NCEA Level 3Definite integrals and areaAS 91579

Definite integrals and area, including between two curves

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

A definite integral

  • Integrate, then F(b) − F(a).
  • The answer is a signed area.
  • Below the x axis counts as negative.
Worked example

∫ from −1 to 2 of (3x² − 2x) dx

  • Integrate each term.
  • Top limit, then bottom.
  • Subtract, with brackets.
  1. ∫₋₁² (3x² − 2x) dx
  2. = [x³ − x²]₋₁²integrate each term
  3. F(2) = 2³ − 2² = 8 − 4 = 4top limit
  4. F(−1) = (−1)³ − (−1)²bottom limit
  5. = −1 − 1 = −2
  6. F(2) − F(−1) = 4 − (−2)brackets
  7. = 6
Worked example

Missing brackets: a 2025 error

  • No brackets: a sign is lost.
  • Put all of F(a) in brackets.
  1. F(2) = 4, F(−1) = −1 − 1
  2. 4 − −1 − 1no brackets
  3. = 4 + 1 − 1 = 4sign lost
  4. 4 − (−1 − 1)all of F(a) in brackets
  5. = 4 − (−2)
  6. = 6
Worked example

y = x² − 4x, from 0 to 5

  • Below the axis from 0 to 4.
  • Above the axis from 4 to 5.
  • Straight across: −25/3.
y = x² − 4x, from 0 to 5
Worked example

Roots, then F at every limit

  • Split at the roots.
  • Evaluate F at 0, 4 and 5.
  1. x² − 4x = x(x − 4) = 0factorise
  2. x = 0 or x = 4
  3. F(x) = x³/3 − 2x²integrate
  4. F(0) = 0
  5. F(4) = 64/3 − 32 = −32/3
  6. F(5) = 125/3 − 50 = −25/3
  7. ∫₀⁵ = F(5) − F(0) = −25/3not the area
Worked example

Split at the roots

  • Integrate each piece.
  • Add the sizes, not signs.
  1. F(0) = 0
  2. F(4) = −32/3
  3. F(5) = −25/3
  4. ∫₀⁴ = F(4) − F(0) = −32/3below the axis
  5. ∫₄⁵ = F(5) − F(4)
  6. = −25/3 − (−32/3) = 7/3above the axis
  7. Area = 32/3 + 7/3add the sizes
  8. = 39/3 = 13
  9. −32/3 + 7/3 = −25/3pieces add up
Key idea

The area between two curves

  • Integrate top minus bottom.
  • Limits: where the curves meet.
  • If they swap over, split there.
Worked example

Between y = x and y = x² − 2

  • They meet at two points.
  • The line is on top.
  • Top minus bottom: no splitting.
Between y = x and y = x² − 2
Worked example

Where they meet, and the integral

  • Set the curves equal.
  • Then top minus bottom.
  1. x² − 2 = xset equal
  2. x² − x − 2 = 0rearrange
  3. (x − 2)(x + 1) = 0factorise
  4. x = −1 or x = 2
  5. A = ∫₋₁² (x − (x² − 2)) dxtop − bottom
  6. = ∫₋₁² (2 + x − x²) dxsimplify
Worked example

Evaluating the area

  • Integrate, then subtract.
  • Area = 9/2.
  1. A = ∫₋₁² (2 + x − x²) dx
  2. = [2x + x²/2 − x³/3]₋₁²integrate
  3. F(2) = 4 + 2 − 8/3 = 10/3
  4. F(−1) = −2 + 1/2 + 1/3
  5. = −7/6
  6. A = 10/3 − (−7/6)brackets
  7. = 20/6 + 7/6 = 27/6
  8. = 9/2
Worked example

Solving for an unknown limit

  • Integrate, then set equal.
  • Reject a root, and say why.
  1. ∫₀ᵏ (2x + 1) dx = 12
  2. [x² + x]₀ᵏ = 12integrate
  3. (k² + k) − 0 = 12substitute k and 0
  4. k² + k − 12 = 0rearrange
  5. (k − 3)(k + 4) = 0factorise
  6. k = 3 or k = −4
  7. k = −4k must be positive
  8. k = 3
  9. 3² + 3 = 12
Worked example

