Area under and between curves
Objective. Split intervals and distinguish signed integral from area.
- 1. Learn
- 2. Worked example
- 3. Practice
- 4. Feedback
- 5. Continue
Learn: the key idea
The area under a curve between two x-values equals a definite integral only when the curve stays above the x-axis; where the curve dips below, the integral gives a negative value that must be made positive before adding to a total area calculation. For the area between two curves y = f(x) and y = g(x), integrate the difference ∫ₐᵇ [f(x) − g(x)] dx between their intersection points, always taking the upper curve minus the lower one. For example, the area between y = x and y = x² from x = 0 to x = 1 is ∫₀¹ (x − x²) dx = [x²/2 − x³/3]₀¹ = 1/6. Splitting the interval at any x-axis crossing keeps regions above and below separately positive.
Worked example
Find the area enclosed between the curves y = x² and y = 2x for x between 0 and 2.
- 1Identify the upper curve: for 0 < x < 2, 2x ≥ x², so area = ∫₀² (2x − x²) dx.
- 2Find the antiderivative: x² − x³/3.
- 3Evaluate at x = 2: 4 − 8/3 = 4/3.
- 4Evaluate at x = 0: 0, so the area is 4/3 − 0 = 4/3.
Answer: 4/3
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Practice
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Start practisingCommon mistake. Students often integrate across an x-axis crossing in one go, letting positive and negative regions cancel so the total understates the true area. Always split at the root and add the absolute value of each part.
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