Calculus Basics 微积分基础

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Notes on basic Calculus concepts.

Study resources

• [x] Khan academy: AP Calculus BC
• [x] Kristan King: Calculus I (Differential Calculus)
• [x] Kristan King: Calculus II (Integral Calculus)
• [x] Kristan King: Calculus III (Multivariable Calculus)
• [x] The Organic Chemistry Tutor: New Calculus Playlist
• [x] Mathispower4u: Calculus playlist
• [x] Math@Xaktly
• [x] Geoff(x): Teach math using Desmos
• [ ] MIT OCW 18.01SC: Single Variable Calculus
• [ ] MIT OCW 18.02SC: Multivariable Calculus
• [ ] Coursera: Mathematics for Machine Learning: Multivariate Calculus

Study Tools

• [x] Symbolab
• [x] Desmos
• [x] GeoGebra
• [ ] MIT Mathlets
  • [ ] TAYLOR POLYNOMIALS
  • [ ] HARMONIC FREQUENCY RESPONSE: VARIABLE INPUT FREQUENCY

Practice To-do List (Linked with Unit tests)

• [x] Limits and continuity
• [x] Differentiation: definition and basic derivative rules
• [x] Differentiation: composite, implicit, and inverse functions
• [x] Contextual applications of differentiation
• [x] Applying derivatives to analyze functions
• [x] Integration and accumulation of change
• [x] Differential equations
• [x] Applications of integration
• [x] Infinite sequences and series

Table of Contents

• ▶ Limit & Continuity
  • Limit properties & Limits of Combined Functions
  • Limits at infinity
  • All types of discontinuities
• ▶ Differential Calculus
  • Differentiability
  • Local linearity & Linear approximation
  • Basic Differential Rules
  • Chain Rule
  • Derivatives of Trig functions
  • Implicit differentiation
  • Higher Order Derivatives
  • Derivative of Inverse functions
  • Derivative of exponential functions
  • Existence Theorems
  • L'Hopital's Rule
  • Critical points
    • Extrema: Maxima & Minima
    • Concavity
    • Inflection Point
  • Second Derivative Test
  • Anti-derivative
  • Analyze Function Behaviors with Derivatives
  • Optimization
  • Applications of Derivatives
    • Motion problems
    • Planar motion
• ▶ Integral Calculus
  • Definite Integrals
  • Antiderivatives
  • Fundamental Theorem of Calculus (FTC)
  • Basic Integral Rules
  • Calculate Integrals
  • Integration using Trig identities
  • Improper Integral
  • U-substitution → Chain Rule
  • Integrate by Parts → Product Rule
  • Partial fractions → Log Rule
  • Trig-substitutions → Trig Rule
  • Average Value of Functions
• ▶ Differential Equations
  • Parametric Equations Differentiation
  • Separable Differential Equations
  • Specific antiderivatives
  • Polar Curve Functions (Differential Calc))
  • Logistic Growth Model
  • Slope Field
  • Euler's Method
• ▶ Applications of definite integrals
  • Area under Rate function
  • Motion problems
  • Planar motion
  • Function Area between curve & axes
  • Area of Polar Curves (Integral Calc)
  • Arc Length
  • Parametric Curve Arc Length
  • Volumes of 3D Solids
  • Disc Method
  • Washer Method
  • Shell Method
• ▶ Series (Calculus)
  • Infinite Seires
  • Infinite Geometric Series
  • Convergence Tests
    • nth Term Test
    • Integral Test
    • p-series Test
    • Comparison Test
    • Ratio Test
    • Root Test
    • Alternating Series Test
  • Absolute vs. Conditional Convergence
    • Error Estimation of Alternating Series
    • Error Estimation Theorem
    • Interval of Convergence
  • Power Series
    • Taylor Series
    • Maclaurin Series
    • Lagrange Error Bound
    • Finding Taylor series for a function
    • Function as a Geometric Series
    • Maclaurin Series of Common functions
    • Euler's Formula & Euler's Identity
• Multivariable functions
  • Parametric Functions
  • Partial derivatives
  • Gradient

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「Limits」 properties

Refer to Khan academy: Limit properties

The limit of a SUM of functions is the SUM of the INDIVIDUAL limits:

Limits of 「Combined Functions」

Refer to Khan academy: Limits of combined functions

Example

Solve:

Example

Solve:

• The limit for each function DOES NOT exists.
• However, the one-side limits of each graph DO exist
• We can use the fact that the limit of a sum of functions is the sum of the individual limits:
• Calculate each side's combined limits:

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Function's Continuity [DRAFT]

Pencil Definition: IF YOU CAN DRAW THE FUNCTION WITH A PENCIL WITHOUT PICKING UP THE PENCIL, THEN THE FUNCTION IS A CONTINUOUS FUNCTION

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Limits

Limits are all about approaching. And the entire Calculus is built upon this concept.

Strategy in finding Limits

Refer to Khan academy.

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Limits at 「infinity」

No matter why kinds of Limits you're looking for, to understand it better, the best way is to read the Step-by-Step Solution from Symbolab: Limit Calculator from Symbolab.

「Rational functions」

The KEY point is to look at the powers & coefficients of Numerator & Dominator. Just the same with Finding the Asymptote.

Refer to previous note on the How to find Asymptote.

Example

Solve:

Quotients with 「square roots」

The KEY point is to calculate both numerator & dominator, then calculate the limit of EACH term with in the square root.

Example

Solve: Refer to Symbolab step-by-step solution.

• Divide by highest dominator power to get:
• Calculate separately the limit of Numerator & Dominator:
• Calculate the Square root: Need to find limits for EACH term inside the square root.
• Then get the result easily.

Quotients with 「trig」

The KEY point is to apply the Squeeze theorem, and it is a MUST.

Example

Solve:

• Know that -1 ≦ cos(x) ≦ 1, so we can tweak it to apply the squeeze theorem to get its limit.
• Make the inequality to: 3/-1 ≦ 3/cos(x)/-1 ≦ 3/1
• Get that right side 3/-1 = -1 and left side 3/1 =1 is not equal.
• So the limit doesn't exist.

Easier solution steps:

• Know the inequality -1 ≦ cos(x) ≦ 1
• Replace cos(x) to ±1 in the equation, 3/±1.
• Calculate limits of two sides.
• If the results are exactly the same, then the limit is the result; Otherwise the limit doesn't exist.

Example

Solve:

• Know that -1 ≦ sin(x) ≦ 1
• Replace sin(x) as ±1
• Left side becomes (5x+1)/(x-5), right side becomes (5x-1)/(x-5)
• Both sides' limits are 5, so the limit exists, and is 5.

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❖ All types of discontinuities

Refer to Mathwarehouse: What are the types of Discontinuities?

「Jump Discontinuities」

A Jump discontinuity occurs at some point, the left side limit is DIFFERNT with the right side limit.

We see that clearly:

「Removable Discontinuities」

A removable discontinuity occurs at some point, both left side limit & right side limit are the SAME.

「Infinite Discontinuities」

An Infinite Discontinuity occurs at some point, both left side & right side are approaching to INFINITY, which means both sides DO NOT have limits.

「Endpoint Discontinuities」

An Endpoint Discontinuity occurs at some point, it DOESN'T HAVE both sides, it only has ONE SIDE.

「Mixed Discontinuities」

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Analyzing functions for discontinuities「algebraic」

If the limits of both side of some point, are EQUAL, then it's continuous at this point.

At a point a, for f(x) to be continuous at x=a, we need lim(x→a)f(x) = f(a).

Example

Solve:

• Got the limit of left side is -1/2
• Got the limit of right side is -1/2
• They're equal, so it's continuous at this point.

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Application of Removable Discontinuities

Example

Solve: It's just the same with calculating the limits of both sides.

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❖ 「Derivative」 Basics

Simply saying, it's just the SLOPE of ONE POINT of a graph (line or curves or anything).

Refer to Mathsisfun: Introduction to Derivatives

A Derivative, is the Instantaneous Rate of Change, which's related to the tangent line of a point, instead of a secant line to calculate the Average rate of change.

“Derivatives are the result of performing a differentiation process upon a function or an expression. ”

Derivative notations

Refer to Khan academy article: Derivative notation review.

「Lagrange's」 notation

In Lagrange's notation, the derivative of f(x) expressed as f'(x), reads as f prime of x.

