IB Maths AA SL Topic 5 — Calculus Paper 1 & 2 ~7 min read

Negative Integrals

A definite integral can come out negative. That’s not an error — it’s the integral telling you the curve dips below the x-axis. But area is always positive, so when the curve goes below, you need an extra step to recover the actual area: either drop a minus sign, or wrap the function in modulus bars.

📘 What you need to know

Definite integrals measure signed area: regions above the x-axis count as positive, regions below count as negative.
If the curve sits entirely below the x-axis, the integral comes out negative — take the absolute value to get the area.
If the curve partly above and partly below, the negative chunk cancels part of the positive — the raw integral is misleading.
The fix: integrate the modulus A = ∫_a^b |y| dx — this reflects negative parts back above the axis.
Or split at the roots of f(x) = 0, integrate each chunk separately, and add their absolute values.

Why does the integral go negative?

When a curve sits below the x-axis, the y-values are negative. So when you sum up all the tiny strips of “height × width” that make up the integral, you’re adding negative numbers — and the result is negative.

Signed area — positive above, negative below

A definite integral is not the same thing as the geometric area — it’s signed. If the question asks for “the value of the integral”, give the signed answer (could be negative). If it asks for “the area”, you must make it positive.

Case 1: curve fully below the x-axis

This is the easy case. Integrate normally — the answer comes out negative — then take the absolute value.

📍

Quick rule

If you compute ∫_a^b f(x) dx and get something like I = −18, the area is just |I| = 18 square units.

Case 2: curve partly above, partly below

This is where students lose marks. If you integrate straight through, the negative bit cancels some of the positive bit and you get the wrong number for area.

The correct formula uses the modulus:

Area between curve and x-axis

A = ∫_a^b |y| dx

✓ in formula booklet

Visually, the modulus reflects any part of the curve that’s below the x-axis, flipping it up so all the area is positive.

Two ways to handle it

method A · GDC Integrate |y| directly

A = ∫_a^b |f(x)| dx

Use on Paper 2 — your GDC has an “Abs” function. One calculation, done.

method B · manual Split at the roots

A = |I₁| + |I₂| + …

Use on Paper 1 — find where f(x) = 0, integrate each chunk, take |·| of any negatives, sum.

🧠

“Reflect the negatives, then add”

The modulus flips below-axis regions up to the positive side. Whether you do that with the |·| symbol on a GDC, or by hand by chopping at the roots and dropping minus signs — the result is the same.

Step-by-step method

Finding total area when the curve crosses the x-axis

Sketch the curve on your GDC and identify the limits (vertical lines or the boundary x-values).
Find the roots in your interval — solve f(x) = 0. These are where the curve switches above/below.
Choose your method. Paper 2: integrate |f(x)| on the GDC. Paper 1: split into chunks at the roots.
Add up the absolute values of each chunk’s integral. The total is the area.

Worked examples

WE 1

Curve fully below the x-axis

Find the area enclosed by the curve y = x² − 4, the x-axis, and the lines x = −2 and x = 2.

step 1 — check position y = x² − 4 has roots at x = ±2 between x = −2 and 2, the curve is below the axisstep 2 — integrate I = ∫(−2 to 2) (x² − 4) dx = [x³/3 − 4x] from −2 to 2 = (8/3 − 8) − (−8/3 + 8) = −16/3 − 16/3 = −32/3step 3 — take absolute value Area = 32/3 square units the integral was negative because the curve sits below the axis — the area is its size, not its sign!

WE 2

Curve crosses the x-axis — split at the root

Find the total area between the curve y = x³, the x-axis, and the lines x = −1 and x = 2.

step 1 — find the root x³ = 0 at x = 0 — curve crosses here below axis: −1 to 0 | above axis: 0 to 2step 2 — integrate each chunk I₁ = ∫(−1 to 0) x³ dx = [x⁴/4] from −1 to 0 = 0 − ¼ = −¼ I₂ = ∫(0 to 2) x³ dx = [x⁴/4] from 0 to 2 = 4 − 0 = 4step 3 — sum absolute values A = |−¼| + |4| = ¼ + 4 = 17/4 Area = 17/4 square units straight ∫(−1 to 2) x³ dx would give 4 − ¼ = 15/4 — wrong! the negative chunk cancelled part of the positive.

WE 3

Trig curve over a full period

Find the total area between the curve y = sin x and the x-axis from x = 0 to x = 2π.

step 1 — locate the root sin x = 0 at x = π in this interval positive: 0 to π | negative: π to 2πstep 2 — integrate each chunk I₁ = ∫(0 to π) sin x dx = [−cos x] from 0 to π = −(−1) − (−1) = 2 I₂ = ∫(π to 2π) sin x dx = [−cos x] from π to 2π = −1 − (−(−1)) = −1 − 1 = −2step 3 — sum absolute values A = |2| + |−2| = 4 Area = 4 square units straight ∫(0 to 2π) sin x dx = 0 — the two halves cancel exactly. always sketch first!

WE 4

Quadratic with the modulus shortcut on GDC

Find the area between the curve y = x² − x − 2 and the x-axis from x = 0 to x = 3.

step 1 — factorise + check sign y = (x + 1)(x − 2) → roots at −1 and 2 on (0, 2) curve is below; on (2, 3) it’s abovestep 2 — GDC method A = ∫(0 to 3) |x² − x − 2| dx type with Abs( ) on GDC GDC: 31/6Area = 31/6 square units GDC + modulus = one-step answer. always faster than splitting on Paper 2!

💡 Top tips

Always sketch first. One look tells you whether the curve crosses the x-axis in your interval.
Read the question carefully. “Value of the integral” can be negative. “Area” is always positive.
Find the roots before splitting — they tell you where to break the integral.
Use modulus on Paper 2. ∫|f(x)| dx on a GDC is one button press.
Check sign at each chunk — drop minus signs only when adding for total area.
Symmetry helps — if the curve is symmetric (like sin x over 0 to 2π), the chunks may be equal in size.

⚠ Common mistakes

Reporting a negative area. Area is never negative — that’s the integral, not the area.
Not splitting when the curve crosses the x-axis. Negative and positive chunks cancel.
Forgetting modulus bars in the formula ∫|y|dx when writing it down.
Skipping the sketch — leads to picking the wrong limits or missing a sign change.
Adding signed values instead of absolute values. |I₁| + |I₂|, not I₁ + I₂.
Wrong roots when factorising — always double-check on the GDC.

Next up: Area Between a Curve and a Line — when the boundary of the region isn’t just the x-axis but also a straight line. The same ideas extend, with one extra step: subtract the line’s area from the curve’s area (or vice versa).

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