minumum values of f on
[a,b], then
min 𝑓 ∗ (𝑏 − 𝑎) ≤
𝑓(𝑥)𝑑𝑥 ≤ max 𝑓 ∗ (𝑏 − 𝑎)
7. Domination: 𝑓(𝑥) ≥ 𝑔(𝑥) [𝑎, 𝑏] → ∫ 𝑓(𝑥)𝑑𝑥 ≥ ∫ 𝑔(𝑥)𝑑𝑥 ,
𝑓(𝑥) ≥ 0 [𝑎, 𝑏] → ∫ 𝑓(𝑥)𝑑𝑥 ≥ 0
Average (Mean) Value
If f is integrable on [a,b], its average (mean) value on [a,b] is
𝑎𝑣(𝑓) =
Ex: 𝑓(𝑥) = 4 − 𝑥 [0,3] →
1
𝑏−𝑎
∫ (4 − 𝑥 )𝑑𝑥 →
𝑓(𝑥) 𝑑𝑥
4𝑥 −
= 1
22
5.4 Fundamental Theorem of Calculus
FTC: Antiderivative part
o If f is continous on [a,b], then the accumulator function
𝐹(𝑥) =
𝑓(𝑡) 𝑑𝑡
Has a derivative at every point x in [a,b], and
𝑑𝐹
𝑑
=
𝑑𝑥 𝑑𝑥
𝑓(𝑡)𝑑𝑡 = 𝑓(𝑥)
Or
( )
𝑑
𝑑𝑥
Ex:
∫
𝑓(𝑡)𝑑𝑡 = 𝑓 𝑢(𝑥) ∗ 𝑢′(𝑥)
(𝑡 − 9)𝑑𝑡 = ((cos 𝑥) − 9)(− sin 𝑥)
FTC: Evaluation Part [Intergral Evaluation Theorem]
If f is continuous at every point of [a,b], and if F is any antiderivative of f on [a,b], then
𝑓(𝑥)𝑑𝑥 = 𝐹(𝑏) − 𝐹(𝑎)
Ex: ∫ 𝑥 𝑑𝑥 =
→
→
−
=
5.5 The Trapezoidal Rule
The Trapezoidal Rule:
To approxomate ∫ 𝑓(𝑥)𝑑𝑥, use
𝑇=
ℎ
(𝑦 + 2𝑦 + 2𝑦 + ⋯ + 2𝑦
2
+𝑦 )
Where [a,b] is particianed into n subintervals of equal length ℎ =
𝑇=
𝐿𝑅𝐴𝑀 + 𝑅𝑅𝐴𝑀
2
(
)
, Equivalently,
23
Unit 6: Differential Equasions and Mathematical Modeling
6.1 Slope Fields and Euler’s Method
Differential Equasion: an equasion involving a derivative is a differential equasion. The order of
a differential equasion is the order of the highest derivative involved in the equasion.
Slope field:
6.2 Antidifferentiation by Substitution
Indefinite Intergrals:
1. Power Formulas
Ex: ∫ 4𝑥 𝑑𝑥 =
+ 𝐶, ∫
𝑢 𝑑𝑢 =
𝑢
+ 𝐶 𝑤ℎ𝑒𝑛 𝑛 ≠ −1
𝑛+1
𝑢
1
𝑑𝑢 = 𝑙𝑛|𝑢| + 𝐶
𝑢
𝑑𝑢 =
𝑑𝑥 = ln|𝑥| + 𝐶
2. Exponential and Logarithmic Formulas
a. ∫ 𝑒 𝑑𝑢 = 𝑒 + 𝐶
Ex: ∫ 𝑒 𝑑𝑢 = 𝑒 + 𝐶
b. ∫ 𝑎 𝑑𝑢 =
+𝐶
Ex: ∫ 5 𝑑𝑥 → ∫ 5 ∗ 𝑑𝑢 =
+𝐶
c. ∫ ln 𝑢 𝑑𝑢 = 𝑢 ln 𝑢 − 𝑢 + 𝐶
Ex: ∫ ln(5𝑥) → ∫ ln 𝑢 ∗ 𝑑𝑢 =
(
)
(
) (
)
= 𝑥 ln 5𝑥 − 𝑥 + 𝐶
24
