The constant of integration: find c, never assume it

A family of curves: y = x² − 2x + c

• Each c shifts the curve up or down.
• Same gradient at every x.
• The point (3, 5) picks c = 2.

The curve through (3, 5)

• Integrate, keep + c.
• Substitute the point.

1. y = ∫ (2x − 2) dx
2. = x² − 2x + ckeep + c
3. 5 = 3² − 2(3) + cx = 3, y = 5
4. 5 = 3 + c
5. c = 2
6. y = x² − 2x + 2

Finding c from a point

• Integrate, keep + c.
• Substitute the point.
• Then write the curve.

1. dy/dx = 6x² − 4x, through (1, 5)
2. y = 6x³/3 − 4x²/2 + cintegrate, + c
3. y = 2x³ − 2x² + c
4. 5 = 2(1)³ − 2(1)² + cx = 1, y = 5
5. 5 = 2 − 2 + c, so c = 5
6. y = 2x³ − 2x² + 5

Acceleration to velocity

• v = 3, s = 0 when t = 0.
• Integrate a to get v.

1. a = 6t − 4
2. v = ∫ (6t − 4) dt
3. = 3t² − 4t + ckeep + c
4. v(0) = 3: 3 = 0 − 0 + c
5. c = 3
6. v = 3t² − 4t + 3

Velocity to displacement

• Integrate v to get s.
• A new letter: d.

1. v = 3t² − 4t + 3
2. s = ∫ (3t² − 4t + 3) dt
3. = t³ − 2t² + 3t + da new constant, d
4. s(0) = 0: 0 = 0 + d
5. d = 0
6. s = t³ − 2t² + 3t
7. s(2) = 8 − 8 + 6 = 6t = 2
8. s = 6 m at t = 2

v = 8e^(−2t), s = 0 at t = 0

• Integrate, keep + c.
• e⁰ = 1, not 0.
• So c = 4, not 0.

1. s = ∫ 8e^(−2t) dt
2. = 8 × (1/(−2))e^(−2t) + c÷ −2
3. = −4e^(−2t) + c
4. 0 = −4e⁰ + ct = 0, s = 0
5. 0 = −4 + ce⁰ = 1
6. c = 4
7. s = 4 − 4e^(−2t)

Does the answer fit?

• c = 0 starts at −4.
• The correct s starts at 0.
• It approaches 4 m.

1. c = 0: s = −4e^(−2t)
2. s(0) = −4e⁰ = −4 ≠ 0does not fit
3. s(0) = 4 − 4e⁰ = 0fits s(0) = 0
4. s(1) = 4 − 4e^(−2)
5. ≈ 3.46 mcalculator
6. t → ∞: e^(−2t) → 0
7. s → 4 m

Cos does it too: v = 2sin t

• tⁿ, sin t: 0 at t = 0.
• eᵏᵗ, cos t: 1 at t = 0.
• Substitute, never assume.

1. s = ∫ 2sin t dt
2. = −2cos t + csin to −cos
3. 0 = −2cos 0 + ct = 0, s = 0
4. 0 = −2 + ccos 0 = 1
5. c = 2
6. s = 2 − 2cos t

The answers: questions 1 and 2

1. 1. y = x³ + 2x + cintegrate 3x² + 2
2. 6 = 1 + 2 + c, so c = 3point (1, 6)
3. y = x³ + 2x + 3
4. 2. v = ∫ 12t dt = 6t² + c
5. v(0) = 2: 2 = 0 + c, c = 2
6. v = 6t² + 2
7. s = ∫ (6t² + 2) dt = 2t³ + 2t + d
8. s(0) = 1: 1 = 0 + d, d = 1
9. s = 2t³ + 2t + 1

The answers: questions 3 and 4

1. 3. s = ∫ 6e^(3t) dt
2. = 2e^(3t) + c÷ 3
3. 0 = 2e⁰ + c = 2 + ct = 0, s = 0
4. c = −2, s = 2e^(3t) − 2
5. 4. s = ∫ 4sin 2t dt
6. = −2cos 2t + csin to −cos, ÷ 2
7. 0 = −2cos 0 + c = −2 + ct = 0, s = 0
8. c = 2, s = 2 − 2cos 2t