「Leibniz's」 notation

In this form, we write dx instead of Δx heads towards 0. And the derivative of is commonly written as:

For memorizing, just see d as Δ, reads Delta, means change. So dy/dx means Δy/Δx. Or it can be represent as df / dx or d/dx · f(x), whatever.

How to understand 「dy/dx」

Refer to Khan academy from Differential Equation section: Addressing treating differentials algebraically

This is a review from "the future", which means while studying Calculus, you have to come back constantly to review what the dy/dx means. ---- It's just so confusing. Without fully understanding the dy/dx, you will be lost at topics like Differentiate Implicit functions, Related Rates, Differential Equations and such.

「Tangent line」 & 「Secant line」

• The secant line is drawn to connect TWO POINTS, and gets us the Average Rate of Change between two points.
• The Tangent line is drawn through ONE POINT, and gets us the Rage of change at the exact moment.

As for the secant line, its interval gets smaller and smaller and APPROACHING to 0 distance, it actually is a process of calculating limits approaching 0, which will get us the tangent line, that been said, is the whole business we're talking about: the Derivative, the Instantaneous Rate of Change.

「Secant line」

Example

Solve:

• What it's asking is the Slope of its secant line:
• which could be applied with this simple formula:
• the result is ( f(3)-f(1) )/ (3-1) = -1/12

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❖ Differentiability

"If the point of a function IS differentiable, then it MUST BE continuous at the point."

Example of NOT differentiable points:

You can see, if the point DOES NOT have limit, it's NOT DIFFERENTIABLE. In another word, the point is not CONTINUOUS, it's Jump Discontinuity, or Removable Discontinuity, or any type of discontinuities.

「NOT」 differentiable situations

• Vertical Tangent (∞)
• Not Continuous
• Two sides' limits are different

「Vertical Tangent」

We know that the Slope of Vertical Tangent is UNDEFINED, on the contrary: IT IS A VERTICAL TANGENT, IF:

• The derivative dy/dx = undefined, or
• The denominator of derivative's expression = 0.

「Horizontal Tangent」

It's a Horizontal Tangent, if:

• dy/dx = 0.

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❖ Derivative equation

The idea of derivative equation is quite simple: The LIMIT of the SLOPE.

The slope is equal to change in Y / change in X. So for a point a, we IMAGINE we have another near point which lies on the SAME LINE with a, and since we have TWO POINTS now, we can then let their Y-value Change divided by their X-value Change to get the slope.

There're two equations for calculating derivative at a point, and the only different thing is how to express the IMAGINARY POINT with respect to the point a, it could either be x or a+h :

or:

How to calculate 「derivative」

Strategy:

• Determine if it's CONTINUOUS at this point, by:
  • See if the point is defined in the interval
  • Calculate LIMITS of both RIGHT SIDE and LEFT SIDE of the point.
  • If two sides' limits are the same, then it's continuous. Otherwise it's discontinuous.
• Determine if it's DIFFERENTIABLE (Actually is the process of getting its derivative):
  • Apply Derivative equation to get both RIGHT SIDE LIMIT and LEFT SIDE LIMIT.
  • If two sides' limits are the same, then that value is the Derivative at the point. If not, then it's NOT DIFFERENTIABLE.

Example

Solve:

• See that the point 3 is defined in the interval.
• Left side limit of the point, is using the first equation, and gets the lim g(x) = -7
• Right side limit of the point, is using the second equation, and gets the lim g(x) = -7
• Limits of both sides are the SAME, so it's continuous, and let's see if it's differentiable.
• Apply the derivative equation for both Left side & Right side:
• Both sides' limits exists but not that same, so it's not differentiable.

Example

Solve:

• See that the point -1 is defined in the interval.
• Left side limit of the point, is using the first equation, and gets the lim g(x) = 1
• Right side limit of the point, is using the second equation, and gets the lim g(x) = 4
• Limits of both sides are NOT SAME, so it's not continuous, then of course not differentiable.

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「Local linearity」 & 「Linear approximation」

Local linearity is for approximating of a point's value by its near known point.

Just think of a curve, a good way to approximate its Y-value, is to find another known point near it, and make a line connecting two points, then gets the value by linear equation.

Refer to Khan academy lecture.

Example

Solve:

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Basic 「Differential Rules」

Jump over to Basic Integral Rules

▼Refer to Math is fun: Derivative Rules

Rules	Function	Derivative
Constant	c	0
With constant	c·f	c·f’
Power Rule	xⁿ	n·xⁿ⁻¹
Sum Rule	f + g	f’ + g’
Difference Rule	f - g	f’ − g’
Product Rule	f · g	fg’ + f’ g
Quotient Rule	f / g	(f’g − g’f ) / g²
Reciprocal Rule	1 / f	−f’ / f²
Chain Rule	f(g(x))	f’(g(x)) · g’(x)
Exponent Rule	eˣ	eˣ
	aˣ	aˣ · ln(a)
Log Rule
Natural Log Rule	ln(x)	1/x
Exponential Rule
Trig Rules	sin(x)	cos(x)
	cos(x)	−sin(x)
	tan(x)	sec²(x) = 1/cos²(x)
	sec(x)	sec(x)·tan(x)
	csc(x)	-csc(x)·cot(x)
	cot(x)	−csc²(x) = -1/sin²(x) = -(1+cot²(x))
Inverse Trig Rules	arcsin(x)	1/√(1−x²)
	arccos(x)	−1/√(1−x²)
	arctan(x)	1/(1+x²)

▼Refer to Wiki: Differentiation rules

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Chain Rule

One of the core principles in Calculus is the Chain Rule.

Refer to Khan academy article: Chain rule ▶ Proceed to Integral rule of composite functions: U-substitution

It tells us how to differentiate Composite functions.

It must be composite functions, and it has to have inner & outer functions, which you could write in form of f(g(x)).

Common 「mistakes」

• Not recognizing whether a function is composite or not
• Wrong identification of the inner and outer function
• Forgetting to multiply by the derivative of the inner function
• Computing f(g(x)) wrongly:

How to identify 「Composite functions」

Seems a basic algebra101, but actually a quite tricky one to identify.

Refer to Khan lecture: Identifying composite functions

The core principle to identify it, is trying to re-write the function into a nested one: f(g(x)). If you could do this, it's composite, if not, then it's not one.

Examples

It's a composite function, which the inner is cos(x) and outer is x².

It's a composite function, which the inner is 2x³-4x and outer is sin(x).

It's a composite function, which the inner is cos(x) and outer is √(x).

Two forms of 「Chain Rule」

The general form of Chain Rule is like this:

But the Chain Rule has another more commonly used form:

Their results are exactly the same. It's just some people find the first form makes sense, some more people find the second one does.

Example

Solve: Refer to Symbolab worked example.%5Cright))

Chain rule for 「exponential function」

Formula:

Because:

Example

Solve:

• Apply the Log power rule to simplify the exponential function:
• Differentiate both sides:

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Derivatives of 「Trig functions」

Reminder of 「Trig identities」 & 「Unit circle」 values

# Reciprocal and quotient identities
tan(θ) = sin(θ) / cos(θ)
csc(θ) = 1/sin(θ)
sec(θ) = 1/cos(θ)
cot(θ) = 1/tan(θ)

Refer to previous note of all trig identities.

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❖ Implicit Differentiation

Bit hard to understand it in the first place.

What is 「Implicit」 & 「Explicit Function」

Refer to video by Krista King: What is implicit differentiation?

• Explicit function: it's the normal function we've seen a lot before, which's in the form of y = x....
• Implicit function: it't NOT YET in the general form of a function and not easily separated, like x² + y² = 1

So knowing how to differentiate an implicit function is quite helpful when we're dealing with those NOT EASILY SEPARATED functions.

How to Differentiate 「Implicit function」

Refer to video: Use implicit differentiation to find the second derivative of y (y'') (KristaKingMath) Refer to video by The Organic Chemistry Tutor: Implicit Differentiation Explained - Product Rule, Quotient & Chain Rule - Calculus

Refer to Symbolab: Implicit Derivative Calculator

Assume you are to differentiate Y WITH RESPECT to X, written as dy/dx:

• Differentiate terms with X as normal
• Differentiate terms with Y as the same to X, BUT multiply by (dy/dx)
• Differentiate terms MIXED with X & Y by using Product Rule, then differentiate each term.

