IGCSE, Edexcel, IB and AP Science Courses | Dr Dabour Hub Academy

Overview

Evaluate definite integrals exactly using the fundamental theorem of calculus.
Use definite integrals to find the area under a curve and the area between two curves.
Find the volume of a solid of revolution about the x-axis and the y-axis.

🌳 Concept tree

Definite Integration and Applications

├── The definite integral

│ ├── Notation and the fundamental theorem

│ └── Evaluating with limits

├── Properties

│ ├── Additivity over intervals

│ └── Reversal and zero-width limits

├── Area

│ ├── Area under a curve

│ └── Area between two curves

└── Volumes of revolution

├── About the x-axisV = π∫y²dx

└── About the y-axisV = π∫x²dy

Definite Integration, Areas and Volumes of Revolution · 1 / 9

The definite integral

Definition and notation

Comparison aspect	Main idea	Keywords
•	The definite integral of f from a to b is written ∫_a^b f(x) dx, where a is the lower limit and b is the upper limit.	definiteintegralwrittenlower
Fundamental theorem	if F is any antiderivative of f, then ∫_a^b f(x) dx = F(b) − F(a).	antiderivativethen
The constant of integration cancels	(F(b) + C) − (F(a) + C) = F(b) − F(a). So never write + C in a definite integral	neverwritedefiniteintegral

∫₀² (3x² + 2x) dx = [x³ + x²]₀² = (8 + 4) − 0 = 12
∫₁⁴ x^−1/2 dx = [2√x]₁⁴ = 4 − 2 = 2
∫₀^π/2 cos x dx = [sin x]₀^π/2 = 1 − 0 = 1
∫₀¹ e^2x dx = [½e^2x]₀¹ = ½(e² − 1) ≈ 3.19

Either change the limits to the new variable, or substitute back before applying the original limits.
Example ∫01 2x(x² + 1)³ dx. Let u = x² + 1, du = 2x dx; limits become 1 and 2: ∫12 u³ du = [u⁴/4]12 = 4 − ¼ = 15/4.

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Properties of the definite integral

Property	Statement	Use
Zero width	∫_a^a f(x) dx = 0	Limits equal → the integral vanishes
Reversal	∫_a^b f = −∫_b^a f	Swapping limits changes the sign
Additivity	∫_a^b f + ∫_b^c f = ∫_a^c f	Split at a point where the rule changes (e.g. \|x\|)
Constant multiple	∫ k·f = k∫ f	Take constants outside
Linearity	∫ (f ± g) = ∫ f ± ∫ g	Integrate term by term
Even and odd	∫_−a^a f = 0 if f is odd; 2∫₀^a f if f is even	Saves a full evaluation

∫₋₁² |x| dx = ∫₋₁⁰ (−x) dx + ∫₀² x dx = ½ + 2 = 5/2.
Splitting at x = 0 is compulsory; integrating |x| as if it were x gives the wrong value.

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Area under and between curves

Area under a curve

If y = f(x) is above the x-axis on [a, b], the area is A = ∫_a^b y dx.
If the curve dips below the x-axis, integrate the absolute value, or split the interval and take the magnitude of the negative part.
Example the area under y = x² from x = 0 to x = 3 is ∫03 x² dx = [x³/3]03 = 9.

Definite Integration, Areas and Volumes: a glass tank filled with sand under the curve y = x² shows definite integration and areas: the area is 8/3.

Area between two curves

Comparison aspect	Main idea	Keywords
•	A = ∫_a^b (upper curve − lower curve) dx, where a and b are the x-coordinates of the intersection points.	uppercurvelowercoordinates
Example	the area between y = x and y = x². Intersections: x = x² → x = 0 or x = 1. On (0 x > x² so A = ∫01 (x − x²) dx = ½ − ⅓ = 1/6	areabetweenIntersections
Example	between y = 4 − x² and the x-axis. Intersections at x = ±2: A = ∫−22 (4 − x²) dx = [4x − x³/3]−22 = 32/3	betweenaxisIntersections

Choosing the variable

If the region is easier to describe as x as a function of y, use A = ∫ (right curve − left curve) dy.
Example between x = y² and x = y + 2 → intersections y = −1, y = 2 → A = ∫−12 (y + 2 − y²) dy = 9/2.

