IB Maths AI HL
Integration
Paper 1 & 2
~6 min read
Finding Areas Using a GDC
The exact area between a curve and the x-axis is a definite integral — an integral with limits. Unlike the indefinite integral, there’s no “+ c” (it cancels out), and your GDC can evaluate it directly. The real skill is reading off the right limits: where the area starts and stops.
📘 What you need to know
- Area under a curve = the region bounded by y = f(x), the x-axis, and lines x = a, x = b.
- Definite integral: A = ∫ab f(x) dx, with a the lower and b the upper limit.
- No “+ c“: in a definite integral the constant cancels.
- Fundamental Theorem: ∫ab f(x) dx = F(b) − F(a).
- Find the limits: from vertical lines, the y-axis (x = 0), or roots of f(x) = 0.
- GDC gives decimals: watch for an exact answer being wanted (e.g. a fraction, not 6.533…).
The definite integral
Area as a definite integral
A = ∫ab f(x) dx = F(b) − F(a)
✓ definite integral notation; the FTC link is standard
🤔 Why is there no “+ c” in a definite integral?
Working it as F(b) − F(a), a “+ c” would appear in both terms — and subtracting cancels it. So the constant never affects a definite integral, which is exactly why a single area comes out as a definite number.
🧠 “Limits replace the plus-c”
Indefinite integral → family of curves, “+ c” unknown. Definite integral → fixed limits a and b, a single number, no “+ c“.
Setting up and evaluating
🧭 Recipe — area with a GDC
- Sketch the curve (use your GDC) and shade the region.
- Find the limits: read vertical lines directly, use x = 0 for the y-axis, or solve f(x) = 0 for roots.
- Write the integral A = ∫ab f(x) dx.
- Evaluate on the GDC — and convert to exact form if the question asks.
Where do the limits come from? A vertical line x = 2 gives a limit directly; the y-axis gives x = 0; and a boundary at the x-axis means solving f(x) = 0 for where the curve crosses.
Worked examples
The region is bounded by y = x4 − 2x2 + 5, the x-axis, and the lines x = 1 and x = 2.
The two vertical lines give the limits directly.
left boundary: x = 1, right boundary: x = 2
a = 1, b = 2
WE 2Write down the integral for the area
Put the function between the limits.
∫12 (x⁴ − 2x² + 5) dx
WE 3Find the exact area, in the form pq
Evaluate; the GDC gives 6.5333…, so convert to a fraction.
GDC: 6.5333… (not exact)
6.5333… = 9815
area = 9815 square units
WE 4A limit from the y-axis: write the integral for the area under y = x2 + 1 from the y-axis to x = 3
The y-axis means x = 0.
left boundary = y-axis → x = 0
∫03 (x² + 1) dx
WE 5Limits from roots: find the area between y = x(4 − x) and the x-axis
No vertical lines given — the boundaries are where the curve meets the x-axis.
solve f(x) = 0: x(4 − x) = 0 → x = 0, x = 4
∫04 (4x − x²) dx
GDC / exact = 323
area = 323 square units
💡 Top tips
- Sketch and shade — mark the region, axes intercepts, and limits on any diagram.
- Find the limits carefully — vertical lines, x = 0 for the y-axis, or roots of f(x) = 0.
- No “+ c“ in a definite integral.
- Exact vs decimal — convert the GDC’s decimal to a fraction or surd when asked.
- Practise your GDC’s integral function before the exam — entry order varies.
- Two-part questions: indefinite first (with “+ c“), then definite later (no “+ c“).
⚠ Common mistakes
- Wrong limits — missing that the y-axis is x = 0, or not solving f(x) = 0 for roots.
- Leaving a decimal when an exact fraction or surd is required.
- Adding “+ c“ to a definite integral — it cancels.
- Swapping the limits — a is the lower (left), b the upper (right).
- Trusting a rough cursor read-off — type exact limits into the GDC where you can.
That wraps up Integration. The unit moved from approximate to exact: you started by estimating areas with the trapezoidal rule, then met integration as antidifferentiation — the reverse of finding a gradient. You learned the power rule (raise and divide), saw how a single point fixes the constant of integration, and finally used the definite integral to find exact areas under a curve, where the limits do the work and the “+ c” disappears. Differentiation breaks a function down to its rate of change; integration builds it back up — and measures the area underneath. Hold those two as mirror images and calculus stays one connected idea.
Need help with Integration?
Get 1-on-1 help from an IB examiner who knows exactly what Paper 1 & 2 are looking for.
Book Free Session →