d. ∫ log 𝑢 𝑑𝑢 = ∫
𝑑𝑢 =
Ex: ∫ log (5𝑥) =
+𝐶
+𝐶
Trigonometric Formulas
o
o
o
o
o
o
∫ cos 𝑢 𝑑𝑢 = sin 𝑢 + 𝐶
∫ sin 𝑢 𝑑𝑢 = − cos 𝑢 + 𝐶
∫ sec 𝑢 𝑑𝑢 = tan 𝑢 + 𝐶
∫ csc 𝑢 𝑑𝑢 = − cot 𝑢 + 𝐶
∫ sec 𝑢 tan 𝑢 𝑑𝑢 = sec 𝑢 + 𝐶
∫ csc 𝑢 cot 𝑢 𝑑𝑢 = − csc 𝑢 + 𝐶
U-Substitution:
Example 1:
𝑥
5 + 2𝑥 𝑑𝑥 →
𝑥 (5 + 2𝑥 ) 𝑑𝑥
Substitute u and find derivative: 𝑢 = (5 + 2𝑥 ), 𝑑𝑢 = 6𝑥 ∗ 𝑑𝑥 → 𝑑𝑥 =
𝑥 (𝑢) ∗
1
𝑑𝑢 →
6𝑥
𝑥
1
(𝑢) 𝑑𝑢 →
(𝑢)
6𝑥
6
Find the antiderivative and plug back in u
1 2
1
∗ ∗ 𝑢 + 𝐶 = (5 + 2𝑥 ) + 𝐶
6 3
9
Example 2:
tan 𝑥 sec 𝑥 𝑑𝑥
Substitute u and find du: 𝑢 = tan 𝑥 𝑑𝑥, 𝑑𝑢 = sec 𝑥 𝑑𝑥
1
𝑢 sec 𝑥
𝑑𝑢 →
sec 𝑥
𝜋
𝜋
𝑢 3 (tan 𝑥) 3
𝑢𝑑𝑢 =
→
2 0
2
0
Evaluate the Function
𝜋
(tan )
3 − (tan 0) = √3 − 0 = 3
2
2
2
2
2
𝑑𝑢
25
6.3 Antidifferentiation by Parts
Intigration by parts formula:
𝑢 𝑑𝑣 = 𝑢𝑣 −
𝑣 𝑑𝑢
Example 1:
𝑥 𝑒 𝑑𝑥
Let 𝑢 = 𝑥 , 𝑑𝑣 = 𝑒 , 𝑑𝑢 = 2𝑥𝑑𝑥, 𝑣 = ∫ 𝑑𝑣 → 𝑒
𝑥 𝑒 𝑑𝑥 = 𝑥 𝑒 −
𝑒 2𝑥 𝑑𝑥 → 𝑥 𝑒 − 2
𝑒 𝑥 𝑑𝑥
Use Intigration by parts again: 𝑢 = 𝑥, 𝑑𝑢 = 𝑑𝑥, 𝑑𝑣 = 𝑒 , 𝑣 = 𝑒
𝑒 𝑥 𝑑𝑥 = 𝑥𝑒 −
𝑒 𝑑𝑥 = 𝑥𝑒 − 𝑒 + 𝐶
Therefore
𝑥 𝑒 −2
𝑒 𝑥 𝑑𝑥 → 𝑥 𝑒 − 2(𝑥𝑒 − 𝑒 + 𝐶) → 𝑥 𝑒 − 2𝑥𝑒 − 2𝑒 + 𝐶
Example 2:
𝑦=
𝑥 ln(𝑥) 𝑑𝑥
Let 𝑢 = ln 𝑥 , 𝑑𝑣 = 𝑥 𝑑𝑥, 𝑑𝑢 = 𝑑𝑥, 𝑣 =
ln(𝑥) 𝑥
−
2
𝑥
ln(𝑥) 𝑥
𝑑𝑥 →
−
2𝑥
2
𝑥
𝑑𝑥
2
ln(𝑥) 𝑥
−
2
𝑥
ln(𝑥) 𝑥
𝑥
𝑑𝑥 =
− +𝐶
2
2
4
Intigrate
6.4 Exponential Growth and Decay
Seperable Differentiable Equasions:
Example 1:
𝑥+4
𝑥𝑦
=
𝑑𝑦
𝑑𝑥
26
Separate x with dx and y with dy
𝑥+4
𝑑𝑥 = (𝑦) 𝑑𝑦
𝑥
Intigrate both sides
(𝑥 + 4)(𝑥
) 𝑑𝑥 =
𝑦 𝑑𝑦 →
𝑥 + 4𝑥
𝑑𝑥 =
𝑦 𝑑𝑦
2𝑥
2𝑦
+ 8𝑥 =
+𝐶
3
3
Example 2:
𝑑𝑦
=
𝑑𝑥
4 − 𝑦 cos √𝑥
𝑦 √𝑥
Separate
𝑦
4−𝑦
𝑑𝑦 =
cos √𝑥
√𝑥
𝑑𝑥 → 𝑦 (4 − 𝑦) 𝑑𝑦 = cos(𝑥 )𝑥
𝑑𝑥
Intigrate
𝑦 (4 − 𝑦) 𝑑𝑦 =
cos 𝑥
𝑥
𝑑𝑥
u-substitution: 𝑢 = 4 − 𝑦, 𝑑𝑢 = −𝑑𝑦; Therefore 𝑦 = (𝑢 − 4)
and 𝑢 = 𝑥 , 𝑑𝑢 =
𝑑𝑥
− (𝑢 − 4) 𝑢
𝑑𝑢 = 2
cos 𝑢 𝑑𝑢 → − (𝑢 − 8𝑢 + 16)𝑢 𝑑𝑢 = 2 sin √𝑥