How to differentiate 「Y with respect to X」

How to differentiate 「term MIXED with both X & Y」

Example

Solve: Refer to Symbolab: Implicit Derivative Calculator

• Treat y as y(x)
• Apply the Sum Rule:
• Apply the normal rules to X term, and
• Apply the Product Rule to the Mixed term, and
• Apply the Chain Rule to the Y term:
• Operate the equation and solve for dy/dx, and get:

Example

Solve:

• First thing we need to find the RIGHT equation of Chain rule. Since it's asking us to find dy/dt, so we will re-write it to this one to form an equation:
• Then since we've given the dx/dt = -3, we only need to find out the dy/dx to get the result.
• We've got an equation of x & y, regardless whom it's respecting to. So we can do either Implicit or Explicit differentiation to the equation y²=7x+1, with respect to y:
• Use the implicit differentiation method, we got the dy/dx = 7/2y
• And since y=6, so 7/2y = 7/12
• Back to the Chain Rule equation, we get dy/dt = 7/12 · (-3) = -7/4 = -1.75

Example

Solve:

• Remind you that, in this problem, it's NOT respecting to x anymore, so you need to change mind before getting confused.
• First thing we need to find the RIGHT equation of Chain rule. Since it's asking us to find dx/dt, so we will re-write it to this one to form an equation:
• Then since we've given the dy/dt = -0.5, we only need to find out the dx/dy to get the result.
• We've got an equation of x & y, regardless whom it's respecting to. It seems easier to differentiate explicitly:
• Then we use d/dx to differentiate the equation to get: dx/dy = y⁻² = (0.2)⁻² = 25
• Back to the Chain Rule equation, we get dx/dt = dx/dy · dy/dt = 25 * (-0.5) = -12.5.

Example

Solve (Same with above examples):

• Form an equation:
• dx/dt has been given equals to 5, so just to find out dy/dx:
• And get:
• Now let's see what is sin(x) equal to:
• All done.

「Vertical & Horizontal Tangents」 of 「Implicit Equations」

► Jump over to Khan academy for practice.

Example

Solve:

• Plug in y = 0 into the equation and get that x = -6, which is the answer.

Example

Solve:

• To have a Vertical Tangent, we have to let the derivative become Undefined,
• which in this case is to let the denominator equal to zero:
• Solve this equation out we get that x = 3y², which means this relationship is true at the point of vertical tangent line.
• Plug that back to the original function to get y = -1, which means the vertical tangent goes through this point.
• Substitute y back and get x = 3
• The answer is (3, -1).

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❖ Related Rates

Just so you know, related rates is actually the Application of Implicit Differentiation by using Chain Rule in the form of dy/dx = dy/du * du/dx.

Btw, at Khan academy it's called the Differentiate related functions.

Refer to Khan lecture. Refer to video by KristaKingMath: Related rates Refer to video by The Organic Chemistry Tutor: Introduction to Related Rates

Strategy:

• Abstract every given condition algebraically, e.g. s(t), s'(t), V(t), V'(t)...
• Find a proper equation to CONNECT all these given conditions
• Differentiate both sides of equation, with using the Chain rule.
• Take back the given values to the Differentiated equation to get the result.

Example: 「Change of volumes」

Refer to previous note of Implicit Differentiation.

Solve:

• Write out all conditions algebraically:
  • r'(t)=1
  • r(t)=5
  • h'(t)=-4
  • h(t)=8
  • V(t) = π·r²·h
• So they're asking for V'(t), it then becomes V'(t) = π·d/dt (r²·h)
• Apply Product rule & Chain rule then substitute: V'(t) = π((r²)'·h + r²h') = π(2r'·r + r²·h') = -20π Notice that: (r²)' is a composite function, so you want to apply the Chain rule, =2r·r'

Example: 「Change of volumes」

Solve:

• From the given conditions, we got that r'(t)=-12, r(t)=40 and h=2.5.
• We know the equation of cylinder's volume is: V= π·[r(t)]²·h
• Differentiate both side of the equation to get:
• Take back all the known values into the equation get V'(t)=-2400π

Example: Change of area

Solve:

• Write out all conditions algebraically:
  • d₁'(t) = -7
  • d₁(t) = 4
  • d₂'(t) = 10
  • d₂(t) = 6
  • A(t) = (d₁·d₂)/2
• So it's asking for instantaneous rate of change, then it is A'(t) = (d₁'·d₂ + d₁·d₂')/2
• Apply the Product rule only, to get A'(t) = -1.

Example: 「Pythagorean Theorem」

Refer to Khan academy lecture: Related rates: Approaching cars

Solve:

• It's a problem to calculate two objects' dynamic absolute distance: it's getting closer and closer until distance become 0, which means they crashes at 0.
• Write down all the conditions algebraically:
  • c₁'(t) = -10
  • c₁(t) = 4
  • c₂'(t) = -6
  • c₂(t) = 3
  • D(t) = √(c₁²+c₂²)
• So D'(t) = d/dt · (√(c₁²+c₂²)) = (2c₁'c₁+2c₂'c₂)/(2√(c₁²+c₂²))

Example: 「Sliding ladder」

Example: 「multiple composite」

Solve:

• Write down all the conditions algebraically:
  • S'(t) = π/2
  • S(t) = 12π
  • S = 2π·r, which means r = S/2π
  • A(t) = π·r²
• So A'(t) = d/dt · (π·r²) = 2π·r·r' = 2π·6·(S/2π)' = 6S' = 3π

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❖ Higher Order Derivatives

• First order derivative: f'(x) or dy/dx, also called "First Derivative"
• Second order derivative: f''(x) or d²y/dx², also called "Second Derivative"
• ...

「Second derivatives」

The second derivative of a function is simply the derivative of the function's derivative.

Notation: Leibniz's notation for second derivative is:

Second derivatives of 「implicit equations」

▶ Jump back to previous note on Implicit Differentiation ▶ Practice at Khan academy: Second derivatives (implicit equations)

▶ Online calculator for 2nd Derivative of Implicit equations

Example

Solve:

Example

Solve:

• The first derivative is:
• The second derivative is:
• Plug in y=1, and get:

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Derivative of Inverse functions

IT'S DERIVED FROM THE CHAIN RULE:

Derivative of 「Inverse Trig functions」

Example

Solve:

• For solving this derivative of an inverse function, we'd use the identity:
• Apply it we get:

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Derivative of 「exponential functions」

It's save a lot of time of life not to dig in how mathematicians developed these formulas. If you do want to, refer to Khan's lecture: Exponential functions differentiation intro

Reminder: Don't forget it's a composite function and you need to apply the chain rule.

Derivative of 「log functions」

Examples

Find the derivative of: Solve:

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❖ Existence Theorems (IVT, EVT, MVT)

Existence theorems includes 3 theorems: Intermediate Value Theorem, Extreme Value Theorem, Mean Value Theorem.

Refer to Khan academy: Existence theorems intro

「Intermediate Value Theorem」 (IVT)

The IVT is saying:

When we have 2 points connected by a continuous curve: one point below the line, the other point above the line, then there will be at least one place where the curve crosses the line!

Refer to Maths if fun: Intermediate Value Theorem Refer to video: Intermediate Value Theorem Explained

Find 「roots」 by using IVT

IVT is often to find roots of a function, which means to find the x value when f(x)=0. So for finding a root, the definition will be:

If f(x) is continuous and has an interval [a, b], which leads the function that f(a)<0 & f(b)>0 , then it MUST has a point f(c)=0 between interval [a,b], which makes a root c.

Example

Tell whether the function f(x) = x² - x - 12 in interval [3,5] has a root. Solve:

• We got that at both sides of intervals: f(3)=-6 < 0, and f(5)=8 > 0
• So according to the Intermediate Value Theorem, there IS a root between [3,5].
• Calculate f(c)=0 get the root c=4.

「Extreme Value Theorem」 (EVT)

The EVT is saying:

There MUST BE a Max & Min value, if the function is continuous over the closed interval.

Refer to Khan lecture: Extreme value theorem Refer to video: Extreme Value Theorem

「Mean Value Theorem」 (MVT)

Refer to Khan academy article: Establishing differentiability for MVT

The MVT is saying:

There MUST BE a tangent line that has the same slope with the Secant line, if the function is CONTINUOUS over [a,b] and DIFFERENTIABLE over (a,b).

Which also means that, if the conditions are satisfied, then there MUST BE a number c makes the derivative is equal to the Average Rate of Change between the two end points.

Conditions for applying MVT:

• Continuous over interval (a, b)
• Differentiable over interval [a, b]

Example

Solve:

• He's totally right.

