IB Maths AA HL Topic 5 — Calculus Paper 1 & 2 ~9 min read

Negative Integrals

When a curve dips below the x-axis, the definite integral over that region is NEGATIVE. But area is always positive — so to find the actual area you have to take the modulus (absolute value). Either compute the signed integral and flip the sign, or use the formula A = ∫|y| dx (in the formula booklet) which a GDC handles directly.

📘 What you need to know

Below x-axis → negative integral: ∫f(x) dx is negative if f(x) < 0 on the interval.
Area is always positive: take the modulus of any negative integral that represents an area.
Modulus formula (formula booklet): A = ∫_a^b |y| dx — works for fully below, fully above, or mixed regions.
If the curve crosses the x-axis inside [a, b], the regular integral CANNOT be used for total area — positive and negative pieces would cancel out.
Split the integral at every x-axis crossing: integrate each piece separately, take absolute value of each, then sum.
GDC can compute ∫|y| dx directly using the absolute value (Abs) function — no splitting needed.
Sketch first: always sketch (or use GDC graphing) to identify where the curve crosses the x-axis within the interval.
“Area enclosed” always means TOTAL geometric area — always positive, always uses modulus.

Why integrals can be negative

Visualise the integral as a signed area. Above the x-axis = positive; below the x-axis = negative. If a curve dips below the axis, the integral over that region contributes a NEGATIVE value. Subtracting positives and negatives gives the net signed area, NOT the total geometric area.

Green region above the x-axis: ∫_a^c f dx is positive. Red region below: ∫_c^b f dx is negative — take the modulus. Total area = positive piece + |negative piece|, or just A = ∫_a^b|y| dx.

The two methods

Method 1 — split at crossings

A = ∑ |∫ f dx|
(piece by piece)

find crossings, integrate each region, take absolute value, sum

Method 2 — modulus formula

A = ∫_a^b |y| dx

in formula booklet; GDC computes directly

Which method when? For exact (Paper 1) work where you must show steps, use Method 1 — split the integral, do each piece by hand. For Paper 2 where decimals are fine, just use Method 2 with the GDC’s |Abs| function. They give the same answer.

🧭 Recipe — area when curve goes below the x-axis

Sketch the curve (or use GDC graphing) — identify where it crosses the x-axis within [a, b].
If curve stays entirely below: compute ∫_a^b f dx (will be negative); area = |result|.
If curve crosses inside [a, b]: find crossing point c by solving f(x) = 0. Split: ∫_a^c + ∫_c^b.
Integrate each piece separately, take the absolute value of any negative pieces.
Sum all absolute pieces for the total area. (Alternative: A = ∫|y| dx via GDC.)

Worked examples

WE 1

Area enclosed by a parabola below the x-axis

Find the exact area enclosed by the curve y = x² − 5x + 6 and the x-axis.

Step 1 — find x-intercepts x² − 5x + 6 = 0 → (x − 2)(x − 3) = 0 → x = 2, x = 3 Step 2 — check whether curve is above or below at x = 5/2: y = 25/4 − 25/2 + 6 = −1/4 < 0 (below axis on (2, 3)) Step 3 — integrate over [2, 3] ∫₂³ (x² − 5x + 6) dx = [x³/3 − 5x²/2 + 6x]₂³ F(3) = 9 − 45/2 + 18 = 54/2 − 45/2 = 9/2 F(2) = 8/3 − 10 + 12 = 8/3 + 2 = 14/3 ∫₂³ = 9/2 − 14/3 = 27/6 − 28/6 = −1/6 Step 4 — take absolute value Area = |−1/6| = 1/6 Area = 1/6 square units the integral came out negative because the curve is below the x-axis on (2, 3) — take |·| for the geometric area

WE 2

Large parabola fully below axis

Find the exact area enclosed by y = x² − 9 and the x-axis.

Step 1 — find x-intercepts x² − 9 = 0 → x = ±3 Step 2 — check sign on (−3, 3) at x = 0: y = −9 < 0 (curve below axis between roots) Step 3 — integrate over [−3, 3] ∫₋₃³ (x² − 9) dx = [x³/3 − 9x]₋₃³ F(3) = 9 − 27 = −18 F(−3) = −9 + 27 = 18 ∫₋₃³ = −18 − 18 = −36 Step 4 — take absolute value Area = |−36| = 36 Area = 36 square units x² − 9 is a parabola opening upward with vertex at (0, −9) — entirely below the x-axis between its roots

WE 3

Cubic crossing the x-axis — split into pieces

Find the total area enclosed between y = x³ − 4x and the x-axis from x = −2 to x = 2.

Step 1 — find x-intercepts x³ − 4x = x(x² − 4) = x(x − 2)(x + 2) = 0 → x = −2, 0, 2 (crossing at x = 0, inside the interval) Step 2 — determine sign on each piece on (−2, 0): at x = −1, y = −1 + 4 = 3 > 0 (above) on (0, 2): at x = 1, y = 1 − 4 = −3 < 0 (below) Step 3 — split and integrate F(x) = x⁴/4 − 2x² ∫₋₂⁰ (x³ − 4x) dx = [x⁴/4 − 2x²]₋₂⁰ = 0 − (4 − 8) = 4 ∫₀² (x³ − 4x) dx = [x⁴/4 − 2x²]₀² = (4 − 8) − 0 = −4 Step 4 — sum absolute values Total area = |4| + |−4| = 4 + 4 = 8 Total area = 8 square units if you’d computed ∫₋₂² (x³ − 4x) dx directly you’d get 0 — the +4 and −4 cancel. That’s the NET signed area, NOT the geometric area

WE 4

Trig curve — multiple sign changes

Find the area between y = sin x and the x-axis on the interval 0 ≤ x ≤ 2π.

