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DABOUR HUB / LESSON 2026-2027

Definite Integration, Areas and Volumes of Revolution

Course: Mathematics Topic: Calculus — the definite integral and its applications

🧭 Lesson points — pick a point and study it with its question bank

1. The definite integral1. The definite integralExplanation2. Properties of the definite integral2. Properties of the definite integralExplanation3. Area under and between curves3. Area under and between curvesExplanation4. Volumes of revolution4. Volumes of revolutionExplanation5. Practice with full answer key5. Practice with full answer keyExplanation
  • 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-axis: V = π∫y²dx └── About the y-axis: V = π∫x²dy

01 The definite integral

Definition and notation

  • The definite integral of f from a to b is written ∫ab f(x) dx, where a is the lower limit and b is the upper limit.
  • Fundamental theorem: if F is any antiderivative of f, then ∫ab f(x) dx = F(b) − F(a).
  • The constant of integration cancels: (F(b) + C) − (F(a) + C) = F(b) − F(a). So never write + C in a definite integral.

Worked examples

  1. ∫02 (3x² + 2x) dx = [x³ + x²]02 = (8 + 4) − 0 = 12
  2. ∫14 x−1/2 dx = [2√x]14 = 4 − 2 = 2
  3. ∫0π/2 cos x dx = [sin x]0π/2 = 1 − 0 = 1
  4. ∫01 e2x dx = [½e2x]01 = ½(e² − 1) ≈ 3.19

Integration by substitution in a definite integral

  • 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.

02 Properties of the definite integral

1Zero width
Statement
  • ∫aa f(x) dx = 0
Use
  • Limits equal → the integral vanishes
2Reversal
Statement
  • ∫ab f = −∫ba f
Use
  • Swapping limits changes the sign
3Additivity
Statement
  • ∫ab f + ∫bc f = ∫ac f
Use
  • Split at a point where the rule changes (e.g.
  • |x|)
4Constant multiple
Statement
  • ∫ k·f = k∫ f
Use
  • Take constants outside
5Linearity
Statement
  • ∫ (f ± g) = ∫ f ± ∫ g
Use
  • Integrate term by term
6Even and odd
Statement
  • ∫−aa f = 0 if f is odd; 2∫0a f if f is even
Use
  • Saves a full evaluation

Worked example with an absolute value

  • ∫−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.

03 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 = ∫ab 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.

Area between two curves

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

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.

04 Volumes of revolution

About the x-axis

  • V = π ∫ab y² dx, where y is the height of the curve above the axis of rotation.
  • Example: y = x² rotated about the x-axis from x = 0 to x = 1: V = π ∫01 x⁴ dx = π/5.
  • Example: y = √x from x = 0 to x = 4: V = π ∫04 x dx = π[ x²/2 ]04 = 8π.

About the y-axis

  • V = π ∫cd 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.

Volume between two curves

  • 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.

05 Practice with full answer key

  1. ∫01 (6x² − 4x + 1) dx → 2 − 2 + 1 = 1
  2. ∫1e (1/x) dx → [ln x]1e = 1
  3. ∫01 x ex² dx → u = x², du = 2x dx → ½(e − 1)
  4. ∫−21 |x| dx → 2 + ½ = 5/2
  5. Area between y = x² and y = 2x → intersections 0 and 2, A = ∫(2x − x²) = 4/3
  6. Area bounded by y = √x, y = 0, x = 4 → ∫04 √x dx = 16/3
  7. Volume of y = x³ about the x-axis, 0 ≤ x ≤ 2 → π∫x⁶ dx = 128π/7
  8. Volume of y = 1/x about the x-axis, 1 ≤ x ≤ 3 → π(1 − ⅓) = 2π/3
  9. ∫0π sin x dx → [−cos x]0π = 2
  10. If ∫03 f(x) dx = 7 and ∫35 f(x) dx = 2, find ∫05 f = 9 and ∫50 f = −9

06 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.