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❖ L'Hopital's Rule [DRAFT]

LHopital's Rule helps us to find the limit of an Undefined limits, like 0/0, ∞/∞ and such. It's quite simple to apply and very convenient to solve some problems.

Refer to L'Hôpital's rule

▶ Jump back previous note on: Asymptote of Rational Expressions ▶ Practice at Khan academy: Disguised derivatives

From my experience, the L'Hopital's Rule is so often been used that we didn't even realize. Actually it's been used almost every time when we are to evaluate the LIMITS OF RATIONAL EXPRESSIONS.

Example of 「0/0」

Example of 「∞/∞」

1. Find the limit: Solve:

2. Find the limit: Solve:

Example of 「1^∞」

L'Hopital Rule for 「Composite functions」

Example of composite exponential function

Solve:

• Direct plug in the x=2 and get limit = 1^∞, which is an indeterminate form
• So we're gonna apply the Log Power Rule to take down the power:
• We use Natural Log for doing this:
• And now we can apply the L'Hopital Rule for ln(y):
• From ln(y) we could get limit of y:

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❖ Critical points

Refer to PennCalc Main/Optimization

For analyzing a function, it's very efficient to have a look at its Critical points, which could be classified as Extrema, Inflection, Corner, and Discontinuity.

How to find 「critical points」

Strategy:

• Knowing that f(x) has critical point c when f'(c) = 0 or f'(c) is undefined
• Differentiate f(x) to get f'(x)
• Solve c for f'(c)=0 & f'(c) undefined

Refer to Symbolab's step-by-step solution.

Example

Solve:

• See that original function f(x) is undefined at x = 2 or -2
• Differentiate f(x) to get f'(x):
• Solve f'(x)=0 only when x=0.
• f'(x) is undefined when x=2 or -2, as the same with f(x) so it's not a solution.

Example

Solve: Refer to Symbolab step-by-step solution.

• Differentiate f(x) to get f'(x):
• f'(x) is undefined when x > 4
• Solve f'(x)=0 get x = 8/3
• So under the given condition, only x=8/3 is the answer.

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❖ Extrema: 「Maxima」 & 「Minima」

Extrema are one type of Critical points, which includes Maxima & Minima. And there're two types of Max and Min, Global Max & Local Max, Global Min & Local Min. We can all them Global Extrema or Local Extrema.

And actually we can call them in different ways, e.g.:

• Global Max & Local Max or in short of glo max & loc max
• Absolute Max & Relative Max or in short ofabs max & rel max

How to identify 「Extrema」

We need two kind of conditions to identify the Max or Min. Now If we have a Non-Endpoint Minimum or Maximum point at a, then it must satisfies these conditions:

• Geometric condition: (It should be understood in a more intuitive way)
  • in the interval [a-h, a+h] there's no point above or below f(a) or
  • f'(a-h) & f'(a+h) have different sign, one negative another positive.
• Derivative condition:
  • f'(a) = 0 or
  • f'(a) is undefined

How to find 「Extrema」

Refer to Khan academy lecture: Finding critical points

We just need to assume f'(x) = 0 or f'(x) is undefined, and solve the equation to see what x value makes it then.

「Increasing」 & 「Decreasing Intervals」

We can easily tell at a point of a function, it's at the trending of increasing or decreasing, by just looking at the instantaneous slope of the point, aka. the derivate. If the derivative, the slope is positive, then it's increasing. Otherwise it's decreasing.

Finding the 「trending」 at a point

Just been said above, we assume at point a, it's value is f(a). So the slope of it is f'(a). And if f'(a) < 0, then it's decreasing; If f'(a) > 0, then it's increasing.

Finding a 「decreasing or increasing interval」

Refer to Khan lecture.

It's just doing the same thing in the opposite way. For find a decreasing interval, we assume f'(x) < 0, and by solving the inequality equation we will get the interval.

Strategy:

• Get Critical points.
• Separate intervals according to critical points, undefined points and endpoints.
• Try easy numbers in EACH intervals, to decide its TRENDING (going up/down):
  • If f'(x) > 0 then the trending of this interval is Increasing.
  • If f'(x) < 0 then the trending of this interval is Decreasing.

Example

Solve:

• Set f'(x) = 0 or undefined, get x = -2 or -1/3
• Separate intervals to (-∞, -2, -1/3, ∞)
• Try some easy numbers in each interval: -3, -1, 0:

How to find 「Relative Extrema」

Remember that an Absolute extreme is also a Relative extreme.

Refer to khan: Worked example: finding relative extrema Refer to Khan Academy article: Finding relative extrema

Strategy:

• Get Critical points.
• Separate intervals according to critical points, undefined points and endpoints.
• Try easy numbers in EACH intervals, to decide its TRENDING (going up/down).
• Decide each critical point is Max, Min or Not Extreme.

Refer to an awesome article: Using calculus to learn more about the shapes of functions

Example

Solve:

• Set f'(x)=0 or undefined, get x=0 or -2 or 1
• Separate intervals to (-∞, -2, 0, 1, ∞)
• Try easy number in each interval: -3, -1, 0.5, 2 and get the trendings:
• Identify critical points' concavity:

How to find 「Absolute Extrema」

Refer to Khan academy article: Absolute minima & maxima review Refer to Khan academy lecture: Finding absolute extrema on a closed interval

Strategy:

• Find all Relative extrema
  • Get Critical points.
  • Separate intervals according to critical points & endpoints.
  • Try easy numbers in EACH intervals, to decide its TRENDING (going up/down).
  • Decide each critical point is Max, Min or Not Extreme.
• Input all the extreme point into original function f(x) and get extreme value.

Example

Solve:

• Set g'(x) = 0 or undefined get x=0
• Separate intervals to [-2, 0, 3] according to the critical point & endpoints of the given condition.
• Try some easy numbers of each interval -1, 2 into g'(x)
• Get the trending of each interval: (+, +)
• So the minimum must be the left endpoint of the given condition, which is x=-2.

Example

Solve:

• Differentiate to set f'(x)=0 and got x=0, -1, 1
• Separate intervals to (-∞, -1, 0, 1, +∞)
• Try some easy numbers in each interval for f'(x) and got the signs: -, +, -, +
• According to the signs we know that -, + gives us a Relative maximum
• But at the end it's + again, and the interval is going up to +∞, means f(x) will go infinitely high.
• So there's NO Absolute maximum.

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Concavity

Refer to Khan academy: Concavity introduction

Two types of concavity: Concave Up & Concave Down.

Understand function by 「1st & 2nd Derivative」

Identify 「concavity」 by using 「Second Derivative」

Example

Solve:

• f'(x) > 0 means we're to find the INCREASING interval on the graph
• f''(x) < 0 means we'll be focusing on the CONCAVE UP shape only
• So it's about the interval of(-4, -3) and (3, 4).

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❖ Inflection Point

An inflection point is a point where the graph of the function changes CONCAVITY (from up to down or vice versa).

It could be seen as a Switching point, which means the point that the Slope of function switch from increasing and decreasing. e.g., the function might be still going up, but at such a point it suddenly increases slower and slower. And we call that point an inflection point.

Refer to Khan academy video for more intuition rapidly: Inflection points from graphs of function & derivatives

Algebraically, we identify and express this point by the function's First Derivative OR Second Derivative.

Example

Intuitive way to solve:

• Draw a tangent line in imagination and move it on the function from left to right
• Notice the tangent line's slope, does it go faster or slower or suddenly change its pace at a point?
• We found it suddenly changed at point c.

More definitional way to solve:

• Looking for the parts of concavity shapes
• Seems that B-C is a part of Concave Down, and C-D is a part of Concave Up
• So C is a SWITCHING POINT, it's a inflection point.

Example

Solve:

• Looking for the parts of concavity shapes
• There's no changing of concavity shapes, there's only one shape: Concave down.

Example

Solve:

• Notice that's the graph of f'(x), which is the First Derivative.
• Checking Inflection point from 1st Derivative is easy: just to look at the change of direction.
• Obviously there're only two points changed direction: -1 & 2

Example

Solve:

• Mind that this is the graph of f''(x), which is the Second derivative.
• Checking inflection points from 2nd derivative is even easier: just to look at when it changes its sign, or say crosses the X-axis.
• Obviously, it crosses the X-axis 5 times. So there're 5 inflection points of f(x).

Example: Finding Inflection points

Solve:

• Function has POSSIBLE inflection points when f''(x) = 0.
• Set f''(x) =0 and solve for x, got x=-3.
• We now know the possible point, but don't know its CONCAVITY. This need to try some numbers from its both sides:
• So it didn't change the concavity at point -3, means there's no inflection point for function.

