Key Idea: A definite integral measures signed area under a curve between two limits — it turns a graph (or a rate of change) into an exact number. It shows up on both papers: by hand on Paper 1, and as a fast GDC numerical integral on Paper 2.
∫ Evaluating a definite integral
- any antiderivative of f(x) (no +C needed — it cancels)
- the lower and upper limits of integration
Zero width: $\intₐᵃ f = 0$. Swap the limits → flip the sign: $\intₐᵇ f = -\intbᵃ f$. Adjacent intervals join: $\intₐᵇ f + \intbᶜ f = \intₐᶜ f$ (add or subtract pieces that share a limit). And trig limits are in radians (e.g. $\int₀\ᵖⁱ/²\cos x\,dx$) — keep your GDC in radian mode.
📐 Finding areas
| Situation | What to compute | Watch out for |
|---|---|---|
| Curve above the x-axis | $\intₐᵇ f(x)\,dx$ — the integral is the area. | Straightforward. |
| Curve below the x-axis | The integral is negative; the area is its magnitude $\left|\intₐᵇ f\right|$. | If it crosses the axis, split at the intercepts and add the magnitudes. |
| Area between two curves | $\intₐᵇ (y\ₜₑₓₜ{ₜₒₚ} - y\ₜₑₓₜ{bₒₜₜₒₘ})\,dx$. | Limits come from solving top = bottom (the intersections). |
✏️ IB-style worked examples
IB-style question — evaluate with a property (Paper 1)
Given $\int₁⁴ g(x)\,dx = 9$ and $\int₁⁶ g(x)\,dx = 14$, find $\int₄⁶ g(x)\,dx$.
Step by step:
Adjacent intervals join: ∫₁⁴ + ∫₄⁶ = ∫₁⁶.
Subtract the given values.
∫₄⁶ g(x) dx = 5.
IB-style question — area where the curve dips below the axis (Paper 1)
Find the area enclosed between $y = x² - 9$ and the x-axis from x = 0 to x = 3.
Step by step:
On [0, 3] the parabola is below the axis, so integrate first.
A negative integral just means it's below the axis — the area is the magnitude.
Area = 18 (the integral is −18 because the region lies below the axis).
IB-style question — area enclosed between two curves (Paper 1)
Find the area enclosed between $y = 4x - x²$ and $y = x$.
Step by step:
Limits = intersections. Solve top = bottom.
On (0, 3) the parabola is on top. Integrate (top − bottom) = 3x − x².
Area = 9/2 = 4.5.
🔒 GDC walkthrough
Step through the exact calculator keystrokes, screen by screen, in study mode.
Important: Area is never negative. A negative integral only tells you the region is below the axis (or that you put the curves the wrong way round). Take the magnitude, and if the curve crosses the axis, split at the intercepts and add the positive pieces — don't integrate straight through, or the parts cancel.
Tap each card to reveal the answer.
What does ∫ₐᵇ f = F(b) − F(a) give you? The signed area under f between a and b — top value minus bottom value of an antiderivative.
∫₂⁷ f = 11 and ∫₂⁴ f = 4. Find ∫₄⁷ f. 7 — adjacent intervals join, so ∫₄⁷ = ∫₂⁷ − ∫₂⁴ = 11 − 4.
Evaluate ∫₀π sin x dx. 2 — [−cos x]₀^π = −cos π − (−cos 0) = 1 + 1 (radian mode).
An integral comes out as −12. What is the area? 12 — the region is below the axis; the area is the magnitude.
How do you find the limits for an enclosed region? Solve top = bottom (or f(x) = 0 for a curve and the axis) — the x-values are your limits.
Area between y = x² and y = 2x? 4/3 — meet at x = 0, 2; ∫₀² (2x − x²) dx = [x² − x³/3]₀² = 4 − 8/3.
Exam Tips
- Evaluate as F(b) − F(a) — drop the +C, it always cancels.
- Properties: ∫ₐᵃ = 0; swapping limits flips the sign; adjacent intervals join.
- Below the axis → integral is negative; the area is its magnitude. Crosses the axis → split and add.
- Area between curves = ∫(top − bottom); find the limits by solving top = bottom.
- Paper 2: use MATH → 9: fnInt( for any definite integral, and ∫(rate) dt = total change.
- Trig limits are in radians — set your GDC to radian mode.