Sharpened Methods Year 12 PDF

Preview

Summary

It is what it is...

Description

ATAR Mathematics Methods Units 3 & 4 Exam Notes for Western Australian Year 12 Students Created by Anthony Bochrinis Version 3.0 (Updated 05/01/20)

ATAR Mathematics Methods Units 3 & 4 Exam Notes Created by Anthony Bochrinis Version 3.0 (Updated 05/01/20) Copyright © ReviseOnline 2020

► About the Author - Anthony Bochrinis Hello! My name is Anthony and I graduated from high school in 2012, completed a Bachelor of Actuarial Science in 2015, completed my Graduate Diploma in Secondary Education in 2017 and am now a secondary mathematics teacher! My original exam notes (created in 2013) were inspired by Severus Snape’s copy of Advanced Potion Making in Harry Potter and the HalfBlood Prince; a textbook filled with annotations containing all of the pro tips and secrets to help gain a clearer understanding. Thank you for being a part of my journey in realising that teaching is my lifelong vocation!

► ReviseOnline. The Power Behind Sharpened. ReviseOnline will revolutionise the way you prepare for tests and exams. It will exceed all the revision materials that you have ever used. No more revision manuals that clog up your desk space, no more outdated books and no more wasting time! It is all here, in one easy to use program that generates topic tests and practice exams, with a click of a button. And, it goes wherever you go.

Thousands of combinations of tests and exams

Revise a huge range of ATAR courses

Monitors your progress

Sets a time for each test and exam you start

► Keep Up to Date with Revise Online! To be kept in the loop with all the latest resource releases, updates, new, tips and more, be sure to follow us on our social media and bookmark our website! Instagram @reviseonline

Facebook @reviseonlinewace

Website reviseonline.com

IN V ER S E F U NC T IO N S

I N DI C E S AN D S U R D S IN D EX A ND SU R D L AW S Index Laws

𝒂𝒎 × 𝒂𝒏 = 𝒂𝒎+𝒏 ( 𝒂𝒎 )𝒏 = 𝒂𝒎×𝒏

𝒎 𝒂𝒏

= √𝒂𝒎 = ( √𝒂 ) 𝟏 𝒂−𝒎 = 𝒂𝒎 Surd Laws 𝒏

𝒏

𝒎

𝒂 𝒎 ÷ 𝒂 𝒏 = 𝒂𝒎 − 𝒏 𝒂𝟎 = 𝟏

( 𝒂𝒃 ) 𝒎 = 𝒂 𝒎 × 𝒃 𝒎 (

√𝒂 × √𝒃 = √𝒂𝒃

𝒂 𝒃

𝒎

) =

𝒂 𝒃𝒎 𝒎

√𝒂 × √𝒂 = 𝒂

𝒎√𝒂 ± 𝒏√𝒂 = (𝒎 ± 𝒏)√𝒂

𝒂 √𝒂 ÷ √𝒃 = √ 𝒃

Rationalising a Surd • Removes surd in denominator of a fraction. 𝟏

𝟏

𝟏

√𝒂 ×𝟏= = × = 𝒂 √𝒂 √𝒂 √𝒂 √𝒂 √𝒂

Determining the Inverse of a Function Step 1 Step 2

Rearrange the function to make 𝑥 the subject instead of 𝑦.

Swap the variables 𝑥 and 𝑦, this is the inverse function, 𝑓 −1 (𝑥).

Exponential vs. Logarithmic Function

Let 𝑦 = 𝑎𝑥 log𝑎 𝑦 = log 𝑎 𝑎 𝑥

LO G AR IT H M IC F UN C T IO N

Logarithmic 𝒚 = 𝒍𝒐𝒈𝒂 (𝒙 − 𝒃) + 𝒄 Domain= {𝑥 ∈ ℝ: 𝑥 > 0} Range= {𝑦 ∈ ℝ}

L O G A R I T H M L AW S The Concept of the Logarithm ( logs ) • The power to which a number (i.e. base) must be raised to produce a given number. 𝒂𝒙 = 𝒚 → 𝒙 = 𝒍𝒐𝒈 𝒂 (𝒚)

• 𝒂 : base number • 𝒙 : exponent • 𝒚 : solution

𝑙𝑜𝑔𝑎 𝑦 = 𝑥 𝑙𝑜𝑔𝑎 𝑎 𝑙𝑜𝑔𝑎 𝑦 = 𝑥 ∴ 𝑦 = 𝑙𝑜𝑔𝑎 𝑥

Logarithmic Function Transformations

L O G AR I T H M S

If 23 = 8 then the matching logarithm is 𝑙𝑜𝑔2 8 = 3

Vertical: 𝑥=𝑏

Important Co-ordinates

𝑥-intercept: (𝑎−𝑐 + 𝑏, 0)