Step 1 — sin(x) crosses x-axis at x = 0, π, 2π on (0, π): sin x > 0 (above axis) on (π, 2π): sin x < 0 (below axis) Step 2 — split at x = π and integrate ∫₀^π sin x dx = [−cos x]₀^π = −cos(π) − (−cos(0)) = −(−1) − (−1) = 2 ∫_π^(2π) sin x dx = [−cos x]_π^(2π) = −cos(2π) − (−cos(π)) = −1 − 1 = −2 Step 3 — sum absolute values Total area = |2| + |−2| = 4 Total area = 4 square units sin x is symmetric about (π, 0) — the area above (0, π) exactly equals the area below (π, 2π), so total is 2 × 2 = 4

WE 5

Mixed region using the modulus formula

Find the exact value of ∫₋₂³ |x² − 4| dx.

Step 1 — find where x² − 4 changes sign x² − 4 = 0 → x = ±2 on (−2, 2): x² − 4 < 0 (below) on (2, 3): x² − 4 > 0 (above) Step 2 — split |x² − 4| into two pieces |x² − 4| = −(x² − 4) on (−2, 2) |x² − 4| = (x² − 4) on (2, 3) Step 3 — integrate each piece F(x) = x³/3 − 4x ∫₋₂² (x² − 4) dx = (8/3 − 8) − (−8/3 + 8) = 16/3 − 16 = −32/3 ∫₂³ (x² − 4) dx = (9 − 12) − (8/3 − 8) = −3 + 16/3 = 7/3 Step 4 — combine using modulus ∫₋₂³ |x² − 4| dx = |−32/3| + 7/3 = 32/3 + 7/3 = 39/3 = 13 ∫₋₂³ |x² − 4| dx = 13 in modulus, “below axis” pieces become positive — that’s why we just flip the sign on the negative piece and add

WE 6

Cubic with a region above AND below — exam-style

The region R is bounded by the curve f(x) = (x + 1)(x − 2)(x − 5), the x-axis, the y-axis, and the line x = 3. Find the exact area of R.

Step 1 — find x-intercepts and sketch on [0, 3] f(x) = 0 at x = −1, 2, 5 inside [0, 3], the curve crosses at x = 2 only Step 2 — check sign on each piece f(0) = (1)(−2)(−5) = 10 > 0 (above axis on (0, 2)) f(3) = (4)(1)(−2) = −8 < 0 (below axis on (2, 3)) Step 3 — expand and integrate f(x) = (x+1)(x-2)(x-5) = x³ − 6x² + 3x + 10 F(x) = x⁴/4 − 2x³ + 3x²/2 + 10x F(0) = 0 F(2) = 4 − 16 + 6 + 20 = 14 F(3) = 81/4 − 54 + 27/2 + 30 = 81/4 + 54/4 − 96/4 = 39/4 Step 4 — compute each piece R₁ = ∫₀² f dx = 14 − 0 = 14 (above — positive) R₂ = ∫₂³ f dx = 39/4 − 14 = 39/4 − 56/4 = −17/4 (below — negative) Step 5 — sum absolute values Area = |14| + |−17/4| = 14 + 17/4 = 56/4 + 17/4 = 73/4 Area of R = 73/4 square units (= 18.25) GDC shortcut: A = ∫₀³ |(x+1)(x−2)(x−5)| dx gives 18.25 directly — no splitting needed

💡 Top tips

Always sketch first — even a quick GDC graph confirms where the curve crosses the x-axis and which pieces are above vs below.
Method 1 (split) for exact answers; Method 2 (∫|y| dx) for GDC speed. Use both to cross-check.
“Area enclosed” or “total area” always means GEOMETRIC area — must be positive. Treat the integral with caution and take |·| as needed.
If the integral gives a negative answer for area, you forgot the absolute value somewhere — go back and fix.
If the integral gives ZERO for what should be a positive area, you almost certainly have a curve that crosses the axis and pieces cancelled — split the interval.

⚠ Common mistakes

Reporting a negative number as an area — areas can never be negative. Always take |·| on the final value.
Not splitting when the curve crosses the x-axis — the integral will give the NET signed area (positive parts minus negative parts), not the geometric total.
Forgetting one of the crossing points — always solve f(x) = 0 to find ALL roots inside [a, b], not just the endpoints.
Putting the modulus on the wrong side: ∫|y| dx NOT |∫y dx|. These are different! The modulus goes INSIDE the integral, applied to the integrand.
Using a GDC without checking the curve — if you don’t verify where the curve is positive/negative, you might end up computing the signed integral by mistake.

Up next: Area Between a Curve and a Line. The natural extension — the bounded region might be enclosed by both a curve AND a non-vertical line. The total area is sometimes the sum (∫curve + triangle) and sometimes the difference (∫curve − triangle), depending on the shape. Sketching is everything.

Need help with Calculus?

Get 1-on-1 help from an IB examiner who knows exactly what Paper 1 & 2 are looking for.

Book Free Session →