Anti-differentiation: 7 exam questions solved step by step
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.
7 questions, one method
- 1. Raise each power by one.
- 2. Divide by the new power.
- 3. Always add + c.
- 4. A point on the curve finds c.
Question 1 · Achieved
- Find f(x) if f′(x) = 8x³ − 6x² + 2x − 5
Q1: term by term
- Raise the power, divide.
- Do not forget + c.
- 8x³ → 8x⁴/4 = 2x⁴power 3 to 4, ÷ 4
- −6x² → −6x³/3 = −2x³power 2 to 3, ÷ 3
- 2x → 2x²/2 = x²power 1 to 2, ÷ 2
- −5 → −5xa constant k → kx
- f(x) = 2x⁴ − 2x³ + x² − 5x + c
- f′(x) = 8x³ − 6x² + 2x − 5differentiate back
Question 2 · Achieved
- f′(x) = 4x − 3 and f(2) = 9.
- Find f(x).
Q2: f′(x) = 4x − 3, f(2) = 9
- Anti-differentiate, + c.
- Substitute the point.
- f(x) = 4x²/2 − 3x + canti-differentiate
- = 2x² − 3x + c
- f(2) = 2(2)² − 3(2) + c = 9f(2) = 9
- 8 − 6 + c = 9
- 2 + c = 9, so c = 7
- f(x) = 2x² − 3x + 7
- f(2) = 8 − 6 + 7 = 9
Question 3 · Merit
- A curve has gradient dy/dx = 3x² − x + 2
- and passes through (2, 4).
- Find the equation of the curve.
Q3: through (2, 4)
- x is x¹: power to 2.
- Then use the point.
- 3x² → 3x³/3 = x³power 2 to 3, ÷ 3
- −x → −x²/2power 1 to 2, ÷ 2
- 2 → 2xa constant k → kx
- y = x³ − x²/2 + 2x + cadd + c
- 4 = 2³ − 2²/2 + 2(2) + cthrough (2, 4)
- 4 = 8 − 2 + 4 + c
- 4 = 10 + c, so c = −6
- y = x³ − x²/2 + 2x − 6
- check: 3x² − x + 2differentiate back
Question 4 · Merit
- f′(x) = (x + 3)(x − 1) and f(3) = 10.
- Find f(x).
- Careful: you cannot do each bracket.
Q4: brackets first
- Expand before you start.
- Then find c.
- f′(x) = x² − x + 3x − 3expand
- = x² + 2x − 3collect
- f(x) = x³/3 + x² − 3x + canti-differentiate
- f(3) = 27/3 + 9 − 9 + c = 10
- 9 + c = 10, so c = 1
- f(x) = x³/3 + x² − 3x + 1
- check: f′(x) = x² + 2x − 3
Question 5 · Merit, in context
- A sunflower in a school garden grows at
- dh/dt = 12 − 0.8t cm per week.
- It is 60 cm tall after 5 weeks.
- How tall was it at planting, and at 10 weeks?
Q5: the sunflower
- c is the height at t = 0.
- Answer with units.
- h = 12t − 0.8t²/2 + canti-differentiate
- = 12t − 0.4t² + c
- 60 = 12(5) − 0.4(5)² + ch = 60 at t = 5
- 60 = 60 − 10 + c, so c = 10
- h = 12t − 0.4t² + 10
- planted at 10 cmt = 0: h = c
- h = 120 − 40 + 10 = 90t = 10
- 90 cm tall after 10 weeks
Question 6 · Excellence
- A curve has gradient dy/dx = 3x² − 27.
- Its minimum point has y-value −50.
- Find the equation of the curve.
Q6: which point is the minimum?
- Two stationary points.
- Test the signs.
- 3x² − 27 = 0stationary
- x² = 9, so x = 3 or x = −3
- dy/dx at 2: 12 − 27 = −15
- dy/dx at 4: 48 − 27 = 21
- − then +: minimum at x = 3
- x = −3 is the maximumnot the minimum
Q6: find c at the minimum
- Anti-differentiate, + c.
- Use the point (3, −50).
- minimum at x = 3
- y = 3x³/3 − 27x + canti-differentiate
- = x³ − 27x + c
- −50 = 3³ − 27(3) + cthe minimum (3, −50)
- −50 = 27 − 81 + c
- −50 = −54 + c, so c = 4
- y = x³ − 27x + 4
- y(3) = 27 − 81 + 4 = −50
Question 7 · Excellence
- A curve has gradient dy/dx = kx − 6.
- It has a turning point at x = 3
- and passes through (1, 3).
- Find the curve and where it cuts the x-axis.
Q7: find k, then c
- Turning point: gradient 0.
- Then y = 0 for the axis.
- k(3) − 6 = 0, so k = 2turning point at x = 3
- dy/dx = 2x − 6
- y = x² − 6x + canti-differentiate
- 3 = 1² − 6(1) + cthrough (1, 3)
- 3 = −5 + c, so c = 8
- y = x² − 6x + 8
- x² − 6x + 8 = 0cuts the x-axis
- (x − 2)(x − 4) = 0factorise
- cuts at x = 2 and x = 4
The method that solved all seven
- Term by term, + c, then a point to find c.
| Step | How |
|---|---|
| Power | raise by one, divide by it |
| Constant | k becomes kx |
| Always | add + c |
| Brackets | expand first |
| A point | substitute it to find c |
| Check | differentiate back |
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