Definite Integration, Areas and Volumes of Revolution · 4 / 9

Volumes of revolution

About the x-axis

Comparison aspect	Main idea	Keywords
•	V = π ∫_a^b y² dx, where y is the height of the curve above the axis of rotation.	heightcurveaboveaxis
Example	y = x² rotated about the x-axis from x = 0 to x = 1: V = π ∫01 x⁴ dx = π/5	rotatedaxis
Example	y = √x from x = 0 to x = 4: V = π ∫04 x dx = π[ x²/2 ]04 = 8π

V = π ∫_c^d x² dy; the curve must be written as x = g(y).
Example x = y² from y = 0 to y = 2: V = π ∫02 y⁴ dy = 32π/5.

V = π ∫ (R² − r²) dx, where R is the outer radius and r the inner radius.
Example the region between y = x and y = x² for 0 ≤ x ≤ 1 rotated about the x-axis: V = π ∫01 (x² − x⁴) dx = π(⅓ − ⅕) = 2π/15.

Definite Integration, Areas and Volumes of Revolution · 5 / 9

Practice with full answer key

∫₀¹ (6x² − 4x + 1) dx → 2 − 2 + 1 = 1
∫₁^e (1/x) dx → [ln x]₁^e = 1
∫₀¹ x e^x² dx → u = x², du = 2x dx → ½(e − 1)
∫₋₂¹ |x| dx → 2 + ½ = 5/2
Area between y = x² and y = 2x → intersections 0 and 2, A = ∫(2x − x²) = 4/3
Area bounded by y = √x, y = 0, x = 4 → ∫₀⁴ √x dx = 16/3
Volume of y = x³ about the x-axis, 0 ≤ x ≤ 2 → π∫x⁶ dx = 128π/7
Volume of y = 1/x about the x-axis, 1 ≤ x ≤ 3 → π(1 − ⅓) = 2π/3
∫₀^π sin x dx → [−cos x]₀^π = 2
If ∫₀³ f(x) dx = 7 and ∫₃⁵ f(x) dx = 2, find ∫₀⁵ f = 9 and ∫₅⁰ f = −9

Definite Integration, Areas and Volumes of Revolution · 6 / 9

Exam traps and key takeaways

💡 High-yield traps

Forgetting to substitute the new limits after a u-substitution — the single most common lost mark.
Reversing limits changes the sign; a negative area is a sign error, not a negative area.
Area is ∫(upper − lower); volume uses y² (or x²), never y — squaring is what makes a solid.
A curve crossing the x-axis splits the area into separate integrals; integrating straight across gives a signed sum.
In volume between curves use R² − r², not (R − r)².
Write the final answer exactly (π/5, 32/3); a decimal-only answer loses the exact-mark.

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Lesson 1 – Integration Techniques and Definite Integrals and Volumes of Revolution

★ Comprehensive Study & Detailed Pedagogical Guide ★
Comprehensive Explanation Advanced Integration Techniques (Substitution and By Parts), Definite Integrals, and Volumes of Revolution
🎯 Targeted Learning Outcomes & Pedagogical Objectives:
• Apply the Integration by Parts formula (∫ u dv = uv – ∫ v du) strategically using the LIATE rule.
• Solve integrals using trigonometric and algebraic substitution methods.
• Evaluate definite integrals and apply the Fundamental Theorem of Calculus.
• Calculate exact areas between curves and volumes of solid revolution about the x-axis and y-axis.
📖 Core Terminology, Scientific Definitions & Glossaries:

Comparison aspect	Main idea	Keywords
🔹 Antiderivative (Indefinite Integral)	The inverse operation of differentiation a family of functions F(x) + C such that F'(x) = f(x)	inverseoperationdifferentiationfamily
🔹 Integration by Parts	A powerful integration technique derived from the product rule of differentiation: ∫ u dv = uv – ∫ v du	powerfultechniquederivedproduct
🔹 Definite Integral	The net signed area between the graph of f(x) and the x-axis over an interval $[a b]$ defined as the limit of Riemann sums: ∫ab f(x) dx = F(b) – F(a)	signedareabetweengraph
🔹 Solid of Revolution	A three-dimensional solid figure obtained by revolving a two-dimensional planar region 360° about an axis of rotation (e.g x-axis or y-axis)	threedimensionalfigureobtained
🔹 LIATE Rule	A priority mnemonic for selecting u in integration by parts: Logarithmic Inverse trigonometric Algebraic Trigonometric Exponential	prioritymnemonicselectingintegration