Another point: (𝑎1−𝑐 + 𝑏, 1)

• Impact on changing function variables:

Condition and Description

• Adding and Subtracting Logarithms:

b Adds b to all x-values

• Index Laws of Logarithms:

c Adds c to all y-values

b > 0 Translate horizontally b units to the left b < 0 Translate horizontally b units to the right Translate vertically c c>0 units upwards Translate vertically c c 0 ∴ 𝑚𝑖𝑛 𝑎𝑡 (0,0) ▪ 𝑓(−2) = 8, 𝑓 ′′ (−2) < 0 ∴ 𝑚𝑎𝑥 𝑎𝑡 (−2,8) ▪ 𝑓(−1) = 4, 𝑓 ′′ (−1) = 0 ∴ 𝑣𝑒𝑟𝑡𝑖𝑐𝑎𝑙 𝑝𝑜𝑖𝑛𝑡 𝑜𝑓 𝑖𝑛𝑓𝑙𝑒𝑐𝑡𝑖𝑜𝑛 𝑎𝑡 (−1,4) 𝑎𝑠 𝑓′(−1) ≠ 0 ▪ Long term behaviour as 𝑥 tends toward ±∞: 𝑥 → +∞, 𝑦 → +∞ 𝑥 → −∞, 𝑦 → −∞ ▪ Sketching function:

*Plot points

ATAR Math Methods Units 3 & 4 Exam Notes Copyright © ReviseOnline 2020 Created by Anthony Bochrinis More resources at reviseonline.com

Page: 1 / 4 Version: 3.0

IN C RE M EN T AL F OR M UL A

OP T IM I S AT I ON

Small Change and Approximation • Calculates the approximate change in a dependent variable 𝑦 from a small change in the matching independent variable 𝑥. 𝜹𝒚 ≈

𝒅𝒚

𝒅𝒙

× 𝜹𝒙

𝜹𝒚

𝒅𝒚 𝜹𝒙 ≈ × 𝒚 𝒅𝒙 𝒙

• 𝜹𝒚 or 𝜹𝒙 : small change in 𝑥 or 𝑦 (must be small for an accurate approximation). • 𝜹𝒚/𝒚 or 𝜹𝒙/𝒙 : % change in 𝑥 or 𝑦.

Incremental Formula Examples (Q1) Find the change in 𝑦 when 𝑥 changes from 3 to 2.98 in the equation: 𝑦 = 3𝑥 2 − 2𝑥 𝛿𝑦 ≈ 𝑑𝑦/𝑑𝑥 × 𝛿𝑥 𝑑𝑦 = 6𝑥 − 2 𝛿𝑦 ≈ (6𝑥 − 2) × 𝛿𝑥 𝑑𝑥 𝛿𝑦 ≈ (6(3) − 2) × (−0.02) 𝛿𝑦 ≈ −0.32 ∴ decrease by 𝟎. 𝟑𝟐 (Q2) Radius of a sphere increases from 15𝑐𝑚 to 15.1𝑐𝑚, what is the increase in surface area? 𝛿𝑆 ≈ 𝑑𝑆/𝑑𝑟 × 𝛿𝑟 𝑑𝑆 = 8𝜋𝑟 𝛿𝑆 ≈ (8𝜋𝑟) × 𝛿𝑟 𝑆 = 4𝜋𝑟2 → 𝑑𝑟 𝛿𝑆 ≈ (8𝜋(15)) × (0.01) 𝟐 𝛿𝑆 ≈ 3.77 ∴ increase by 𝟑. 𝟕𝟕𝒄𝒎

𝛿𝑦 ≈ 2 cos(2) + 3𝑒 3(1) × (0.1) +3𝑒 3𝑥 𝛿𝑦 ≈ 5.94 ∴ increase by 𝟓. 𝟗𝟒 (Q4) The radius of a sphere increases by 2%, find the percentage increase in the volume. 4 𝑉 = 𝜋𝑟3 𝛿𝑟 𝛿𝑉 ≈ 4𝜋𝑟2 × 𝛿𝑟 𝛿𝑉 3 ≈3× 𝛿𝑉 4𝜋𝑟2 𝛿𝑟 𝑟 𝑑𝑉 𝑉 ≈ = 4𝜋𝑟2 𝛿𝑉 𝑉 𝑉 𝑑𝑟 ≈ 3 × 2% 2 4𝜋𝑟 𝛿𝑟 𝑉 𝛿𝑉 𝑑𝑉 𝛿𝑉 ≈ 𝛿𝑉 ≈ 4𝜋𝑟3 /3 𝑉 𝛿𝑟 𝑑𝑟 ≈ 3 × 6% 𝑉 𝑑𝑉 𝛿𝑉 3𝛿𝑟 × 𝛿𝑟 ≈ 𝛿𝑉 ≈ ∴ 𝟔% increase 𝑑𝑟 𝑟 𝑉 GR OW T H AN D D E C AY Growth and Decay Formulae