− (𝑢 − 8𝑢 + 16𝑢 )𝑑𝑢 = 2 sin √𝑥 → −
𝑢 𝑑𝑢 + 8
𝑥 𝑑𝑢 − 16
𝑢
𝑑𝑢 = 2 sin √𝑥
Finish intigration
−
2𝑢
16𝑢
+
− 32(𝑢) = 2 sin √𝑥 + 𝐶
5
3
Undo u-substitution
−
2 4 − 𝑦(3𝑦 + 16𝑦 + 128)
2(4 − 𝑦)
16(4 − 𝑦)
+
− 32(4 − 𝑦) → −
= 2sin √𝑥 + 𝐶
5
3
15
27
Law of Exponential Change:
If
= 𝑘𝑦, and if 𝑦 = 𝑦 at t=0, then
𝑦 = 𝐶𝑒
Continuously Compunding Interest
𝑟
𝐴(𝑡) = 𝑃(1 + )
𝑛
Compounding continuously:
𝐴(𝑡) = 𝑃𝑒
Example 1: deposite $800 in an account that pays 6.3% annual interest. How much will you have
8 years later if the interest is a) compounding continuously b) compounding quarterly?
A) 𝐴(8) = 800𝑒 ( . )( ) = 1324.26
.
B) 𝐴(8) = 800(1 +
)( )( ) = 1319.07
Radioactive decay (Half-life) equasion
ℎ𝑎𝑙𝑓 − 𝑙𝑖𝑓𝑒 =
ln 2
𝑘
Example: Using Carbon-14 dating
Carbon-14 half life of 5700 years. Find the age of a sample in which 10% of the radioactive
nuclei originally present have decayed
1
( )
2
= 0.9 →
𝑡
1
ln(0.9)
ln = ln 0.9 → 𝑡 = 5700
→ 𝑡 = 866
5700 2
ln(0.5)
Unit 7: Applications of Definite Integrals
7.1 Accumulation and Net Change
Intigrals can be used to model area, and other products.
Example 1:
The velocity of a particle is
= 𝑣(𝑡) = 𝑡 −
(
)
, find the displacement of the particle if
the initial possition is s(0)=9 at t=1 sec.
8
(𝑡 + 1)
1 8
11
= + −8= −
3 2
3
𝑣(𝑡)𝑑𝑡 = 𝑑𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑚𝑒𝑛𝑡 →
𝑡 −
𝑑𝑡 →
𝑡
8
1
+
3 𝑡+1 0
28
Hooke’s Law:
𝐹 = 𝑘𝑥
Example 2: Work
It takes 10N to stretch a spring 2m, how much work is done stretching it 4m?
𝐹(2) = 10 = 𝑘 ∗ 2 → 𝑘 = 5 →
5𝑥 𝑑𝑥 =
5𝑥 4
= 40𝑁 ∗ 𝑚
2 0
7.2 Areas in the Plane
Area Between Curves:
If f and g are continuous with 𝑓(𝑥) ≥ 𝑔(𝑥)[𝑎, 𝑏], then the area between the curves y=f(x) and
y=g(x) is the intigral of [f-g] from a to b.
