NCEA Level 2Anti-differentiationAS 91262

Anti-differentiation and why you need + c

What this lesson covers

  • Anti-differentiation reverses differentiation, and every answer has a family of curves, so + c is needed.
  • A known point on the curve fixes the value of c.
Key idea

Going backwards

  • Differentiating takes a function to its gradient.
  • Anti-differentiating goes back the other way.
  • Given the gradient, find the function.
Key idea

The rule, reversed

  • Raise the power by one.
  • Divide by the new power.
  • Then add + c.
TermDoGet
xⁿraise the power by 1xⁿ⁺¹
divide by the new powerxⁿ⁺¹/(n + 1)
a constant kbecomeskx
alwaysadd+ c
Worked example

Anti-differentiate 6x² − 4x + 3

  • Term by term.
  • Three terms in, three out.
  1. 6x² → 6x³/3 = 2x³power 2 to 3, ÷ 3
  2. −4x → −4x²/2 = −2x²power 1 to 2, ÷ 2
  3. 3 → 3xa constant k → kx
  4. 2x³ − 2x² + 3x + cadd + c
  5. check: 6x² − 4x + 3differentiate back
Worked example

Fractions, and checking

  • Same rule for fractions.
  • Differentiate back to check.
  1. x² → x³/3 + cpower 2 to 3, ÷ 3
  2. 3 − x/2 = 3 − ½xx/2 is ½x
  3. 3 → 3xa constant k → kx
  4. −½x → −½ × x²/2power 1 to 2, ÷ 2
  5. = −x²/4½ ÷ 2 = ¼
  6. 3x − x²/4 + c
  7. check: 3 − 2x/4 = 3 − x/2differentiate back
Key idea

Why + c: a family of curves

  • x², x² + 2 and x² − 2 all have gradient 2x.
  • At x = 1 they are all equally steep.
  • The gradient cannot tell them apart.
Why + c: a family of curves
Worked example

A point picks one curve

  • Suppose the curve passes through (1, 3).
  • 1 + c = 3, so c = 2.
  • Only y = x² + 2 goes through it.
A point picks one curve
Worked example

Finding c, written out

  • Anti-differentiate, with + c.
  • Substitute the point.
  • Solve for c.
  1. f′(x) = 2x + 5
  2. f(x) = x² + 5x + canti-differentiate
  3. f(1) = 1² + 5(1) + c = 10through (1, 10)
  4. 6 + c = 10, so c = 4
  5. f(x) = x² + 5x + 4
  6. check: f′(x) = 2x + 5
  7. check: f(1) = 1 + 5 + 4 = 10
Key idea

The two ways marks go

  • Leaving out + c.
  • Dropping a term from a long expression.
  • Count terms in, count terms out.
Worked example

In context: filling a tank

  • It starts with 50 litres.
  • V(0) = 50 fixes c.
  1. dV/dt = 20 − 2tlitres a minute
  2. V = 20t − 2t²/2 + canti-differentiate
  3. = 20t − t² + c
  4. V(0) = 0 − 0 + c = 5050 L at the start
  5. c = 50, V = 20t − t² + 50
  6. V(5) = 20 × 5 − 5² + 50
  7. = 100 − 25 + 50 = 125 L
Key idea

Where the marks are

  • Achieved: a general anti-derivative with + c.
  • Merit: c found from a point.
  • Excellence: used in a real problem.
Summary

What to take away

  • Raise, divide, + c, then let a point fix c.
Anti-differentiateraise the power, divide by it
Constant kbecomes kx
Every time+ c
Whya family of curves, same gradient
A pointfixes c
Checkdifferentiate your answer back
Your turn

Your turn: four questions

  • Pause here and answer each one.
  • Every answer needs + c, or a value for it.
  • Check by differentiating back.
Question
1Anti-differentiate 4x³ + 2x.
2f′(x) = 6x and f(2) = 15. Find f(x).
3Anti-differentiate x²/2.
4Gradient 2x − 3, through (0, 4). Find the curve.
Show answer

1. x⁴ + x² + c. 2. f(x) = 3x² + c; 12 + c = 15, so c = 3 and f(x) = 3x² + 3. 3. x³/6 + c. 4. y = x² − 3x + c; at (0, 4), c = 4, so y = x² − 3x + 4.

Worked example

The answers: questions 1 and 2

  1. 1. 4x³ → 4x⁴/4 = x⁴power 3 to 4, ÷ 4
  2. 2x → 2x²/2 = x²power 1 to 2, ÷ 2
  3. x⁴ + x² + cadd + c
  4. 2. f(x) = 6x²/2 + c = 3x² + canti-differentiate
  5. f(2) = 3 × 2² + c = 15f(2) = 15
  6. 12 + c = 15, so c = 3
  7. f(x) = 3x² + 3
Worked example

The answers: questions 3 and 4

  1. 3. x²/2 → x³/(2 × 3)power 2 to 3, ÷ 3
  2. = x³/6 + cadd + c
  3. 4. 2x → x², −3 → −3xanti-differentiate
  4. y = x² − 3x + c
  5. 4 = 0² − 3(0) + cthrough (0, 4)
  6. c = 4
  7. y = x² − 3x + 4
  8. check: dy/dx = 2x − 3

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