Constant of integration: 7 exam questions solved step by step

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What this lesson covers

  • Every indefinite integral carries a constant, and the question's conditions decide its value.
  • A starting value of zero does not mean the constant is zero.
Key idea

7 questions, one method

  • 1. Integrate: + c every time.
  • 2. Substitute the condition.
  • 3. Solve for c before going on.
  • 4. A new constant for each integral.
Your turn

Question 1 · Achieved

  • dy/dx = 4x³ − 6x + 1
  • The curve passes through (2, 9).
  • Find the equation of the curve.
Worked example

Q1: through (2, 9)

  • Integrate, keep + c.
  • The point finds c.
  1. y = ∫ (4x³ − 6x + 1) dx
  2. = x⁴ − 3x² + x + ckeep + c
  3. 9 = 2⁴ − 3(2)² + 2 + cx = 2, y = 9
  4. 9 = 16 − 12 + 2 + c
  5. 9 = 6 + c, so c = 3
  6. y = x⁴ − 3x² + x + 3
  7. 16 − 12 + 2 + 3 = 9
Your turn

Question 2 · Achieved

  • f′(x) = 6√x and f(4) = 20
  • Find f(x).
Worked example

Q2: f′(x) = 6√x, f(4) = 20

  • Root as a power.
  • c can be negative.
  1. f(x) = ∫ 6x^(1/2) dxroot as a power
  2. = 6x^(3/2) ÷ (3/2) + cpower 1/2 to 3/2
  3. = 4x^(3/2) + c
  4. 4 × 4^(3/2) + c = 20f(4) = 20
  5. 4^(3/2) = (√4)³ = 8root, then cube
  6. 32 + c = 20, so c = −12
  7. f(x) = 4x^(3/2) − 12
Your turn

Question 3 · Merit

  • d²y/dx² = 12x − 4
  • At (1, 2) on the curve, the gradient is 5.
  • Find the equation of the curve.
Worked example

Q3: integrate twice

  • Gradient finds c.
  • Point finds d.
  1. dy/dx = ∫ (12x − 4) dx
  2. = 6x² − 4x + cfirst constant, c
  3. 5 = 6 − 4 + c, so c = 3gradient 5 at x = 1
  4. dy/dx = 6x² − 4x + 3
  5. y = 2x³ − 2x² + 3x + dsecond constant, d
  6. 2 = 2 − 2 + 3 + dthrough (1, 2)
  7. d = −1
  8. y = 2x³ − 2x² + 3x − 1
Your turn

Question 4 · Merit, in context

  • A jet ski moves in a straight line
  • with acceleration a = 4 − 2t m/s².
  • At t = 0, v = 1 m/s and s = 0.
  • Find its displacement after 3 s.
Worked example

Q4: a to v to s

  • v first, find c.
  • Then s, a new constant.
  1. v = ∫ (4 − 2t) dt = 4t − t² + c
  2. 1 = 0 − 0 + c, so c = 1v = 1 at t = 0
  3. v = 4t − t² + 1
  4. s = ∫ (4t − t² + 1) dt
  5. = 2t² − t³/3 + t + da new constant, d
  6. 0 = 0 + d, so d = 0s = 0 at t = 0
  7. s(3) = 2(9) − 27/3 + 3
  8. = 18 − 9 + 3 = 12
  9. 12 m from its start after 3 s
Your turn

Question 5 · Merit, in context

  • A cyclist starts from rest with velocity
  • v = 10 − 10e^(−0.5t) m/s.
  • Her displacement is 0 at t = 0.
  • Find her displacement after 4 s.
  • Careful with the constant.
Worked example

Q5: find c

  • Integrate, keep + c.
  • e⁰ is 1, not 0.
  1. s = ∫ (10 − 10e^(−0.5t)) dt
  2. −10 ÷ (−0.5) = 20÷ a = −0.5
  3. s = 10t + 20e^(−0.5t) + c
  4. 0 = 0 + 20e⁰ + cs = 0 at t = 0
  5. 0 = 20 + c, so c = −20e⁰ = 1, not 0
  6. s = 10t + 20e^(−0.5t) − 20
Worked example

Q5: after 4 seconds

  • Substitute t = 4.
  • Round at the end.
  1. s = 10t + 20e^(−0.5t) − 20
  2. s(4) = 10(4) + 20e^(−2) − 20
  3. = 20 + 20e^(−2)
  4. = 20 + 2.707 = 22.707
  5. about 22.7 m after 4 s
  6. c = 0 gives 42.7 mstarts 20 m out
  7. ds/dt = 10 − 10e^(−0.5t)
Your turn

Question 6 · Excellence, in context

  • A float bobs on a wave, with
  • a = −8cos 2t cm/s² (up positive).
  • At t = 0 it is at rest, with s = 0.
  • Find s, and its lowest point.
Worked example

Q6: v, then s

  • sin 0 = 0, so c = 0.
  • cos 0 = 1, so d ≠ 0.
  1. v = ∫ −8cos 2t dt
  2. = −4 sin 2t + ccos to sin, ÷ 2
  3. 0 = −4 sin 0 + c, so c = 0sin 0 = 0
  4. v = −4 sin 2t
  5. s = ∫ −4 sin 2t dt
  6. = 2cos 2t + dsin to −cos, ÷ 2
  7. 0 = 2cos 0 + d = 2 + dcos 0 = 1, not 0
  8. d = −2
  9. s = 2cos 2t − 2
Worked example

Q6: the lowest point

  • cos is never below −1.
  • Answer in words.
  1. s = 2cos 2t − 2
  2. −1 ≤ cos 2t ≤ 1
  3. lowest when cos 2t = −1
  4. s = 2(−1) − 2 = −4
  5. 2t = π, so t = π/2 ≈ 1.57the first time
  6. 4 cm below its start
  7. first at t = π/2 ≈ 1.57 s
Your turn

Question 7 · Excellence

  • A curve has dy/dx = 3x² + kx
  • and a stationary point at (2, −1).
  • Find the equation of the curve.
Worked example

Q7: find k, then c

  • Stationary: dy/dx = 0.
  • Then the point finds c.
  1. stationary at x = 2: dy/dx = 0
  2. 3(2)² + 2k = 0
  3. 12 + 2k = 0, so k = −6
  4. dy/dx = 3x² − 6x
  5. y = x³ − 3x² + cintegrate, + c
  6. −1 = 2³ − 3(2)² + cthrough (2, −1)
  7. −1 = 8 − 12 + c, so c = 3
  8. y = x³ − 3x² + 3
  9. 8 − 12 + 3 = −1
Summary

The method that solved all seven

  • Integrate with + c, then let the question decide it.
StepHow
Integrate+ c, every time
Find csubstitute the condition
a, v, sa new constant each time
At t = 0e⁰ = 1 and cos 0 = 1
Stationarydy/dx = 0 there
Checkit fits the start

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