Example: Finding Inflection points

Solve:

• Function has POSSIBLE inflection points when f''(x) = 0.
• Set f''(x) =0 and solve for x, got x=0 or 6. Refer to Symbolab for f''(x). Refer to Symbolab for f''(x)=0.
• We now know the possible point, but don't know its CONCAVITY. This need to try some numbers from its both sides:
• So it didn't change the concavity at point 0, means only 6 is the inflection point.

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Second Derivative Test

Example

Solve:

• f'(c) = 0 means c is a critical point, could be max, min, inflection.
• f''(c) < 0 means around point c it's a Downward Concave.
• Conclusion then is that c is a maximum point.

Example

Solve:

• f'(c) = 0 means c is a critical point, could be max, min, inflection.
• f''(c) = 0 means it's either a Concave up or down, we don't know yet.
• So the answer is "No enough information to tell."

Example

Solve:

• It's Concave Down when f''(x) < 0
• Find formula of f''(x) and set inequality equation f''(x)<0 and solve to get x > 2.

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❖ Anti-derivative

Looks like a fancy word, but it just means: Use the Derivative function describe the Original function.

Notice that: There might be multiple possible anti-derivatives.

The 「graphical relationship」 between a function & its derivative

Refer to Khan academy: The graphical relationship between a function & its derivative (Part 1)

How to draw the function's 「Anti-derivative」

Refer to Khan academy: The graphical relationship between a function & its derivative (part 2)

Example

Refer to Khan academy.

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❖ Analyze Function Behaviors with Derivatives

Analyzing function's behaviors is one of the Core Purposes of studying Calculus.

AND THE CORE PURPOSE OF ANALYZING FUNCTION, IS FOR COMPUTER TO UNDERSTAND IT "BLINDLY", OR SAY "ALGEBRAICALLY"! BECAUSE IT CAN'T BE LIKE HUMAN TO "EYE BALL" IT!

First Derivative

• Function has critical points when f'(x) = 0
• Function has relative extrema when f'(x) crosses X-axis:
  • It has relative maxima when f'(x) crosses UP.
  • It has relative minima when f'(x) crosses DOWN.

Second Derivative

• Function has inflection points when f''(x)=0 and f''(x) crosses X-axis.
• It's concave up if f''(x) > 0
• It's concave down if f''(x) < 0

1st & 2nd Derivative

• It has a relative maximum when f'(x)=0 & f''(x) < 0
• It has a relative minimum when f'(x)=0 & f''(x) > 0
• It has an inflection point when f'(x) changes direction, OR f''(x) changes sign.
• It has a possible inflection point when f'(x) = 0 & f''(x) = 0.

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❖ Optimization [DRAFT]

Once you've mastered the Derivatives, you would know the optimization problems are easy. It's just a progress to get the Maximum or Minimum points of a given function.

Example: Max product

Solve:

• It's bit tricky to understand the question.
• Form a function: P(x) = x(a-x) = ax - x²
• To get a maximum, need to find the critical points first.
• Set P'(x) = 0 then get x = a/2.
• To perform a 2nd Derivative Test: P''(x) = -2, means it's Concave Down
• So the critical point a/2 is a maximum.

Example: Largest Area of Trapezoid Inscribed in a Semicircle

Q: What is the area of the largest trapezoid that can be inscribed in a semicircle with radius r = 1?

Refer to Kristaking's video: Largest area of a rectangle inscribed in a semicircle

Understanding:

• For this Trapezoid inscribed in circle problem, you really want to draw it out before anything else.Refer to this animated tool from Geogebra that I created for this problem.
• Remember that a trapezoid has to have TWO BASES to be parallel.
• Know that, a quadrilateral CAN be inscribed in a circle or even a semicircle, which means 4 vertices are all on the circle.
• A trapezoid of maximum area inscribed in the semicircle will have its base on the X-axis. Which means the length of bottom base should be twice radius: b₁ = 2·r
• Since it's a trapezoid, and inscribed in a circle, then IT HAS TO BE A ISOSCELES TRAPEZOID. And it means it's absolute symmetric about the X-axis.
• With knowing all these conditions above, you could start to abstract the information into equations.
• Get some clue from the result first, then see what's missing and find a way to get it.

Solve:

• First form the equation of trapezoid's area: A = 1/2 · (b₁+b₂) · h
• b₁ is the bottom base, which is equal to 2r = 2, because the largest-area-trapezoid inscribed in circle MUST:
  • lay its base on X-axis
  • equal to the Diameter (because it's the longest base's length)
  • symmetric about the Y-axis (because two bases are parallel)
• Assume one of the vertex on the Top base is (x, y).
• Since the two bases are parallel, so the Top Base is symmetric as well, means it has equal distance to both sides' vertices, which could conclude that:
  • The other vertex on the Top base should be(-x, y).
  • The length of Top Base should be b₂ = x - (-x) = 2x
• The h is hight of the trapezoid, which is equal to the top vertex's y value.
• So the equation then becomes: A = 1/2 · (2r + 2x) · y.
• We need to form a function as Area in term of x, means the Area would change with the change of x. So the y has to transform to the term of x.
• Since (x, y) is a point of the circle, so the circle's Standard formula should work: x² + y² = r²
• Then we get: y = √(r²-x²) = √(1-x²)
• So the final function looks like this: A(x) = 1/2 · (2r + 2x) · √(r²-x²) = (1+x)·√(1-x²)
• Back to the beginning, since we are to find the largest area, so it's saying we are to find the Maximum value of the function A(x).
• Set A'(x) = 0 to find the critical point first. After solve the first derivative by applying product rule, we get: x=-1 or 1/2.
• Since x is a length, so it's a positive value, so x = 1/2. Then y = √(1-x²) = √3/2
• Do the Second derivative test to make sure it's a maximum: A''(x) < 0.
• Substitute the values back to the area equation, we get A = 3√3 / 4

Example: Smallest paper

Solve:

• Assume the area of the inside rectangle (the texts) is T = w · h = 150
• So the area of paper should be A = (w+2)·(h+3)
• Let h = T/w = 150/w
• So the function of area should be A(w) = (w+2)·(150/w +3)
• For getting the area's smallest value, we need to find the critical points first:
• Set A'(w) = 0, and differentiate the function to get: A'(w) = 3 - 300/w², and w=10.
• Substitute the value back to get h=15, So the paper should be (10+2) wide, and (15+3) high.

Example: Largest area of rectangle inscribed in triangle

Solve:

• Form the equation of area of the rectangle: A = w · h
• Apply the Similar triangle property from Geometry lessons, telling that the ratio between two sides are the same with its similar triangle's. So use any of the small triangle there, to form the equation: h/(8-w) = 10/8.
• And we make h = 5(8-w)/4, and the area equation then be A(w) = 5/4 (8w - w²)
• ....

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Applications of Derivatives [DRAFT]

Lose of bears

Solve:

• Just take the derivative B'(t) and input t=2 to get the value B'(2) ≈ -361 bears/year.

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Motion problems (Differential calc)

Motion Problems are all about this relationships: Moving position -> Velocity(or speed) -> Acceleration.

These terms are constantly confusing people, especially the follow parts:

• Velocity is NOT the derivative of speed, but only the speed with a direction: s(t) = |v(t)|.
• Velocity IS the derivative of Position: v(t) = p'(t)
• Acceleration is the derivative of the Velocity: a(t) = v'(t)
• Max or Min Position means Velocity = v(t) = p'(t) = 0
• Max or Min Velocity means Acceleration = a(t) = v'(t) = 0
• Max or Min Acceleration means a'(t) = v''(t) = p'''(t) = 0

Jump over here for Khan academy's quizzes.

Example

Solve:

• The tricky part here is the relationships: Position -> Velocity -> Acceleration
  • Position: p(t) = x(t)
  • Velocity: v(t) = x'(t)
  • Acceleration: a(t) = v'(t) = x''(t)
• To conclude, the Max velocity should satisfy this: a(t) = 0 & a'(t) < 0
• Differentiate x(t) twice and set x''(t) = 0, get t = 1.

Example

Solve:

• The velocity is v(t) = x'(t)
• The Acceleration is a(t) = v'(t) = x''(t) = 0, and get t=1
• Substitute to v(1) = 3

Example

►Refer to the note: Related rates.