𝒅𝑨 = 𝒌𝑨𝟎 𝒆𝒌𝒕 = 𝒌𝑨 𝒅𝒕 • 𝑨𝟎 : Initial (starting) amount at time = 0. • 𝒌 : constant of proportionality. ▪ 𝒌 > 𝟎 : represents exponential growth. ▪ 𝒌 < 𝟎 : represents exponential decay. • 𝒕 : time (units differ as per the question). 𝑨 = 𝑨𝟎 𝒆𝒌𝒕

Half Life and Doubling Time • Half Life: decay specific (for 𝑘 < 0). 𝑨 = 𝟎. 𝟓𝑨𝟎

• Time for initial amount to reduce by 50% (halve).

• Doubling Time: growth specific (for 𝑘 > 0). 𝑨 = 𝟐𝑨𝟎

𝑑𝐴 𝑑𝑡

• Time for initial amount to increase by 100% (double).

Derivation of Growth/Decay Formulae

= 𝑘𝐴 (i.e. rate is in direct proportion with 𝑘). 𝑙𝑛(𝐴) = 𝑘𝑡 + 𝑐 *Let 𝑒 𝑙𝑛(𝐴) = 𝑒 𝑘𝑡+𝑐 𝐴 = 𝑒 𝑘𝑡 × 𝑒 𝑐 constant 𝑘𝑡 𝐴 = 𝑒 × 𝐴0 𝑒 𝑐 = 𝐴0 𝐴 = 𝐴0 𝑒 𝑘𝑡 Growth/Decay Examples

𝑑𝐴 = 𝑘𝐴 × 𝑑𝑡 𝑑𝐴 = 𝑘 𝑑𝑡 𝐴 1 ∫ 𝑑𝐴 = ∫ 𝑘𝑑𝑡 𝐴

(Q1) Population of 10000 bacteria is decaying according to time measured in minutes after 7am. The time taken for the population to decrease to half its original size is 7 minutes. (Q1a) Find the constant of proportionality, 𝑘. 𝐴 = 0.5𝐴0 0.5 = 𝑒 7𝑘 𝑘 = 𝑙𝑛(0.5) /7 ∴ 0.5𝐴 0 = 𝐴0 𝑒 7𝑘 ln(0.5) = 7𝑘 𝑘 = −𝟎. 𝟗𝟗 (Q1b) Find the population at 7:05am.

𝐴 = 10000𝑒 −0.99𝑡 → 𝐴 = 10000𝑒 −0.99(5) = 𝟔𝟎𝟗𝟓 (Q1c) When will the population fall below 100? 100 = 10000𝑒 −0.99𝑡 → 𝑡 = 46.507 = 𝟒𝟔𝒎 𝟑𝟏𝒔 (Q1d) What is the rate of change at 7:15am? 𝑑𝐴 = 𝑘𝐴 = 𝑘𝐴0 𝑒 𝑘𝑡 = −0.99 × 10000𝑒 −0.99×15 𝑑𝑡

= −𝟐𝟐𝟒 bacteria per minute (i.e. decreasing). (Q2) If 𝑑𝐴/𝑑𝑡 = 0.252𝐴, find the initial value

for 𝐴 given that amount at time = 10 is 565.

565 = 𝐴0 𝑒 0.252 (10) 𝑙𝑛(565) − 𝑙𝑛(𝐴0 ) = 2.52 565 𝑙𝑛(𝐴0 ) = 𝑙𝑛(565) − 2.52 = 𝐴0 𝑒 0.252 (10) 𝐴0 𝑙𝑛(𝐴0 ) = 3.8168 565 ∴ 𝐴0 = 𝑒 3.8168 = 𝟒𝟓. 𝟒𝟔 ) = 2.52 𝑙𝑛 ( 𝐴0 (Q3) The foam in a glass of soft drink shrinks according to 𝐻 = 20𝑒 −0.005𝑡 where 𝐻 is height of the foam in mm and 𝑡 is time in seconds. (Q3a) Find the average rate of change of the foam height during the second minute. 𝐻(120) − 𝐻(60) 10.98 −14.82 = −𝟎. 𝟎𝟔𝟒𝒎𝒎 = 60 120 − 60 (Q3b) Find the instantaneous rate of change of the height of the foam after 24 seconds. 𝑑𝐻 𝑑𝐻 = −𝟎. 𝟖𝟗𝒎𝒎/𝒔 = −0.1𝑒 −0.005𝑡 → 𝑑𝑡 𝑑𝑡