𝐴=
[𝑓(𝑥) − 𝑔(𝑥)] 𝑑𝑥
Example 1:
Find the area bounded by the equasions 𝑦 = (𝑥 + 1)√𝑥 + 5𝑥 and 𝑦 = 6x sin(𝑥 + 1)
Find the bounds of integration by finding where the formulas are eqaul, [graph]
Find the greater function using the domination rule, plug in a point in both equasions
Two points intersect at [0,0] and [0.59, 1.974]
Set up the intigral:
.
.
(𝑥 + 1) 𝑥 + 5𝑥 𝑑𝑥 −
Solve using your calculator or u substitution
6x sin(𝑥 + 1) 𝑑𝑥
29
1
5
.
.
√𝑢 − 6
sin(𝑢) →
2𝑢
2(𝑥 − 1)
0.59
+ 6 cos 𝑢 →
+ 6 cos(𝑥 + 1)
15
15
0
Example 2: Integrating with respect to y
Find the area bounded by 𝑦 = √𝑥, 𝑦 = 𝑥 − 2, and the x-axis.
𝑦 = 𝑥 − 2 becomes 𝑥 = 𝑦 + 2
𝑦 = √𝑥 becomes 𝑥 = 𝑦 , 𝑦 > 0
Y values measured from 0 to 2, so bounds of intigration is [0,2]
𝑎𝑟𝑒𝑎 =
(𝑦 + 2 − 𝑦 ) 𝑑𝑦 →
𝑦
𝑦
+ 2𝑦 −
2
3
2 10
=
0
2
7.3 Volume of an intergral
Volume of a solid:
The intigral of an area function
𝑉=
𝐴(𝑥) 𝑑𝑥
How to find volume by the method of slicing
1.
2.
3.
4.
Sketch the solid and a typical cross section.
Find a formula for A(x).
Find the limits of intigration.
Integrate A(x) to find the volume.
Example 1:
The region between the graph of 𝑓(𝑥) = 2 + 𝑥 cos 𝑥 and the x-axis over the interval [-2,2] is
revolved about the x-axis to generate a solid. Find the volume of the solid.
Area of a circle is 𝜋𝑟
30
𝐴(𝑥) =
𝐴(𝑥) 𝑑𝑥 →
𝜋(𝑓(𝑥)) 𝑑𝑥 →
𝜋(2 + 𝑥 cos 𝑥) 𝑑𝑥 ≈ 52.429
Example 2:
The region in the first quadrant enclosed by the y-axis and the graphs of 𝑦 = 1 + cos 𝑥 and
𝑦 = 1 + sin 𝑥 is resolved about the x-axis to form a solid. Find its volume.
𝐴(𝑥) = 𝜋(cos 𝑥 + 1) − 𝜋(sin 𝑥 + 1)
𝑉(𝑥) = 𝜋
(cos 𝑥 + 1) − (sin 𝑥 + 1) 𝑑𝑥
(cos 𝑥 + 1) (sin 𝑥 + 𝑥) − (sin 𝑥 + 1) (− cos 𝑥 + 𝑥)
𝜋
3
𝜋
4 ≈ 4.173
0
Example 3:
The region bounded by the graph of 𝑦 = 3
, the x-axis, and the line x=4, is revolved around
the line x=4 to generate a solid. Find the volume of the solid
Radius = 4 - x
𝑦=3
𝑥
4𝑦
→𝑥=
4
9
31
𝐴(𝑦) = 𝜋 4 −
4𝑦
9
→
16𝜋
(9 − 𝑦 )
81
Volume of the solid
𝑉=
16𝜋
16𝜋
(9 − 𝑦 ) 𝑑𝑦 →
81
81
(81 − 18𝑦 + 𝑦 ) 𝑑𝑦
16𝜋
𝑦
81𝑦 − 6𝑦 +
81
5
3 128𝜋
=
0
5
Other Cross Sections:
The base of a solid is bounded by 𝑦 = 1 − 𝑥 and the x-axis. Find the volume of the solid if the
cross sections are equilateral triangles perpendicular to the x-axis.
