IB Maths AI HLFurther IntegrationPaper 1 & 2~6 min read
Negative Integrals
When a curve dips below the x-axis, the definite integral there comes out negative — it’s a signed value, not a true area. To find the actual area you must split at the roots, integrate each piece, and add the magnitudes. Miss this and parts above and below the axis can silently cancel to a wrong answer.
📘 What you need to know
Signed area: a definite integral counts area above the axis as positive and below as negative.
Below the axis: where f(x) < 0, the integral is negative.
Area is positive: a true area can never be negative — take the modulus.
Find the roots: solve f(x) = 0 to locate where the curve crosses the axis.
Split & add magnitudes: integrate each interval separately, then sum the absolute values.
Cancellation trap: integrating straight across a crossing can make regions cancel.
Signed area vs true area
A curve crossing the x-axis
Above the axis the integral is positive; below it the integral is negative. True area adds both as positive amounts.
🤔 Why does “below” give a negative?
The definite integral sums f(x) × (tiny width). Where the curve is below the axis, f(x) is negative, so every strip contributes a negative amount. The integral faithfully reports this signed total — useful for things like displacement, but not for a physical area, which is always positive.
🧠 “Roots, split, modulus, add”
Find the roots, split the integral at each crossing, take the modulus of each piece, then add. Four steps stop the above/below cancellation.
Finding the true area
🧭 Recipe — area when the curve crosses the axis
Solvef(x) = 0 to find the roots in the interval.
Split the region at each root into above- and below-axis parts.
Integrate each part separately.
Take the modulus of any negative piece.
Add the magnitudes for the total area.
GDC shortcut (Paper 2): many calculators integrate |f(x)| directly — entering the absolute value gives the true area in one step, no manual splitting.
Signed integral∫abf dxCan be negative or cancel — gives net signed area.
True area∫ab |f| dxAlways positive — split at roots and add magnitudes.
Worked examples
WE 1
Find the area between y = x − 2, the x-axis, and x = 1 to x = 3
The line crosses at x = 2. Straight integration would cancel — split there.
root: x − 2 = 0 → x = 2∫12 (x−2) dx = −12, ∫23 (x−2) dx = 12area = |−12| + |12|area = 1 square unitstraight ∫13 would give 0 — the trap!
WE 2
Find the area between y = x2 − 4x and the x-axis from x = 0 to x = 3
On 0 < x < 3 the curve is entirely below the axis, so the integral is negative throughout.
Find the area between y = x3 and the x-axis from x = −2 to x = 2
The curve is below on the left, above on the right, symmetric about the origin — straight integration cancels to 0.
root at x = 0; split there∫−20 x³ dx = −4, ∫02 x³ dx = 4area = |−4| + |4|area = 8 square unitsstraight ∫−22 = 0 — must split
WE 5
Find the area enclosed between y = x2 − x − 2 and the x-axis
Between its roots the parabola sits below the axis, so the integral is negative.
roots: x² − x − 2 = (x−2)(x+1) = 0 → x = −1, 2∫−12 (x²−x−2) dx = −92area = |−92|area = 92 square units
💡 Top tips
Sketch first — see where the curve is above or below the axis.
Always find the roots when a curve may cross within the interval.
Split at every crossing, integrate each piece, add magnitudes.
“Enclosed area” with no limits means use the roots as the limits.
Modulus the negatives — a true area is never negative.
GDC: integrate |f(x)| to get the area in one go on Paper 2.
⚠ Common mistakes
Integrating straight across a root so positive and negative parts cancel.
Reporting a negative area instead of taking the modulus.
Missing a root inside the interval and treating it as all one sign.
Forgetting the roots ARE the limits for an “enclosed” region.
Adding signed values rather than magnitudes when summing pieces.
Next up — Areas Between Curves. You now know a definite integral is a signed area, and that true area below the axis needs the modulus and a split at the roots. The next topic extends this to two curves: the area trapped between them is the integral of (top − bottom) across their intersection points — and getting the order right keeps every piece positive.
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