Differentiation and integration
Connect a function with its gradient and use antiderivatives to evaluate signed area.
Exam scope and assessment notes
Pure Mathematics 1 (Paper 1) · Mathematics 9709, 2026–2027. This revision lesson focuses on stationary points and finite polynomial integrals/areas; it does not cover all of sections 1.7–1.8, improper integrals or volumes of revolution. Written solutions use self-check criteria, not an official mark scheme. Specialist accuracy review has not been completed. The diagram illustrates a different cubic from the practice questions; use the function specified in each question.
Before you begin
- Differentiate and integrate polynomial terms, including constants.
- Factor a quadratic and use signs on intervals to interpret a curve.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
A differentiable function has f′(4) = 0, f(4) = −3 and f″(4) = 6. Which description is justified?
What you will learn
- Locate stationary points.
- Evaluate a definite integral with limits.
Differentiate to study change
For a power term axⁿ, the derivative is anxⁿ⁻¹. Differentiate a sum term by term. A stationary point has derivative zero; finding the point also requires substituting x into the original function.
A second derivative can classify a stationary point when it is non-zero: positive indicates a local minimum and negative a local maximum. If it is zero, this test alone is inconclusive.
Integration reverses differentiation
For n≠−1, an antiderivative of axⁿ is axⁿ⁺¹/(n+1). Add a constant for an indefinite integral. For a definite integral, calculate F(upper)−F(lower); the arbitrary constant cancels.
A definite integral gives signed area. Regions below the x-axis contribute negatively. If the question asks for total geometric area, split at roots and account for the signs.
Sources for this lesson
Pause and explain
Between the stated bounds, a graph encloses 3 square units above the x-axis and 8 square units below it. What are the definite integral and the total geometric area?
Worked example
Evaluate the integral of 2x+3 from x=1 to x=4.
Show the worked solution
- An antiderivative is F(x)=x²+3x.
- F(4)=16+12=28 and F(1)=1+3=4.
- Subtract lower from upper: 28−4.
Answer 24
From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
For f(x) = x³ − 3x² − 9x + 6, find every stationary point and classify it. Show full coordinate pairs and how the gradient changes between them.
Give me a hint
Factor f′(x), then evaluate f and f″ separately at each stationary input.
Compare my reasoning
- f′(x) = 3x² − 6x − 9 = 3(x − 3)(x + 1), so the stationary inputs are x = −1 and x = 3.
- Using the original function, f(−1) = 11 and f(3) = −21. The points are (−1, 11) and (3, −21).
- f″(x) = 6x − 6. At −1 this is −12, giving a local maximum; at 3 it is 12, giving a local minimum.
- The derivative is positive for x < −1, negative for −1 < x < 3 and positive for x > 3. These signs confirm both classifications; the factored quadratic gives no other stationary inputs.
Local maximum (−1, 11); local minimum (3, −21).
Look for these in your work
- Solve the derivative equation completely.
- Find heights from the original function, not from its derivative.
- Justify each nature with a non-zero second derivative or gradient signs.
For y = x² − 4 on −1 ≤ x ≤ 3, calculate the definite integral and the total area between the curve and the x-axis. Explain why taking the absolute value of the integral does not give the area.
Give me a hint
Locate roots inside the interval before deciding which pieces need their signs reversed.
Compare my reasoning
- The roots are −2 and 2, but only 2 is inside [−1, 3]. The curve is below the axis on [−1, 2) and above it on (2, 3].
- Use F(x) = x³/3 − 4x. F(−1) = 11/3, F(2) = −16/3 and F(3) = −3. The signed pieces are −9 and 7/3.
- The definite integral is −9 + 7/3 = −20/3. The geometric area is 9 + 7/3 = 34/3 square units. The magnitude 20/3 of the net integral has already lost area through cancellation.
Signed integral −20/3; total area 34/3 square units.
Look for these in your work
- Split only at roots within the given bounds.
- Evaluate an antiderivative at every required endpoint.
- Add signed contributions for the integral and positive magnitudes for area.
For y = 2(x + 1)⁴ − 7, find and classify the stationary point. A learner says that y″ = 0 there means there is no maximum or minimum. Evaluate this claim without assuming the second-derivative test gives an answer.
Give me a hint
Use the sign of y′ on both sides; the original fourth power gives a second way to check the conclusion.
Compare my reasoning
- y′ = 8(x + 1)³ is zero only at x = −1. The original function gives y = −7, so the stationary point is (−1, −7).
- y″ = 24(x + 1)² is zero there. This makes the second-derivative test inconclusive; it does not rule out an extremum.
- For x < −1, y′ is negative; for x > −1, y′ is positive. The curve decreases then increases, so the point is a minimum. Also 2(x + 1)⁴ ≥ 0 for every real x, so −7 is the global minimum value.
Minimum (−1, −7); y″ = 0 makes the second-derivative test inconclusive, not the point unclassifiable.
Look for these in your work
- Find the full stationary coordinate pair.
- State the precise limitation of the second-derivative test.
- Use gradient signs or the non-negative fourth power to justify a minimum.
Common mistakes
- Do not substitute limits into the derivative instead of the antiderivative.
- A stationary x-coordinate is not a full coordinate pair.
What can you explain now?
Find the stationary point of y=x²−6x+2.
Compare with the explanation
(3,−7), a minimum
The derivative 2x−6 is zero at x=3. Substitution gives −7, and the second derivative is 2>0.
After trying it yourself, choose your next review. This is your self-assessment.
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Make it stick
Distinguish the equations f′(x) = 0 and f(x) = 0. Explain what to do when the second-derivative test is inconclusive, then describe why signed integral and total area require different treatment at an axis crossing.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.