Step Draw a diagram of the scenario 1 and define all variables. If there are more than 2 variables, Step reduce the number of variables to 2 2 by substitution and simplification. Step Determine the derivative of 𝑓(𝑥). 3 ClassPad: 𝑑𝑖𝑓𝑓(𝑓 (𝑥))

Make derivative equal to 0 and Step solve for 𝑥 to find turning points. 4 ClassPad: 𝑠𝑜𝑙𝑣𝑒[𝑑𝑖𝑓𝑓 (𝑓(𝑥))] = 0

Find nature of all turning points by Step subbing in 𝑥 co-ord found in step 4 by using the second derivative test. 5 ClassPad: 𝑑𝑖𝑓𝑓[𝑑𝑖𝑓𝑓 (𝑓(𝑥))] Find optimal dimensions and Step maximum or minimum value 6 required according to question.

Optimisation Examples 𝒙 𝒙 (Q1) A rectangular box is made from a sheet 𝒙 𝒙 Box Net of metal with squares (Corners of length 𝑥 to be cut cut out) from the corners and 𝒙 folded. If the sheet of 𝒙 𝒙 𝒙 𝟏𝟎𝒄𝒎 metal is 6cm wide and 10cm long, find 𝑥 that maximises the volume. ▪ Identify all equations relevant to question: 𝑉 = 𝑙𝑤ℎ → 4 variables in this equation. ▪ Reduce to two variables by substitution: 𝑙 = 10 − 2𝑥 , 𝑤 = 6 − 2𝑥 and ℎ = 𝑥 ▪ Find derivative and test all turning points: 𝑉 = (10 − 2𝑥 )(6 − 2𝑥 )𝑥 = 60𝑥 − 32𝑥 2 + 4𝑥 3 𝑑2𝑉 𝑑𝑉 = 24𝑥 − 64 = 12𝑥 2 − 64𝑥 + 60, 𝑑𝑥 2 𝑑𝑥 Solving for when 𝑑𝑣/𝑑𝑥 = 0: 𝑥 = 4.12,1.21 𝑊ℎ𝑒𝑛 𝑥 = 4.12, 𝑑2 𝑉/𝑑𝑥2 = 34.88 ∴ 𝑚𝑖𝑛𝑖𝑚𝑢𝑚 𝑊ℎ𝑒𝑛 𝑥 = 1.21, 𝑑2 𝑉/𝑑𝑥2 = −34.96 ∴ 𝑚𝑎𝑥𝑖𝑚𝑢𝑚 ▪ Find dimensions and maximum volume: Sub 𝑥 = 1.21 to find max 𝑉 = 32.84𝑐𝑚3 ∴ The volume is a max when 𝑥 = 𝟏. 𝟐𝟏𝒄𝒎. (Q2) A cone has a slant height of 2√3 𝑐𝑚. The sloped edge makes 𝒉 an angle 𝜃 where 0 < 𝜃 < 𝜋/2. Find the cone max volume. 𝜽 ▪ Identify all equations: 𝒓 1 2 𝑉 = 𝜋𝑟 ℎ, . ℎ = 2√3𝑠𝑖𝑛𝜃, . 𝑟 = 2√3𝑐𝑜𝑠𝜃 3 ▪ Reduce to two variables by substitution:

𝜋(2√3𝑐𝑜𝑠𝜃 ) (2√3𝑠𝑖𝑛𝜃) = 12√3cos2 𝜃𝑠𝑖𝑛𝜃 3 𝑉 = 12√3(1 − sin2 𝜃)𝑠𝑖𝑛𝜃= 12√3(𝑠𝑖𝑛𝜃 − sin3 𝜃) ▪ Find derivative and test turning point: 2

𝑉=

𝑑𝑉/𝑑𝑥 = 12√3𝑐𝑜𝑠𝜃 − 36√3sin2 𝜃𝑐𝑜𝑠𝜃 Solving for when 𝑑𝑉/𝑑𝑥 = 0: 𝑥 = 0, 0.6155 Disregard 𝜃 = 0 as a possible solution, 𝑊ℎ𝑒𝑛 𝜃 = 0.6155, 𝑑2 𝑉/𝑑𝑥2 = −48 ∴ 𝑚𝑎𝑥𝑖𝑚𝑢𝑚 ▪ Find dimensions and maximum volume: Sub 𝜃 = 0.6155 to find max 𝑉 = 𝟏𝟔. 𝟕𝟔𝒄𝒎𝟑

Indefinite Integrals ∫ 𝒙 𝒅𝒙

∫ 𝑥 𝑑𝑥 =

Sketching Derivative Functions

• All local max/min are 𝑥 -intercepts on 𝑓′(𝑥). • All points where the function is increasing,

𝑓′(𝑥) is above the 𝑥-axis and vice versa. • Where there is a point of inflection on the graph (vertical or horizontal), the derivative has a maximum or minimum turning point. Sketching Derivative Examples