Comparison aspect	Main idea	Keywords
•	⚖️ Governing Scientific Laws Theorems & Mathematical Models	GoverningScientificLawsTheorems
★ Integration by Parts Formula	[ Formula: ∫ u dv = u v – ∫ v du (Select u following LIATE priority and $dv$ as the easily integrable factor). ]	SelectfollowingLIATEpriority
★ Basic Standard Integrals	[ Formula: ∫ xn dx = (xn+1) / (n+1) + C (n ≠ -1) \| ∫ 1 / x dx = ln \|x\| + C∫ ekx dx = 1 / k ekx + C \| ∫ akx dx = akx / (k ln a) + C∫ sec 2 x dx = tan x + C \| ∫ sec x tan x dx = sec x + C. ]	Formula
★ Area Between Two Curves	[ Formula: A = ∫ab \|f(x) – g(x)\| dx (Upper curve minus lower curve over $[a b]$). ]	FormulaUpperminuslower
★ Volume of Revolution Formulas	[ Formula: About x-axis: Vx = π ∫ab [y(x)]2 dx = π ∫ab [(yupper)2 – (ylower)2] dx.About y-axis: Vy = π ∫cd [x(y)]2 dy = π ∫cd [(xright)2 – (xleft)2] dy. ]	axisyupperylowerxright

📌 Exhaustive In-Depth Pedagogical Breakdown:
◄ I. Strategic Execution of Integration by Parts and the LIATE Hierarchy
Integration by Parts arises from integrating the product rule d(uv) = u dv + v du.
Rearranging yields ∫ u dv = uv – ∫ v du.
The success of this technique hinges upon choosing u such that its derivative $du$ is simpler than u, while $dvis readily integrable.
Mathematicians establish the LIATE hierarchy for choosing u 1. L – Logarithmic functions (ln x, log x)2. I – Inverse trigonometric functions (\arcsin x, \arctan x)3. A – Algebraic polynomials (x, x2, x3)4. T – Trigonometric functions (sin x, cos x)5. E – Exponential functions (ex, 2x)For tabular integration (repeated parts), a polynomial u differentiated to zero against an alternating sign sequence (+, -, +, -$) allows rapid evaluation without cumbersome nested brackets.

Figure (integration_areas_and_volumes_of_revolution.png): Figure (6): Disk and washer methods for computing 3D volumes of revolution about coordinate axes
◄ II. Fundamental Properties of Definite Integrals
Definite integrals obey powerful algebraic symmetries 1. Additivity: ∫ab f(x) dx = ∫ac f(x) dx + ∫cb f(x) dx.2. Reversal: ∫ba f(x) dx = -∫ab f(x) dx and ∫aa f(x) dx = 0.3. Even Function Symmetry: If f(-x) = f(x), then ∫-aa f(x) dx = 2 ∫0a f(x) dx.4. Odd Function Symmetry: If f(-x) = -f(x), then ∫-aa f(x) dx = 0 (Net cancellation of symmetric positive and negative lobes).
Recognizing odd symmetry instantaneously evaluates otherwise formidable integrals, such as ∫_-π^π x³ cos x dx = 0.

◄ III. Geometric Determination of Areas and Volumes of Revolution
To find the volume generated by revolving the region under y = f(x) from x = a to x = b through a complete revolution (2π radians) about the x-axis, we slice the solid into infinitesimally thin circular disks of radius r = y and thickness $dx$.
Each disk has differential volume dV = π r2 dx = π y2 dx.
Integrating across the interval gives V = π ∫ab y2 dx.
When the region is bounded between two curves y1 and y2 (y1 gt; y2), revolving about the x-axis creates circular washers with outer radius R = y1 and inner hole radius r = y2.
The volume is given by the Washer Method: V = π ∫ab (y12 – y22) dx.
Be careful: (y12 – y22) ≠ (y1 – y2)2!

Comparison aspect	Main idea	Keywords
•	🧮 Exemplary Step-by-Step Worked Problems & Applied Exercises	ExemplaryStepWorkedProblems
Worked Example 1	Worked Problem 4 – Integration by Parts and Solid Volume of Revolution	ProblemIntegrationPartsSolid
Problem Statement	Problem:1) Evaluate the indefinite integral ∫ x e2x dx.2) Calculate the volume of the solid generated by revolving the region bounded by the parabola y = x2 the line y = 0 and x = 2 about the x-axis through 360°.Detailed Solution:1) Choose u and $dv$ by LIATE (Algebraic before Exponential):Let u = x ⇒ du = dx.Let dv = e2x dx ⇒ v = ∫ e2x dx = 1 / 2 e2x.Applying ∫ u dv = uv – ∫ v du:∫ x e2x dx = x(1 / 2 e2x) – ∫ 1 / 2 e2x dx = 1 / 2 x e2x – 1 / 4 e2x + C = 1 / 4 e2x(2x – 1) + C.2) Volume of revolution about x-axis:Formula: V = π ∫ab y2 dx.Here y = x2 ⇒ y2 = (x2)2 = x4.Bounds are from x = 0 to x = 2:V = π ∫02 x4 dx = π [ x5 / 5 ]02 = π (32 / 5 – 0 ) = 32π / 5 = 6.4π cubic units	EvaluateindefiniteintegralCalculate