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Planar motion (Derivative of vectors)

It's still the Motion problem but the object not only moves on the X-axis but move in a PLANE, with X-coordinate and Y-coordinate. So it becomes differentiation of vectors. But the differentiation steps are almost the same.

Here are some algebraical expressions:

• Position: P(t) = (x, y)
• Velocity: v(t) = P'(t) = (x', y')
• Acceleration: a(t) = v'(t) = P''(t) = (x'', y'')

Jump to do the Khan academy practice.

Example

Solve:

• Write down all the conditions algebraically:
  • Position: P(t) = (x, y) = (-t²+10t, t³-10t)
  • Velocity: v(t) = P'(t) = (x', y') = (-2t+10, 3t²-10)
  • t=4
• Substitute to get v(4) = (2, 38)

Example

Solve:

• P(t) = (2t²-6t, -t³+10t)
• v(t) = P'(t) = (4t-6, -3t²+10)
• v(2) = (2, -2)
• |v(2)| = √(4+4) = 2√2

Example: Motion along a curve

Refer to Khan academy's quizzes for these practices Solve:

• Position: P(t) = (x, y)
• The rate of change means velocity: v(t) = P'(t) = (x', y')
• Since x' = -2, so it becomes v(t) = (-2, y'). How to get the y'?
• We could find an equation x²y²=16 helps us to get y'.
• It's easier to do implicit differentiation than explicit one: (x²y²)'=(16)' -> 2x·x'·y² + x²·2y·y' = 0 -> -4xy² + x²·2y·y' = 0 -> y' = 2y/x
• Substitute (1,4) to the y's rate of change to get y' = 2*4/1 = 8

Example

Solve:

• P(t) = (x, y)
• x' = 1/2
• v(t) = P'(t) = (x', y') = (1/2, y')
• y'(t) = d/dt (-2x⁴+10) = -2·4·x³·x' = -4x³
• y'(x=-1, y=8) = -4(-1)³ = 4
• So v(t) = (1/2, 4)
• |v(x=-1, y=8)| = √(1/4 + 16)

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❖ 「Integral Calculus」 basics

Integral calculus is a process to calculate the AREA between a function and the X-axis (or Y-axis).

Core idea of 「Integral Calculus」

Refer to Khan academy: Introduction to integral calculus

「Riemann Sums」

A Riemann sum is an approximation of the area under a curve by dividing it into multiple simple shapes (like rectangles or trapezoids).

「Riemann Sums」 Notation

Refer to Khan academy: Definite integral as the limit of a Riemann sum

The letter ʃ (reads as "esh" or just "integral") is called the Integral symbol/sign.

Calculate 「Riemann Sums」

Finding 𝚫x: It's meant to get HOW MANY rectangles we're to sum.

Finding indices m & n: It's meant to find the i for Σ sums:

• For Left Sums or Midpoint Sums: i starts from 0 ends with subdivisions - 1
• For Right Sums: i starts from 1 ends with subdivisions

Finding xi: With equally spaced points (left/right/mid), the xi is a Geometric series of those points, which the rate is the 𝚫x. We're gonna find the right pattern/equation for xi, so that we can plug xi into f(x).

Finding f(xi): Just to plug in the Geometric series expression of xi into f(x), and make it as a function in terms of i.

「Left Riemann Sums」 & 「Right Riemann Sums」 Approximation

Refer to Maths is fun: Integral Approximations

• Left Riemann Sum: take the Left boundary value of Δx to be the rectangle's height.
• Right Riemann Sum: take the Right boundary value of Δx to be the rectangle's height.

As you can see, they would be either Over-estimated or Under-estimated. Neither of these approximations would be called a good one, normally.

「Midpoint Sums」 Approximation

It's an enhancement to the Left sums and Right sums, it takes the midpoint value, and sometimes makes better approximation.

Example

Solve:

Example

Solve:

• It's easy to find the Δx=2.
• Then let's find the f(x𝖎). It's actually a progress to find the Arithmetic Sequence.
• So the sequnce is S(𝖎) = a + 𝖎·Δx = 2 + 2𝖎, where a represents the first x value which is 2.
• So x𝖎 = S(𝖎) = 2+2𝖎
• Takes it back to the function and gets: f(x𝖎) = |2+2i-5| = |2i -3|

Example

Solve:

Example

Solve:

How to calculate 「Riemann Sums」

Refer to Khan academy: Rewriting definite integral as limit of Riemann sum

Refer to the Map of Integration: mrozarka.com

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Definite Integrals

DEFINITE means it's defined, means both two boundaries are constant numbers.

Definite integrals 「properties」

Refer to Khan academy article: Definite integrals properties review

「Definite integral」 ←→ 「Limit of Riemann Sum」

Example

Solve:

• It's easy to find Δx = (π-0)/n = π/n
• And x𝖎 = S(𝖎) = a + 𝖎·Δx = 0+𝖎·Δx = 𝖎·π/n
• So the result is:

Example

Solve:

• Look at the boundaries, it's from 0 -> 5,
• So the Δx must be cut to n pieces, whereas Δx = (5-0)/n = 5/n
• From the definition, We know the function f(x) = x+1
• To fill in the x𝖎 in f(x𝖎), we need to figure out the sequence:
• Sequence x𝖎 = S(𝖎) = a+𝖎·Δx, and since a represents the bottom boundary,
• So x𝖎 = 𝖎(0+Δx) = 𝖎·5/n
• Get x𝖎 back in f(x) to have:

Example

Solve:

• We could easily get that the Δx = 5/n
• And the function is f(x) = ln(x)
• Since the Δx comes from Top & Bottom boundaries,
• So Δx = (Top - Bottom)/n = 5/n = (Top - 2)/n,
• And we get the Top = 7, and the Definite Integral is:

Example

Solve:

• See the i=1 means it's using Right Riemann Sum, so the integral would be:
• The Δx = 9/n is easily seen.
• And we need to get the Sequence x𝖎 = S(𝖎) = a + (𝖎-1)·Δx = (𝖎-1)·9/n
• What we got there above, tells us a=0.
• According to thatΔx = (b-a)/n = (b-0)/n = 9/n, we get b = 9
• So the answer is:

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❖ 「Indefinite integrals」 = 「Antiderivatives」

That's easier being said, THE ANTIDERIVATIVES IS THE INDEFINITE INTEGRALS.

Let's make it more intuitive (might not be accurate but good for learning):

• Antiderivatives is the RESULT.
• Indefinite integrals is the OPERATION.

(They're saying the same thing)

And, just for refreshing:

• Anti-derivatives: Means the Original function where the derivative is from.
• Indefinite integrals: Indefinite means not-defined, means both BOUNDARIES are not defined. That's why the symbol is without any number but ʃ alone.

Why is the 「Indefinite Integral」 so confusing

It's a simple reason: Because they use the f(x) in the Integral expression, but actually it means f'(x)!

You all know the expression of indefinite integral is ʃ f(x) dx, But actually it should be ʃ f'(x) dx, which means the function appears in the middle is a derivative, from somewhere. And the mission of that integral, is to find the f(x) the original function of the derivative!

So trust me, the world would be much nicer if you always see it as the expression as below:

Why is 「Antiderivative」 so confusing too

Because your first impression of the antiderivative is that is it anti- something? Anti- is a reverse, derivative is also a transform of something, so putting them together is really a horrible idea because it seems leading to nowhere.

Now here is the mojo, things would be much nicer if you see and call an antiderivative as: The original function f(x)

It may not be accurate, but good enough to proceed to next stage of study.

Review of 「Antiderivatives」

Before you proceed to the next, you really want figure out completely what an antiderivative means with respect to the Integration.

!! Refer to video from The Organic Chemistry Tutor: Antiderivatives

Here are a few examples to quick review what is antiderivative:

How to understand this 「reverse process」

Doing an Integration, is actually to find the antiderivative.

At the example below, you will find it makes so much sense if you FIX YOUR EYES only onto the MIDDLE part of the integration formula, the part between ʃ & d/dx.

Example

Solve:

• First need to clear your mind, in this case:
  • The antiderivative is the "original function" f(x) we have learned.
  • The function f(x) here is actually the derivative, which is f'(x) we've learned.
• Just think it as matching function and its derivative, everything will be tackled real quick.

Example

Solve:

• First need to make sense of the terms:
  • The function appears in the Integral expression, is actually the derivative
  • The "result" is just the Original function where the derivative comes from, or you could call the result is the antiderivative.
• Just think it as matching function and its derivative, everything will be tackled real quick.