• Key: 𝑓(𝑥) 𝑓′(𝑥) Turning Point Analysing Derivative Graphs Example (Q1) Sketch the function on the axes below: −2

−1

0

𝒇′(𝒙)

+

0

−

𝒇(𝒙)

𝒇′′(𝒙) ▪ ▪ ▪ ▪ ▪

− −

+ −

1

2

0

+

0

−

0

+

𝑥 = −2 → increasing 𝑥 = −1 → minimum 𝑥 = 0 → vert. inflection 𝑥 = 1 → maximum 𝑥 = 2 → increasing

+ +

(Q2) Sketch the function on the axes given: 𝑓(𝑥) ≥ 0 𝑓𝑜𝑟 𝑥 ≥ −1 𝑓 ′′(𝑥) > 0 𝑓𝑜𝑟 𝑥 = 2 𝑓 ′ (𝑥) = 0 𝑓𝑜𝑟 𝑥 = 1,2 𝑓 ′′(𝑥) < 0 𝑓𝑜𝑟 𝑥 = 1 ▪ ▪ ▪ ▪ ▪

𝑥 𝑥 𝑥 𝑥 𝑥

< 1 → increasing = 1 → maximum = 2 → minimum > 2 → increasing → ±∞, 𝑓 (𝑥) → ±∞

(Q3) Is it possible for a function to have no max or min points but have an inflection point? ▪ Yes, it is possible (e.g. 𝑦 = 𝑥 3 ) 𝑑2 𝑦 𝑑𝑦 ▪ 𝑑𝑥 = 𝑑𝑥 2 = 0 at only one point, (𝟎, 𝟎).

(Q1) Integrate the function ∫ 𝑒 −6𝑥 + 2√𝑥 − 4𝜋𝑑𝑥

𝑥

2

1

2√𝑥 = 2𝑥2 ∴ ∫ 𝑓(𝑥 )𝑑𝑥 =

+𝑐

2

• Indefinite integrals produce an equation and a constant (+𝑐) as it caters for a constant in the original function 𝑓(𝑥) , which disappears (i.e. becomes 0) after being differentiated. Definite Integrals

2 22 12 𝑥2 ∫ 𝑥 𝑑𝑥 = [ ] = [ − ] = 1.5 2 2 1 2 1

𝒃

∫ 𝒙 𝒅𝒙 𝒂

2

• 𝒂 : integral lower bound (on 𝑥-axis). • 𝒃 : integral upper bound (on 𝑥-axis).

• Definite integrals produce a single number answer (all other variables are eliminated). • Definite integral of a function that is below the 𝑥 -axis results in a negative answer. Common Functions and Integrals Function

Integral 𝑥 𝑛+1 Polynomial ∫ 𝑥 𝑛 𝑑𝑥 +𝑐 𝑛 +1 𝑛+1 [ ] 𝑓(𝑥) Chain ∫ 𝑓 ′ (𝑥)[𝑓(𝑥)]𝑛 𝑑𝑥 +𝑐 Rule 𝑛 +1 𝑓(𝑥) 𝑒 Exponential +𝑐 ∫ 𝑒 𝑓(𝑥) 𝑑𝑥 (Euler) 𝑓 ′ (𝑥)

Equation

𝑓′(𝑥) ∫ 𝑑𝑥 𝑓(𝑥) ∫ 𝑠𝑖𝑛(𝑥) 𝑑𝑥

Reciprocal Sine

−𝑐𝑜𝑠(𝑥) + 𝑐 𝑠𝑖𝑛(𝑥) + 𝑐

𝒂

𝒂

𝒃

∫ 𝒇(𝒙) = − ∫ 𝒇(𝒙) 𝒂

ln(𝑓(𝑥)) + 𝑐

∫ 𝑐𝑜𝑠(𝑥) 𝑑𝑥

Cosine

Integration Laws

∫ 𝒇(𝒙) = 𝟎 𝒂

𝒃

∫ 𝒂 × 𝒇(𝒙) 𝒅𝒙 = 𝒂 × ∫ 𝒇(𝒙) 𝒅𝒙

∫[𝒇(𝒙) ± 𝒈(𝒙)]𝒅𝒙 = ∫ 𝒇(𝒙)𝒅𝒙 ± ∫ 𝒈(𝒙)𝒅𝒙 𝒂

𝒃

𝒂

∫ 𝒇(𝒙)𝒅𝒙 + ∫ 𝒇(𝒙)𝒅𝒙 = ∫ 𝒇(𝒙)𝒅𝒙 𝒄

𝒃

𝒄

I N T E G R AT I O N B Y E S T I M AT IO N Inscribed & Circumscribed Rectangles

Inscribed

Circumscribed

Series of rectangles below a curve

Series of rectangles above a curve

Underestimation & Overestimation Underestimation (U) Adding areas of inscribed rectangles to underestimate area under a curve.

