Integration
Integration is the reverse of differentiation. Indefinite integrals need a constant, + c; definite integrals give a number, which is the area between a curve and the x-axis when the curve is above it.
Part 1 of 3: Learn it
In short
- ∫xⁿ dx = xⁿ⁺¹ ÷ (n + 1) + c, for any n except −1.
- Use a known point on the curve to find the value of c.
- A definite integral: integrate, then substitute the top limit minus the bottom limit.
Where this is in your specification
Spec points: DfE H1 to H3 (AQA 7357 H1 to H3, Edexcel 9MA0 Pure topic 8, OCR A H240 1.08)
| Board | Topic: Integration |
|---|---|
| DfE content | H1-H4 |
| AQA 7357 | H1-H4 |
| Edexcel 9MA0 | Pure topic 8 |
| OCR H240 | 1.08 |
| Higher C847 76 | Calculus skills: integration, area between curves |
Integrating powers
- Raise the power by one, then divide by the new power: 6x² becomes 2x³.
- A constant k integrates to kx.
- Rewrite roots and fractions as powers first: 1/x³ = x⁻³, which integrates to −½x⁻² + c.
- Expand brackets and split fractions before integrating; there is no product rule for integration.
Finding a curve
If you know dy/dx and one point on the curve, integrate to get y with + c, then put in the point's coordinates to find c.
Definite integrals and area
The constant cancels, so leave it out. When the curve is above the x-axis between a and b, the answer is the area under it. When the curve is below the axis, the integral is negative: the area is its size. If the curve crosses the axis, find each part separately and add their sizes.
Find ∫ (6x² − 4x + 5) dx.
Show the answer
2x³ − 2x² + 5x + c.
Part 2 of 3: See it worked
Worked examples
Example 1
Evaluate ∫₁³ (3x² + 2) dx.
- Integrate: [x³ + 2x]
- Top limit: 27 + 6 = 33. Bottom limit: 1 + 2 = 3
- 33 − 3
Answer: 30.
Example 2
dy/dx = 4x − 3, and the curve passes through (2, 5). Find y in terms of x.
- y = 2x² − 3x + c
- 5 = 2(4) − 6 + c = 2 + c
- c = 3
Answer: y = 2x² − 3x + 3.
Example 3
Find the area enclosed by y = x(4 − x) and the x-axis.
- The curve meets the axis at x = 0 and x = 4, and is above it in between
- ∫₀⁴ (4x − x²) dx = [2x² − x³/3]₀⁴
- = 32 − 64/3 = 32/3
Answer: 32/3 square units (about 10.67).
Common mistakes
- Forgetting + c on an indefinite integral.
- Multiplying by the new power instead of dividing by it.
- Integrating a product term by term without expanding first.
- Taking a negative definite integral as the area without making it positive.
Find ∫ x⁻² dx.
Show the answer
−x⁻¹ + c, which is −1/x + c.
Part 3 of 3: Test yourself
Check yourself
Answer each one in your head or on paper first, then open it to check.
Find ∫ (6x² − 4x + 5) dx.
2x³ − 2x² + 5x + c.
Find ∫ x⁻² dx.
−x⁻¹ + c, which is −1/x + c.
Evaluate ∫₀² x³ dx.
[x⁴/4] from 0 to 2 = 16/4 − 0 = 4.
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