Definite integrals: 7 exam questions solved step by step

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What this lesson covers

  • A definite integral gives a signed area, so regions below the axis count as negative.
  • The area between two curves is the integral of the top curve minus the bottom one.
Key idea

7 questions, one method

  • 1. Integrate: show F(x), no + c.
  • 2. F(b) − F(a), with brackets.
  • 3. Areas: split at the roots.
  • 4. Two curves: top minus bottom.
Your turn

Question 1 · Achieved

  • Evaluate ∫₋₁² (4x³ − 6x + 2) dx
Worked example

Q1: ∫₋₁² (4x³ − 6x + 2) dx

  • Integrate each term.
  • Brackets around F(−1).
  1. F(x) = x⁴ − 3x² + 2xintegrate
  2. F(2) = 16 − 12 + 4 = 8top limit
  3. F(−1) = 1 − 3 − 2 = −4bottom limit
  4. F(2) − F(−1) = 8 − (−4)brackets
  5. = 12
  6. 8 − 1 − 3 − 2 = 2brackets dropped
Your turn

Question 2 · Achieved

  • Evaluate exactly
  • ∫ from 0 to π/6 of 4 sin 2x dx
Worked example

Q2: 4 sin 2x, from 0 to π/6

  • Reverse chain rule: ÷ 2.
  • Radians, not degrees.
  1. ∫ sin 2x dx = −(1/2)cos 2xreverse chain rule
  2. ∫ 4 sin 2x dx = −2 cos 2x× 4
  3. = [−2 cos 2x] from 0 to π/6
  4. x = π/6: −2 cos(π/3) = −1top limit
  5. x = 0: −2 cos 0 = −2bottom limit
  6. −1 − (−2) = 1brackets
Your turn

Question 3 · Merit

  • Find the area enclosed by
  • y = x² − 2x − 3 and the x-axis.
Question 3 · Merit
Worked example

Q3: a region below the axis

  • The roots are the limits.
  • An area is positive.
  1. x² − 2x − 3 = (x − 3)(x + 1)factorise
  2. x = −1 or x = 3
  3. F(x) = x³/3 − x² − 3xintegrate
  4. F(3) = 9 − 9 − 9 = −9top limit
  5. F(−1) = −1/3 − 1 + 3 = 5/3bottom limit
  6. ∫₋₁³ = −9 − 5/3 = −32/3below the axis
  7. Area = 32/3 ≈ 10.7 units²
Your turn

Question 4 · Merit

  • y = x³ − 3x² + 2x
  • Find the total area between the curve
  • and the x-axis from x = 0 to x = 2.
  • Careful: the curve crosses the axis.
Question 4 · Merit
Worked example

Q4: roots, then F at each one

  • Split where it crosses.
  • F at 0, 1 and 2.
  1. x³ − 3x² + 2x = x(x − 1)(x − 2)
  2. roots: x = 0, 1, 2
  3. F(x) = x⁴/4 − x³ + x²integrate
  4. F(0) = 0
  5. F(1) = 1/4 − 1 + 1 = 1/4
  6. F(2) = 4 − 8 + 4 = 0
  7. ∫₀² = F(2) − F(0) = 0not the area
Worked example

Q4: add the sizes

  • One integral per region.
  • Add sizes, not signs.
  1. F(0) = 0, F(1) = 1/4, F(2) = 0
  2. ∫₀¹ = 1/4 − 0 = 1/4above the axis
  3. ∫₁² = 0 − 1/4 = −1/4below the axis
  4. Area = 1/4 + 1/4add the sizes
  5. = 1/2 unit²
  6. 1/4 + (−1/4) = 0pieces give ∫₀²
Your turn

Question 5 · Merit, in context

  • A garden bed is the region between
  • y = x² and y = 8 − x², in metres.
  • Find the area of the bed.
Question 5 · Merit, in context
Worked example

Q5: between two curves

  • Limits: where they meet.
  • Top minus bottom.
  1. x² = 8 − x²set equal
  2. 2x² = 8, x² = 4, x = ±2
  3. A = ∫₋₂² ((8 − x²) − x²) dxtop − bottom
  4. = ∫₋₂² (8 − 2x²) dx
  5. = [8x − 2x³/3]₋₂²integrate
  6. F(2) = 16 − 16/3 = 32/3
  7. F(−2) = −16 + 16/3 = −32/3
  8. A = 32/3 − (−32/3) = 64/3
  9. ≈ 21.3 m²
Your turn

Question 6 · Excellence

  • The line y = kx, with k > 0, and
  • the curve y = x² enclose an area
  • of 36 square units. Find k.
Worked example

Q6: y = kx and y = x²

  • The limits depend on k.
  • Area in k, then solve.
  1. x² = kxset equal
  2. x(x − k) = 0, so x = 0 or k
  3. kx ≥ x² for 0 ≤ x ≤ kthe line is on top
  4. A = ∫₀ᵏ (kx − x²) dxtop − bottom
  5. = [kx²/2 − x³/3]₀ᵏk is a constant
  6. = k³/2 − k³/3 = k³/6
  7. k³/6 = 36, so k³ = 216
  8. k = 6
  9. 6³/6 = 216/6 = 36
Your turn

Question 7 · Excellence

  • Find a > 0 so that
  • ∫₀ᵃ (3x² − 6x) dx = 0.
  • Explain what this means for area.
Worked example

Q7: solve for a

  • Integrate, set to 0.
  • Reject with a reason.
  1. [x³ − 3x²]₀ᵃ = 0integrate
  2. a³ − 3a² = 0F(0) = 0
  3. a²(a − 3) = 0factorise
  4. a = 0a > 0, no width
  5. a = 3
Worked example

Q7: what the zero means

  • Find where it crosses.
  • Signed areas cancel.
  1. a = 3
  2. 3x² − 6x = 3x(x − 2): roots 0, 2
  3. ∫₀² = 8 − 12 = −4below the axis
  4. ∫₂³ = 0 − (−4) = 4above the axis
  5. area below = area above = 4
  6. total area 8 units², not 0
Summary

The method that solved all seven

  • Integrate, substitute with brackets, split at roots.
StepHow
Integrateshow F(x), no + c
SubstituteF(b) − F(a), with brackets
Below axissplit at the roots
Add areasadd sizes, not signs
Two curvestop minus bottom
Unknownreject a root with a reason

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