2𝑥

▪ (1 − 4𝑥 2 ) =− 8𝑥 ∴ numerator must be −8𝑥 𝑑𝑥 𝟏 −8𝑥 1 𝑑𝑥 = − 𝒍𝒏(𝟏 − 𝟒𝒙𝟐 ) + 𝒄 − ∫ 4 1 − 4𝑥 2 𝟒 𝑑

(Q4) Integrate the function ∫

𝑎

𝑼 +𝑶 𝟐

𝑨𝒓𝒆𝒂 = ∑ 𝒇(𝒙𝒊 )𝜹𝒙 𝒊 𝒊

• 𝜹𝒙 : interval size (i.e. width of rectangles). • 𝑼 : add the areas of all inscribed rectangles from 𝑥 = 𝑎 to 𝑥 = 𝑏 − 𝛿𝑥 . • 𝑶 : add the areas of all circumscribed rectangles from 𝑥 = 𝑎 + 𝛿𝑥 to 𝑥 = 𝑏 .

Estimating Area Under Curve Examples (Q1) 𝑓(𝑥) is graphed below for −0.5 ≤ 𝑥 ≤ 2.5: 𝒇(𝒙)

-0.5 0 0.5 1 1.5 2 2.5 (Q1a) Estimate 𝒙

𝒇(𝒙)

0 1

2 ∫0

Function Inscribed Rectangles Circumscribed Rectangles

𝒙

𝑓(𝑥)𝑑𝑥 using 𝛿𝑥 = 0.5:

0.5 2

1

2.5

1−12𝑥 2

𝑑𝑥 3𝑥 𝐥𝐧(𝒙) 1 4𝑥 1 = − 𝟐𝒙𝟐 + 𝒄 − 4𝑥𝑑𝑥 = ln(𝑥) − 2 𝟑 3 3𝑥 (Q5) Integrate the function ∫ 2sin(4 − 3𝑥)𝑑𝑥 −2 cos(4 − 3𝑥 ) 𝟐 𝐜𝐨𝐬(𝟒 − 𝟑𝒙) = +𝒄 −3 𝟑 2

∫

(Q5) Integrate the function ∫ 32𝑥 𝑑𝑥 𝑒 2𝑙𝑛 (3)𝑥 𝟑𝟐𝒙 +𝒄 = 2𝑙𝑛(3) 𝟐𝒍𝒏(𝟑)

∫ 𝑒 2𝑙𝑛(3)𝑥 =

2

(Q6) Integrate the function ∫−1(−𝑒 3𝑥 + 1)𝑑𝑥 2 𝑒6 𝑒 −3 𝑒 3𝑥 + 2) − (− = [− + 𝑥] = [(− − 1)] 3 3 3 −1

𝑒 −3 𝒆−𝟑 − 𝒆𝟔 𝑒6 +𝟑 +2 + +1 = 𝟑 3 3 𝜋/4 (Q7) Integrate the function ∫0 (2𝑥 − 𝑐𝑜𝑠𝑥 )𝑑𝑥

=−

𝜋 2 𝝅𝟐 √𝟐 𝜋 = [𝑥 2 − 𝑠𝑖𝑛𝑥]𝜋/4 [( ) − sin ( ) − 0] = − 0 = 𝟐 4 4 𝟏𝟔 Applications of Integration Examples (Q1) If