Comparison aspect	Main idea	Keywords
•	💡 Master Teacher Insights & Critical Exam Cautions	MasterTeacherInsightsCritical
★ Washer Method Alert	In volume of revolution between two curves V = π ∫ (y12 – y22) dx. It is a fatal error to write π ∫ (y1 – y2)2 dx!	volumerevolutionbetweencurves
★ Odd Function Shortcut	Always check if the integrand is odd on a symmetric interval $[-a a]$. If f(-x) = -f(x) the integral is immediately ZERO without calculating!	Alwayscheckintegrandsymmetric
★ Integral of ln x	Remember to use integration by parts with u = ln x and dv = dx: ∫ ln x dx = xln x – x + C	Rememberintegrationparts
•	🌍 Practical Real-World Industrial & Societal Connections	PracticalRealWorldIndustrial
✔ Mechanical Engineering & Fluid Tanks	Calculating exact fluid volumes and capacities for custom parabolic hemispherical and elliptical storage tanks and industrial silos	Calculatingexactvolumescapacities

✔ Naval Architecture & Hydrodynamics→Determining center of buoyancy and submerged hull displacement volumes for supertankers and container ships.

Definite Integration, Areas and Volumes of Revolution · 8 / 9

Lesson summary

Comparison aspect	Main idea	Keywords
•	★ High-Yield Summary Mind Maps & Comparison Tables ★	HighYieldSummaryMind
Summary	Advanced Integration Techniques (Substitution and By Parts) Definite Integrals and Volumes of Revolution	AdvancedIntegrationTechniquesSubstitution
•	🗺️ Core Concept Map & Knowledge Architecture	CoreConceptKnowledgeArchitecture
✔ Integration Techniques	Substitution (u = g(x)) \| By Parts (∫ u dv = uv – ∫ v du LIATE)	SubstitutionPartsLIATE
✔ Standard Forms	∫ u’/u = ln \|u\| \| ∫ u’ eu = eu \| ∫ u’ un = un+1/(n+1)
✔ Definite Integrals	Area under curve \| Properties (additivity reversal even/odd symmetry)	AreaundercurveProperties
✔ Volumes of Revolution	Disk Method Vx = π ∫ y2 dx \| Washer Method V = π ∫ (y12 – y22) dx	DiskMethodWasher
•	📊 Classification & Comparative Summary Tables	ClassificationComparativeSummaryTables
Table	Master Summary Table of Advanced Integration Formulas and Geometric Applications	MasterSummaryAdvancedIntegration

Form / Operation	Mathematical Formula	Conditions & Practical Application Notes
Integral of Derivative Over Function	∫ (f'(x)) / (f(x)) dx = ln \|f(x)\| + C	Applicable whenever the numerator is the exact derivative of the denominator
Integration by Parts	∫ u dv = uv – ∫ v du	Choose u following LIATE (Log, Inverse trig, Algebraic, Trig, Exp)
Area Between Two Curves	A = ∫_a^b [y₁(x) – y₂(x)] dx	Requires determining intersection points a and b; y₁ ≥ y₂ on interval
Volume of Revolution (x-axis)	V_x = π ∫_a^b [y(x)]² dx	Revolving region under curve about x-axis (Disk Method)
Volume Between Curves (Washer)	V_x = π ∫_a^b [y_outer² – y_inner²] dx	Washer method subtract squared radii do NOT square the difference!

💡 Exam Flashcard Capsules & High-Yield Secrets:
★ ∫ tan x dx = ∫ (sin x) / (cos x) dx = -ln |cos x| + C = ln |sec x| + C.
★ ∫_-a^a odd function dx = 0 saves minutes of tedious algebra on exams.
★ When revolving about the y-axis, express x in terms of y and integrate with respect to y: V = π ∫ x² dy.
★ Constant of integration C must be included on all indefinite integrals.

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Curriculum

Mathematics

Unit 1

Unit 2

Unit 3

Unit 4

Unit 5

Unit 6

Unit 7

Definite Integration, Areas and Volumes of Revolution