Area of an equalateral triagle is given by
𝐴(𝑥) =
𝑎 √3
4
The side a is equal to 𝑦 = 1 − 𝑥 , therefore
𝐴(𝑥) =
(1 − 𝑥 ) √3
4
The parabola 𝑦 = 1 − 𝑥 intersects the x-axis at x = -1 and x = 1
𝑉(𝑥) =
𝐴(𝑥) 𝑑𝑥 →
(1 − 𝑥 ) √3
√3
𝑑𝑥 →
4
4
𝑥
𝑥
√3
𝑥− +
4
3
5
1
4√3
=
−1
15
(1 − 𝑥 + 𝑥 ) 𝑑𝑥
32
7.5 Applications from Science and Statistics
Work Revisisted: Work = Force*Distance, 𝑊 = ∫ 𝐹(𝑥) 𝑑𝑥
Example 1: Work Done Lifting
A leaky bucket weighs 22 newtons empty. It is lifted from the ground at a constant rate to a point
20m above the ground by a rope weighing 0.4 N/m. The bucket starts with 70N of water, but it
leaks at a constant rate and just finishes draining as the bucket reaches the top. Find the work
done
a) Lifting the bucket
b) Lifting the water alone
c) Lifting the rope alone
d) Lifting the bucket, water, rope together
(a). Lifting the bucket: 𝑊 = ∫ 22 → 22𝑥
(b). Lifting the water: 𝐹(𝑥) = 70
𝑊=
= 440𝐽
→ 70 1 −
= 70 − 3.5𝑥 𝑁
(70 − 3.5𝑥) 𝑑𝑥 → [70𝑥 − 1.75𝑥 ]
20
→ 1400 − 700 = 700𝐽
0
(c). Lifting the Rope: 𝐹(𝑥) = (0.4)(20 − 𝑥)𝑁
𝑊=
(0.4)(20 − 𝑥) 𝑑𝑥 → [8𝑥 − 0.2𝑥 ]
20
= 160 − 80 = 80𝐽
0
(d). All together: 440 + 700 + 80 = 1220 J
Example 2: Work Done Pumping
A conical tank of radius 5ft and height of 10ft is filled to within 2 ft of the top with olive oil
weighing 57 𝑙𝑏/𝑓𝑡 . How much work is done to pump the oil to the rim of the tank?
Calculate the volume
∆𝑉 = 𝜋𝑟 ∆𝑥 → 𝜋
1
𝑦
2
∆𝑦 →
𝜋
𝑦 ∆𝑦
4
Calculate the force
𝐹(𝑦) = 57∆𝑉 →
57𝜋
𝑦 ∆𝑦
4
Force acts until the oil is at the rim, (10 − 𝑦)
𝑊=
57𝜋
57𝜋
𝑦 (10 − 𝑦) 𝑑𝑦 →
4
4
(10𝑦 − 𝑦 ) 𝑑𝑦 →
57𝜋 10𝑦
𝑦
−
4
3
4
8
≈ 30,561𝐽
0
33
Unit 8: Sequences, L’Hopital’s Rule, and Improper Integrals
8.2 L’Hopital’s rule
If lim 𝑓(𝑥) = lim 𝑓(𝑥) = 0, or lim
→
→
→
( )
( )
is indeterminate, and f and g are differentiable on an