Keep the 1/a factor

  • Reverse chain rule: ÷ 2.
  • Dropping it scored nothing.
  1. ∫ e²ˣ dx, from 0 to ln 2
  2. = [(1/2)e²ˣ] from 0 to ln 2
  3. e^(2 ln 2) = e^(ln 4) = 4top limit
  4. e⁰ = 1bottom limit
  5. = (1/2)(4 − 1) = 3/2
  6. 4 − 1 = 3no 1/2: no credit
Worked example

Trig limits are in radians

  • Integrate: divide by 3.
  • Check the calculator mode.
  1. ∫ cos 3x dx, from 0 to π/6
  2. = [(1/3)sin 3x] from 0 to π/6
  3. 3(π/6) = π/2, sin(π/2) = 1top limit
  4. sin 0 = 0bottom limit
  5. = (1/3)(1 − 0) = 1/3
Key idea

Show the integrated function

  • No credit for a calculator answer alone.
  • Unknown constants stop the calculator.
  • Brackets, substitution, answer.
Key idea

Where the marks are

  • Achieved: evaluate, or solve for an unknown.
  • Merit: areas, including below the axis.
  • Excellence: unfamiliar areas, clear notation.
Summary

What to take away

  • Signed area: split at the roots, and integrate top minus bottom.
EvaluateF(b) − F(a)
Bracketsaround F(a)
Below the axissplit at the roots
Between curvestop − bottom
Unknown limitcheck both roots
Show itevery integration
Your turn

Your turn: four questions

  • Pause here and answer each one.
  • Question 2: find the roots first.
  • Question 3: which curve is on top?
GivenFind
16x² − 4x, x = 1 to 2the integral
2y = x² − 3x, x = 0 to 4total area
3y = 6 − x² and y = xarea enclosed
4∫₀ᵏ e²ˣ dx = 4k, exactly
Show answer

1. [2x³ − 2x²] from 1 to 2 = 8 − 0 = 8. 2. Roots 0 and 3: ∫₀³ = −9/2 and ∫₃⁴ = 11/6, so the area is 9/2 + 11/6 = 19/3. 3. They meet at x = −3 and x = 2; ∫(6 − x − x²) dx from −3 to 2 = 125/6 ≈ 20.8. 4. (e²ᵏ − 1)/2 = 4, so e²ᵏ = 9 and k = ln 3 ≈ 1.10.

Worked example

The answers: questions 1 and 2

  1. 1. [2x³ − 2x²]₁²integrate
  2. = (16 − 8) − (2 − 2) = 8
  3. 2. x(x − 3) = 0, so x = 0 or 3roots
  4. F(x) = x³/3 − 3x²/2integrate
  5. ∫₀³ = F(3) − F(0) = 9 − 27/2
  6. = −9/2
  7. ∫₃⁴ = F(4) − F(3)
  8. = −8/3 − (−9/2) = 11/6
  9. Area = 9/2 + 11/6 = 19/3add the sizes
Worked example

The answers: question 3

  1. 3. 6 − x² = xset equal
  2. x² + x − 6 = 0rearrange
  3. (x + 3)(x − 2) = 0, so x = −3 or 2
  4. A = ∫₋₃² (6 − x² − x) dxparabola on top
  5. = [6x − x³/3 − x²/2]₋₃²integrate
  6. F(2) = 12 − 8/3 − 2 = 22/3
  7. F(−3) = −18 + 9 − 9/2 = −27/2
  8. A = 22/3 − (−27/2) = 125/6
  9. ≈ 20.8
Worked example

The answers: question 4

  1. 4. ∫₀ᵏ e²ˣ dx = [(1/2)e²ˣ]₀ᵏreverse chain rule
  2. = (1/2)(e²ᵏ − 1) = 4substitute k and 0
  3. e²ᵏ − 1 = 8, so e²ᵏ = 9multiply by 2
  4. 2k = ln 9ln both sides
  5. k = (1/2)ln 9 = ln 3ln 9 = 2 ln 3
  6. ≈ 1.10

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