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❖ Fundamental Theorem of Calculus (FTC)

This is somehow dreaded and mind-blowing. But it's the only thing to relate the Differential Calculus & Integral Calculus.

It's so much clearer if you see the function in the middle of integration as a derivative.

Notice that: In this theorem, the lower boundary a is completely "ignored", and the unknown t directly changed to x.

►Refer to Khan academy: Fundamental theorem of calculus review ►Jump over to have practice at Khan academy: Contextual and analytical applications of integration (calculator active).

「1st FTC」 & 「2nd FTC」

The Fundamental Theorem of Calculus could actually be used in two forms. They have different use for different situations.

(Notice that boundaries & terms are different)

How to 「Differentiate Integrals」

We could CONVERT the integral formula to Differential formula, by using the fundamental theorem of calculus, and use the Rules we've learnt to solve the differential equations.

Refer to video from Krista King: PART 2 OF THE FUNDAMENTAL THEOREM OF CALCULUS!

We got different strategies for different boundaries situation:

• A variable and a number.
• A function and a number.
• Two functions.

▼ Here is formulas for different boundaries of integration:

Example

Solve:

• It's to apply the boundary situation strategy of A variable & a number: G'(x) = g(x)
• Assume the function in the middle of integral is G'(x) = 3x²+4x
• Since it's asking for g'(x), so it's differentiate the Integral: d/dx ʃ G'(x) dt expression
• So g'(2) = G'(x) = 3x²+4x = 20

Example

Solve:

• According to the different Boundary situation strategies, here we apply the A function & a number strategy: F'(x) = f[g(x)] · g'(x)
• So F'(x) = √(15 - 2x) · (2x)' = 2√(15-2x)

Example

Solve:

• It's asking you to apply the FTC in form of d/dx ʃ f'(x) dx = f(b) - f(a)
• So it becomes calculating F(3) - F(0) = 125 - 1 = 124

Example

Solve:

• We could use the Second Fundamental Theorem of Calculus:
• which in this case is:
• And we move the known terms to one side and keep the asking term at another side:

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Basic 「Integral Rules」

Remember there're a bunch of Differential Rules for calculating derivatives. And for integration we need to reverse them.

Refer to Lamar's math book: Common Derivatives and Integrals [PDF] Refer to Khan academy article: Common integrals review

Jump back to review the basic differential rules

Basic Rules

Reversed 「Polynomial Rules」

Reversed 「Exponential / Log Rules」

Reversed 「Trig Rules」

Reversed 「Inverse Trig Rules」

「Special Rules」

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Calculate Integrals [DRAFT]

Calculate 「Indefinite Integrals」

Example

Solve:

Integration using 「completing the square」

Example

Solve:

• Rewrite it to this form:
• Use the special arctan rule to solve it, and get:

Example

Solve: Solve:

• Rewrite it to this form:
• Use the special arctan rule to solve it, and get:

Calculate 「Definite Integrals」

Example

Solve: Don't get confused when you see the Upper boundary is smaller than the Lower one.

Example (piecewise integration)

Solve: The key is to seperate the intervals and integrate them piece by piece.

Example (Absolute value)

Solve:

• Splitting up the absolute value, and notice that the absolute value function is a piecewise function. Here we have that:
• Splitting up the definite integral:
• Evaluating each piece:
• Answer is 20.

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Integration using 「Trig identities」

The goal is to convert the multiplication of terms to be addition of simple terms.

▶ Cheatsheet on previous note: Basic Integral Rules ▶ Cheatsheet on previous note:Basic Differential Rules ▶ Cheatsheet on previous note: All trig identities ▶ Jump back to previous note on: U-substitution → Chain Rule

Refer to Khan academy: Integrating using trigonometric identities

Example

Solve:

• We can identify there's a Integral trig rule we can apply on this problem:
• Therefore we're gonna organize it a bit with u-substitution and get the answer:

Example

Solve:

• This integrand has an odd power of sin(x), so we're gonna leave sin(x) alone and make an even power for sin(x), then make u-substitution upon it:
• And also we need to change the boundaries for the integral:
• Now we can start to integrate:

Example

Solve:

• We're gonna use a trig-integral-rule and a trig identity:
• Then apply them:

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Improper Integral

After learning Definite Integral, Indefinite Integral, now it's Improper Integral. The major difference between them is their Boundaries.

The improper integral means the integral's boundary or boundaries are infinite, ∞ (or -∞).

Refer to Khan academy: Introduction to improper integrals Refer to Improper Integrals (KristaKingMath)

It looks so fearful yet not too hard to understand.

「Types」 of Improper Integral

Refer to video from ProfRobBob: Improper Integrals 5 Examples

There're 6 cases of different improper integral:

• Case 1: From a constant to positive infinity.
• Case 2: From negative infinity to a constant.
• Case 3: From negative infinity to positive infinity.
• Case 4: From 0 to e.
• Case 5: From a constant to a constant, but has an infinite discontinuity.
• Case 6:

「Convergent」 & 「Divergent」

We can call an improper integral:

• Divergent: When the limit of the improper integral DOES NOT EXIST.
• Convergent: When the limit of the improper integral EXISTS.

Solve 「Improper Integrals」

Basic Strategy:

• Replace the infinite as a variable, etc. t
• Rewrite the expression as taking the limit of the Integral, whereas the t → ∞
• Calculate the Integral with a normal variable first, and gets the result function.
• Calculate the limit of the function

「Type 1」

Solve:

「Type 2」

Solve:

• Rewrite the improper integral to limit form:
• Do the basic calculation.
• The key point is:

「Type 5」

Solve:

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❖ 「U-substitution」 → Chain Rule

The u-substitution is to solve an integral of composite function, which is actually to UNDO the Chain Rule.

▶ Back to previous note on: Chain Rule

Compare how we handle the composite functions with derivatives & integrals:

• For taking the derivative of a COMPOSITE function, we apply the Chain rule.
• For taking the integral of a COMPOSITE function, we apply the u-substitution.

Refer to Khan academy: 𝘶-substitution: defining 𝘶

We use u-substitution when we need to integrate an expression of the form of:

Strategy:

• Find a function as u
• Find or MAKE an u' at the outside so that you can pair u' with dx
• Replace u' · dx with du, because u' = du/dx
• Rewrite the Integral in term of u, and calculate the integral
• Back substitute the function of u back to the result.

How to select u

Selecting u is the most tricky part here.

Example

Solve:

• Apparently, we ignore the wrapper sin() here.
• We notice that the derivative of -x+2 is -1 which we could find it at outside.
• So let u = -x+2 and u' = -1
• So rewrite the integral to ʃ sin(u) · u' · dx = ʃ sin(u) · du
• It looks quite neat, so the u = -x+2 is alright.

Example

Solve:

• Apparently it's in form of ʃ u'/u · dx
• So that we can make u'·dx = du and the integral becomes ʃ 1/u · du
• Quite nice, so the answer would be out of there.

How to calculate 「Indefinite Integral」 with u-substitution

Example

Solve:

• With a real quick eyeballing, we see it's in form of ʃ u' · u⁶ · dx
• So with u' · dx = du we will get the simplified form ʃ u⁶ · du = u⁷/7
• Back substitute function of u back to get the result:

How to calculate 「Definite Integral」 with u-substitution

Example

Solve:

Example (self-made u')

Solve:

Example (Inverse Trig Rule)

Solve:

• Notice this radical form should directly use the Reversed Inverse Trig Rule:
• So that we assume a = 1 & u = 3x.
• Since u' = 3 so we need to make a 3 from nowhere.
• Rewrite the formula to: 1/3 ʃ 3/(1+u²) ·dx = 1/3 ʃ 1/(1+u²) ·du
• Apply the Reversed Inverse Trig Rule to get: 1/3 arctan(u) + C
• Back substitute 3x to u and the boundaries back to x get the result π/6.

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「Integrate by Parts」 → Product Rule

It's the Reverse Product Rule. And here is the formula to solve the integration by parts:

Refer to Khan academy: Integration by parts intro

Trick & Strategy:

• Recognize it's an integral with functions' product:
• Carefully choose which function to be f(x) and the other to be g'(x).
• Better to choose the easier one as f(x) when taking derivative.

Example

Solve: Refer to Symbolab

Example

Solve:

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「Partial fractions」 → Log Rule

A technique for integrating Rational functions.

▶ Jump back to previous note on Partial fractions.