open interval I containing a, then:
𝑓(𝑥)
𝑓′(𝑥)
= lim
→ 𝑔(𝑥)
→ 𝑔′(𝑥)
lim
List of Indeterminate Forms:
0 ∞
, , ∞ − ∞, 0 , 1 , ∞ , 0 ∗ ∞, 1 ∗ ∞
0 ∞
Example 1:
lim
→
lim
→
ln 𝑥
2√𝑥
𝑥
= lim
→
√𝑥
→ lim
→
ln 𝑥
2√𝑥
→
∞
∞
2√𝑥
√𝑥
= lim
→
𝑥
1
→ lim
→
1
2 √𝑥
→
1
2√∞
=0
Example 2:
𝑓(𝑥) = lim √𝑥 → lim 𝑥 → ∞
→
lim ln 𝑓(𝑥) = lim
→
→
→
ln 𝑥
𝑥
= lim ln 𝑓(𝑥) = lim
→
→
𝑥
1
lim 𝑓(𝑥) = 𝑒
→
→𝑒 =1
→ lim 𝑓(𝑥) = lim 𝑒
→
→
34
Main Ideas and Things to Memorise
Derivative → ← Intigral
cos 𝑥 → − sin 𝑥
sin 𝑥 → cos 𝑥
cot 𝑥 → − csc 𝑥
csc 𝑥 → csc 𝑥 cot 𝑥
sec 𝑥 → sec 𝑥 tan 𝑥
Inverse Trig:
𝑑
sin
𝑑𝑥
𝑥=
1
√1 − 𝑥
𝑑
cos
𝑑𝑥
𝑥=
𝑑
tan
𝑑𝑥
𝑥=
𝑑
cot
𝑑𝑥
−1
𝑥=
1+𝑥
𝑑
sec
𝑑𝑥
𝑥=
𝑑
csc
𝑑𝑥
𝑥=
−1
√1 − 𝑥
1
1+𝑥
1
|𝑥|√𝑥 − 1
−1
|𝑥|√𝑥 − 1
Exponential Functions:
𝑑
𝑑𝑢
(𝑒 ) = 𝑒
𝑑𝑥
𝑑𝑥
𝑑
𝑑𝑢
(𝑎 ) = 𝑎 ln 𝑎
𝑑𝑥
𝑑𝑥
𝑑
𝑒 ( ) = 𝑒 ( ) 𝑔 (𝑥)
𝑑𝑥
𝑎 ( ) = 𝑎 ( ) ln 𝑎 𝑔 (𝑥)
Equasion of a line:
𝑦 = 𝑚(𝑥 − 𝑥 ) + 𝑦
Liniar Approxomation:
𝑓 (𝑥 )(𝑥 − 𝑥 ) + 𝑓(𝑥)
Logarithmic Functions:
𝑑
1
(ln 𝑥) = , 𝑥 > 0
𝑑𝑥
𝑥
𝑑
𝑔 (𝑥)
ln 𝑔(𝑥) =
𝑑𝑥
𝑔(𝑥)
𝑑
1
log 𝑥 =
,𝑥 > 0
𝑑𝑥
𝑥 ln 𝑥
𝑑
𝑔′(𝑥)
log 𝑔(𝑥) =
𝑑𝑥
𝑔(𝑥) ln 𝑎
Fundemental Theory:
( )
𝑑
𝑓(𝑡) 𝑑𝑡
𝑑𝑥
= 𝑓 𝑢(𝑥) 𝑢 (𝑥)
𝑓(𝑥) 𝑑𝑥 = 𝐹(𝑏) − 𝐹(𝑎)
Power Formula
𝑢 𝑑𝑢 =
Product rule:
𝑑
[𝑓(𝑥)𝑔(𝑥)] = 𝑓(𝑥)𝑔 (𝑥)
𝑑𝑥
+ 𝑔(𝑥)𝑓′(𝑥)
𝑢
=
𝑒 𝑑𝑢 = 𝑒 + 𝐶
𝑑
𝑥 = 𝑛𝑥
𝑑𝑥
𝑎 𝑑𝑢 =
Chain Rule:
𝑓 𝑔(𝑥) = 𝑓 𝑔(𝑥) 𝑔 (𝑥)
Definition of a Derivative
𝑓(𝑥 + ℎ) − 𝑓(𝑥)
→
ℎ
lim
Mean Value Theorm:
𝑓(𝑏) − 𝑓(𝑎)
𝑏−𝑎
Mean Value Theorm for
definite intigrals:
1
𝑏−𝑎
1
= ln|𝑥| + 𝑐
𝑢
Exponential and Logs
Power Rule:
𝑓(𝑥) =
𝑢
+𝐶
𝑛+1
𝑓(𝑥) 𝑑𝑥
Second Derivative Test:
𝑓 (𝑥) > 0 = 𝑐𝑜𝑛𝑐𝑎𝑣𝑒 𝑢𝑝
𝑓 (𝑥) < 0 𝑐𝑜𝑛𝑐𝑎𝑣𝑒 𝑑𝑜𝑤𝑛
𝑎
+𝐶
ln 𝑎
ln 𝑢 = 𝑢 ln 𝑢 − 𝑢 + 𝐶
log 𝑢 =
𝑢 ln 𝑢 − 𝑢
+𝐶
ln 𝑎
Exponential Change, half-life
𝑦 = 𝐶𝑒 , 𝑡 =
ln 2
𝑘
Volume of a Solid
𝑉(𝑥) =
𝐴(𝑥) 𝑑𝑥
Area between curves:
𝐴=
[𝑓(𝑥) − 𝑔(𝑥)] 𝑑𝑥
L’Hopital’s Rule:
𝑓(𝑥)
𝑓′(𝑥)
= lim
→ 𝑔(𝑥)
→ 𝑔′(𝑥)
lim
35