𝑑

𝑑𝑥

(𝑥𝑒 𝑥 ) = 𝑥𝑒 𝑥 + 𝑒 𝑥 , determine ∫0 𝑥𝑒 𝑥 𝑑𝑥 . 1

1

1

[𝑥𝑒 𝑥 ]10 = ∫ 𝑥𝑒 𝑥 𝑑𝑥 + ∫ 𝑒 𝑥 𝑑𝑥 0

1

1

0

∫ 𝑥𝑒 𝑥 = [𝑥𝑒 𝑥 ]01 − ∫ 𝑒 𝑥 𝑑𝑥 0

0 1

∫ 𝑥𝑒 𝑥 = [𝑥𝑒 𝑥 ]10 − [𝑒 𝑥 ]10 0

*Create reverse equation and rearrange to find the integral

= [𝑒 1 − 0] − [𝑒 1 − 𝑒 0 ] = 𝑒 − 𝑒 − −1 = 𝟏

(Q2) If 𝑑𝑥 = 𝑎𝑥 2 − 12𝑥 and 𝑃(𝑥) has a stationary 𝑑𝑃

point at (4,8), determine the value of 𝑃(10). ▪ Use stationary point to determine 𝑎: 0 = 16𝑎 − 48 → 16𝑎 = 48 → 𝑎 = 3 ▪ Integrate to find 𝑃(𝑥) and solve for 𝑐 : 𝑃 = 𝑥 3 − 6𝑥 2 + 𝑐 8 = 64 − 96 + 𝑐 8 = 43 − 6(42 ) + 𝑐 𝑐 = 8 − 64 + 96 = 40 ▪ Find the value of 𝑃(10) using equation: 𝑃(10) = 103 − 6(102 ) + 40 = 𝟒𝟒𝟎 (Q3) The graph of the function 𝑓(𝑥) is shown: Area 𝐴 = 4 units2 Area 𝐵 = 1 units2 𝑨 𝑪 Area 𝐶 = 3 units2 Roots are 𝑥 = −10, −5, 0 & 9.

(Q4a) Determine ∫−10 𝑓(𝑥)𝑑𝑥 9

∫−10 𝑓(𝑥)𝑑𝑥= 𝐴 − 𝐵 + 𝑐 = 𝟔 9

(Q4b) Determine ∫0 3𝑓(𝑥)𝑑𝑥 9

∫0 3𝑓(𝑥)𝑑𝑥 = 3 × 𝐶 = 𝟗 9

(Q4c) Determine ∫9 𝑓(𝑥)𝑑𝑥 9 − ∫−5 𝑓(𝑥)𝑑𝑥 = −𝐶 + 𝐵 = −𝟐 −5

(Q4d) Determine ∫−10 𝑓(𝑥) − 2𝑑𝑥 −5

= ∫−10 𝑓(𝑥)𝑑𝑥 − −5

−5

∫−10 2𝑑𝑥

= 𝐴 − [2𝑥]−10 = 4 − 10 = −𝟔 −5

AR E A U N DE R A C UR V E

• Overestimation & Underestimation Average: ∫ 𝑓(𝑥)𝑑𝑥 ≈

𝟑𝟐

(Q3) Integrate the function ∫ 1−4𝑥 2 𝑑𝑥

Overestimation (O) Adding areas of circumscribed rectangles to overestimate area under a curve. 𝑏

−𝒆−𝟔𝒙

𝟒𝒙 − 𝟒𝝅𝒙 + 𝒄 + 𝟑 𝟔 (Q2) Integrate the function ∫ 4(3𝑥 − 2)5 𝑑𝑥 4(3𝑥 − 2)6 4(3𝑥 − 2)6 𝟐(𝟑𝒙 − 𝟐)𝟔 +𝒄 = = = 𝟗 3 ×6 18

𝑩

S K E T CH I N G D E R I V A T I V E S

𝒙

Definite & Indefinite Integral Examples

I N T E G R A L L AW S

𝟔𝒄𝒎

(Q3) Find the change in 𝑦 when 𝑥 changes from 1 to 1.1 in the equation: 𝑦 = 𝑠𝑖𝑛(2𝑥 ) + 𝑒 3𝑥 𝛿𝑦 ≈ 𝑑𝑦/𝑑𝑥 × 𝛿𝑥 𝑑𝑦 = 2 cos(2𝑥) 𝛿𝑦 ≈ (2 cos 2𝑥 + 3𝑒 3𝑥 ) × 𝛿𝑥 𝑑𝑥

IN T EG R AT I ON A LG E BR A

I N T EG R AT I O N

Optimising Dimensions of a Scenario

1.5 2.8

2 3

𝑈 = 0.5(1 + 2 + 2.5 + 2.8) = 0.5 × 8.3 = 4.15 𝑂 = 0.5(2 + 2.5 + 2.8 + 3) = 0.5 × 10.3 = 5.15 2 𝑈 + 𝑂 4.15 5.15 ≈ ∫ 𝑓(𝑥)𝑑𝑥 ≈ 𝟒. 𝟔𝟓 ≈ 𝐴𝑟𝑒𝑎 ≈ 2 2 0

(Q1b) If 𝑓(𝑥) = (4𝑥 + 1)/(𝑥 + 1), what is the margin of error in your prediction in part (a)? 24𝑥 + 1 ∫ 𝑑𝑥 = 4.7042 ∴ 4.7042 − 4.65 = 𝟎. 𝟎𝟓 0 𝑥 +1 (Q1c) How can the accuracy of the estimate of the area under curve in part (a)? be increased? Reduce interval size 𝛿𝑥 (i.e. smaller than 0.5)

Area Underneath a Curve 𝒃

∫ |𝒇(𝒙)|𝒅𝒙 𝒂

• |𝒇(𝒙)| : absolute value (i.e. change the number inside from negative to positive).

Negative Area Underneath a Curve Determine roots of the function (i.e. factorise and solve for when 𝑦 = 0).

Step 1

Create and add separate integrals that are above and below x-axis.