▶Refer to Khan academy: Partial fraction expansion to evaluate integral

Example

This process is to break down the Rational Function to some simple fractions, which assume there are A & B leads to a system of equation:

• (A+B)·x + (B-A) = 1·x + (-4)
• So (A+B) = 1 and (B-A) = -4, which gets us A = 5/2 & B = -3/2

Strategy:

• Look at the Nominator & Dominator's degrees.
• If the dominator's degrees are higher or equal than the nominator, we do Long division of polynomial to downgrade it.
• Try to factorize the dominator if you can.
  • Assume two variables A & B
  • Apply the Partial Fraction Expansion technique.
• Apply the basic Log Rule to solve the parts.

Example

Solve:

Example

Solve: Refer to Symbolab.

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「Trig-substitution」 → Trig Rule

The goal is to simplify expression by CONVERTING terms of x into simplified expression in terms of θ.

How to do this? We can identify some trigonometric patterns in the expression and apply the Pythagorean Theorem.

▶ Cheatsheet on previous note: Basic Integral Rules ▶ Cheatsheet on previous note: Basic Differential Rules ▶ Cheatsheet on previous note: All trig identities ▶ Back to previous note on: U-substitution → Chain Rule

▶ Practice at Khan academy: Trigonometric substitution

Refer to Khan academy: Introduction to trigonometric substitution

Example

Solve:

• In this expression, we can easily identify there is a "pythagorean-like terms":
• We can see x and 2 as two sides of a triangle:
• Once we identify the pattern, we can easily get some information out of it in terms of θ:
• From the information tan(θ)=x/2 we can get the relationship between x & θ, which is: x=2tan(θ)
• And now we can convert the original expression to be in terms of θ:

Example

Solve: ...

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❖ 「Average Value」 of Functions

Refer to Khan academy: Average value over a closed interval Refer to video: Average Value of a Function on an Interval

Calculating Favg is just to get the actual area of the function, and then "reform" it to a rectangle, then divide it by its width, then you get the height.

Strategy:

• Calculate the function's integral, which is the Actual area of the function
• Calculate the Interval, which is the imaginary rectangle's width.
• Divide the area by width to get the Average value of function, which is the height.

「Mean value theorem」 for integrals

It actually IS the Average Value of Functions

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Differential Equations Intro

Example

Solve:

Example

Solve:

• Just for reminder: the Inversely proportional means y = k/x where k is constant. Jump back to previous note: Proportional Relationship.
• Assume the function of distance is S(t) = v · t.
• So the speed must be the rate of change of distance, so the speed is v = S'(t)
• Since the speed is inversely proportional to distance's square, so it means v = S'(t) = k/S²

Example

Solve:

• Assume the amount of medication is M(t).
• Now all the informations we have are:
  • M(0) = 150
  • M(13) = 150/2 = 75
  • M' = dM/dt = k · M because they're Proportional.
  • The problem is asking M(8) = ?.
• So change a bit of M' to 1/M · dM = k · dt.
• Take integral of each side to get ln(M) = k · t +C, and further M = C · eᵏᵗ
• By introduce the initial condition, we get M(0) = 150 = C · e⁰ = C
• By another information, we get M(13) = 150 · e¹³ᵏ = 75
• And further, 13k = ln(1/2), so k = ln(0.5)/13
• And now we get everything of the function M(t), let's solve for M(8)
• `M(8) = 150 · e^(8 · ln(0.5)/13) ≃ 97.9

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「Parametric Equations」 Differentiation

►Refer to Khan academy: Parametric equations differentiation ►Jump over to have some practice at Khan academy.

Why can we operate 「dy/dx」 algebraically?

Refer to Khan academy: Addressing treating differentials algebraically

Jump back to previous note: How to understand dy/dx

Differentiate 「Parametric Equations」

▼How to take derivative of a parametric differential equation?

Example

Solve:

• Let's do some trick to dy/dt and use this one instead:
• Substitute the derivatives back:
• Plug in t=1 to get the answer:

「Second derivatives」 of Parametric functions

Example

Solve:

• Follow the rule, we got the first derivative:
• And let's apply the rule for second derivative:

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Separable 「Differential Equations」

This section is an essential method for solving differential equations. Especially about the initial condition, it is the critical information for getting the original function.

Example

Solve:

• No we can't. Because:

Example

Solve:

• First to transfer same terms to the same side.
• Then take integral of each side
• Operate to get y

Example

Solve:

• We could easily get the derivative of second equation is y' = -2/3.
• Let's see if two of the derivatives are equal by substitute back the y expression:
• Clearly they're equal. So the answer is YES.

Example

Solve:

• Move the same terms to each side:
• Take integral of both side:
• Get that:
• Plug in y(0) = 3 to get C=4, so the equation then be:
• Set y=1 and get t = ln(1/2) = -ln(2)

Exponential model equations

►Jump to Khan academy for practice

►Refer to Khan academy: Worked example: exponential solution to differential equation

Example

Solve:

• Rewrite the equation, and take integral of both side:
• And we get:
• Let's plug in g(3)=2 to solve for C:
• Take C back and get the equation for g(x):

Example

Solve:

• We are told that the rate of change of P is proportional to P, which means in Math is:
• It's clear that is a Differential Equation, and we rewrite them and take integral of both side to get:
• Solve for C:
• Solve for k:
• get the k:
• Now we have the full equation, and get the result:

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Specific antiderivatives

Normally the antiderivative is in form of f(x) +C. But actually we could use some additional information to get the C and get the function only in terms of x. And we often call the "additional information" as Initial Conditions, or f₀(x).

Example

Solve:

• We could Integrate the f'(x) to get f(x) = 9eˣ + C.
• Since f(8) = 9e⁸ - 8, we could easily see that C = -8
• So the function under this condition would be: f(x) = 9eˣ -8
• Then we will get the result f(0) = 9*1 - 8 = 1.

Example

Solve:

• We could integrate f'(x) to get f(x) = x³ - x² + 7x +C
• Since f(6)=200, so we could substitute 6 into f(x):
• f(6) = 6³ - 6² + 7*6 +C = 200 which results C = -22
• So the function under this condition should be f(x) = x³ - x² + 7x - 22
• And f(1) = 1³ - 1² + 7*1 - 22 = -15

Example (Separable equations with specific solutions)

Solve:

• It looks confusing, but let's assume f(x) as y so dy/dx = 2y
• Separate differential equations to get dy/y = 2dx
• Take integral of both sides:
• Since f(1) = 5, so:
• And f(3) = 5e⁴, which means m = 5, n = 4

Example

Solve: Hint: f(0) = 2

Example

Solve Hint: Don't need to solve y completely.

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「Polar Curve」 Functions (Differential Calc)

▶ Practice at Khan academy.

Refer to Khan academy: Polar functions derivatives

▶ Proceed to: Area of Polar Curves (Integral Calc)

In the Polar World, instead of the relationship between y & x, the function is now representing the relationship between Radius & Angle, which could be presented as:

Finding the right 「boundaries」

The most tricky part in Polar system, is finding the right boundaries for θ, and it will be the first step for polar integral as well.

Differentiate 「Polar Functions」

Taking derivative of Polar function is actually DIFFERENTIATING PARAMETRIC FUNCTION. To take the derivative we need to:

• Convert the Polar function in terms of x & y:
• Take derivative of the parametric function.

Example

Solve:

• Since it's asking for the Rate of change of y-coordinate, so we convert the polar function to rectangular function:
• And we take the derivative dy/dΘ:
• Plug in the point Θ=π and get:

「Tangents」 to Polar curves

Steps:

• Find the slope dy/dx
• Convert the polar function to get the x(θ) and y(θ) parametric equations
• Solve dy/dx and get the slope
• Plug in the point's information to solve for x & y
• Get the equation of the line (tangent).

Example

Solve:

• To find the tangent line, we need to get the slope first, which is dy/dx.
• And dy/dx would be a parametric problem:
• Plug in the Θ value, to evaluate the slope:
• Find the x & y value according to the Θ:
• Now we got everything to form the equation for the tangent line:

Example

Solve:

• First, we need to convert the polar function to x(θ) & y(θ):
• And we need to find the slope dy/dx:
• Since it's a horizontal tangent, so Slope =0, which means dy/dx =0. But dx is dominator can't be zero, so we can set dy = 0 and solve for θ:
• So the answer is:

Example

Solve:

• To find a vertical tangent, we have to set the dominator of the slope as 0, that's the only thing makes it undefined.
• The slope is dy/dx, so we set dx = 0.
• The equation for x is: ....