Below the axis → the integral is negative: Where a curve is below the x-axis, the definite integral is negative.
The area is the magnitude: take the absolute value (or split at the x-intercepts and add the magnitudes).
IB-style question — area below the axis
Find the area between y = x² − 4 and the x-axis from x = 0 to x = 2.
Step by step
- Integrate (the curve is below the axis here).
- Area is the magnitude.
Final answer
Area = 16/3 (the integral is −16/3 because the region is below the axis).
Area is never negative: A negative integral just means the region is below the axis — report the positive area.
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Integrate (top − bottom): The area between two curves from x = a to x = b is ∫ₐᵇ (top − bottom) dx, where 'top' is the upper curve.
This works even below the x-axis, because you subtract the lower curve.
The region between two curves — its area is the integral of (upper curve − lower curve) over the interval.
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IB-style question — between two curves
Find the area enclosed between y = x and y = x² (from x = 0 to x = 1).
Step by step
- On [0,1], y = x is above y = x². Integrate the difference.
- Evaluate.
Final answer
Area = 1/6.
Decide which curve is on top: Test a value between the limits (or sketch) to see which curve is higher — that's the 'top'.
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Limits come from intersections / intercepts: If the region is enclosed, the limits are where the curves meet — solve top = bottom (or, for a curve and the axis, solve f(x) = 0).
Then integrate (top − bottom) between those x-values.
IB-style question — find limits then area
Find the area enclosed between y = 6x − x² and y = x.
Step by step
- Intersections: 6x − x² = x ⇒ 5x − x² = 0 ⇒ x(5 − x) = 0 ⇒ x = 0, 5.
- Top is 6x − x²; integrate the difference 5x − x² over [0,5].
Final answer
Area = 125/6 ≈ 20.8.
Solve top = bottom first: Always find the intersection x-values before integrating an enclosed region — they are your limits.