Step 2

Area Between Curves Examples

(Q1) Find the area between 𝑦 = 𝑒 0.5𝑥 and the 𝑥-axis between the lines 𝑥 = −2 and 𝑥 = 2.

Integral exists above 𝑥 -axis. 2

∫ 𝑒 0.5𝑥 𝑑𝑥 = [𝑒 0.5𝑥 ]2−2 −2

= [𝑒 1 − 𝑒 −1 ] = 𝟐. 𝟑𝟓

(Q2) Find the area between 𝑦 = 𝑥 2 − 1 and the 𝑥-axis between the lines 𝑥 = 0 and 𝑥 = 2. 𝑨

𝑩

𝑦 = 𝑥 2 − 1 = (𝑥 + 1)(𝑥 − 1) ∴ Root at 𝑥 = 1 which means that integral is above & below 𝑥 -axis. ∴ Must add 2 separate integrals.

𝐴𝑟𝑒𝑎 = 𝐴 + 𝐵 = ∫1 𝑥 2 − 1𝑑𝑥 + ∫0 |𝑥 2 − 1|𝑑𝑥 = [𝑥 2 − 1]12 + [|𝑥 2 − 1|]10 = 3 − 0 + 0 − |−1| = 𝟐 2

1

ATAR Math Methods Units 3 & 4 Exam Notes Copyright © ReviseOnline 2020 Created by Anthony Bochrinis More resources at reviseonline.com

Page: 2 / 4 Version: 3.0

AR E A B ET W E EN C UR V ES

FI N AN C I AL C A L CU L U S

Area Between Curves Formulae ∫ (

𝒅 𝒍𝒆𝒇𝒕 𝒓𝒊𝒈𝒉𝒕 ) −( ) 𝒅𝒚 ∫ ( 𝒇𝒖𝒏𝒄𝒕𝒊𝒐𝒏 𝒇𝒖𝒏𝒄𝒕𝒊𝒐𝒏

• Revenue, Profit and Average Cost:

• Upper and Lower Bounds on the 𝑥 -axis: 𝒖𝒑𝒑𝒆𝒓 𝒍𝒐𝒘𝒆𝒓 ) 𝒅𝒙 𝒇𝒖𝒏𝒄𝒕𝒊𝒐𝒏) − ( 𝒇𝒖𝒏𝒄𝒕𝒊𝒐𝒏 • Upper and Lower Bounds on the 𝑦-axis: 𝒂

𝒃

Area Between Curves Examples (Q1) Find an expression for finding the shaded area between the two functions 𝑓(𝑥) and 𝑔(𝑥): (Q1a) 𝑓(𝑥) (0,0) 𝑔(𝑥) (𝑎, 𝑏) 𝑎 ∫ 𝑓(𝑥) − 𝑔(𝑥)𝑑𝑥 *Subtract 𝑔(𝑦) from 𝑓(𝑦) and simplify first. 0 (𝑏, 𝑐) (Q1b) 𝑓(𝑥) (0, 𝑒) 𝑔(𝑥) (𝑎, 0) (𝑑, 0) 𝑏

𝑎

𝑑

∫ 𝑓(𝑥 )𝑑 +∫ 𝑓(𝑥) −𝑔(𝑥)𝑑𝑥 +∫ 𝑔(𝑥 ) −𝑓 (𝑥)𝑑𝑥 0

𝑎

(Q1c) (0, 𝑎)

𝑏

𝑓(𝑥) 𝑔(𝑥)

(0, 𝑏) 𝑎

*Rearrange 𝑓(𝑦) and 𝑔(𝑦) to make 𝑥 the subject. (Q2) Find the area between the two different functions 𝑓(𝑥) = ln(𝑥) and 𝑔(𝑥) = (𝑥 − 4)2 ▪ Find intersection points between curves: 𝑓(𝑥) = 𝑔(𝑥) 𝑓(4) = 1.39 and 𝑔(4) = 0 ln(𝑥) = (𝑥 − 4)2 ∴ 𝑓(𝑥) is upper function 𝑥 = 2.96, 5.29 and 𝑔(𝑥) is lower function. ▪ Find area between curves: 𝑏 𝑏 ∫𝑎 𝑢𝑝𝑝𝑒𝑟 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛 𝑑𝑥 − ∫𝑎 𝑙𝑜𝑤𝑒𝑟 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛 𝑑𝑥 ∫ 𝑓(𝑦) − 𝑔(𝑦)𝑑𝑦 𝑏

𝑏

𝑪(𝒙) 𝒙

Financial Calculus Examples

(Q1) The marginal cost of producing 𝑥 units is 𝐶 ′ (𝑥) = 0.3𝑥 2 − 0.2𝑥 + 100 dollars per unit. (Q1a) Calculate ...