IB Math AA SL Topic 5.11: Definite integrals & areas | Notes & Flashcards | Aimnova

IB Math AA SL — Definite integrals & areas

Topic 5.11 of IB Mathematics: Analysis and Approaches covers Definite integrals & areas, which is part of Unit 5: Calculus. Students explore key concepts including Definite integrals, Area between curves. A strong understanding of definite integrals & areas is essential for IB Math AA SL exams and builds the foundation for connected topics across the syllabus.

Key concepts in Definite integrals & areas

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

\int_a^b f(x)\,dx = \big[F(x)\big]_a^b = F(b) - F(a)

$F(x)$: any antiderivative of f(x) (no +C needed — it cancels)
$a, b$: 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

✏️ 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.
$\int_4^6 g = \int_1^6 g - \int_1^4 g$
Subtract the given values.
$= 14 - 9 = 5$

Final answer:

∫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.
$\int_0^3 (x^2 - 9)\,dx = \left[\tfrac{x^3}{3} - 9x\right]_0^3 = 9 - 27 = -18$
A negative integral just means it's below the axis — the area is the magnitude.
$\text{Area} = |{-18}| = 18$

Final answer:

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.
$4x - x^2 = x \;\Rightarrow\; 3x - x^2 = 0 \;\Rightarrow\; x = 0,\, 3$
On (0, 3) the parabola is on top. Integrate (top − bottom) = 3x − x².
$\int_0^3 (3x - x^2)\,dx = \left[\tfrac{3x^2}{2} - \tfrac{x^3}{3}\right]_0^3 = \tfrac{27}{2} - 9 = \tfrac{9}{2}$

Final answer:

Area = 9/2 = 4.5.

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.

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.

IB Math AA SL — Definite integrals & areas

Key concepts in Definite integrals & areas

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

\int_a^b f(x)\,dx = \big[F(x)\big]_a^b = F(b) - F(a)

$F(x)$: any antiderivative of f(x) (no +C needed — it cancels)
$a, b$: 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

✏️ 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.
$\int_4^6 g = \int_1^6 g - \int_1^4 g$
Subtract the given values.
$= 14 - 9 = 5$

Final answer:

∫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.
$\int_0^3 (x^2 - 9)\,dx = \left[\tfrac{x^3}{3} - 9x\right]_0^3 = 9 - 27 = -18$
A negative integral just means it's below the axis — the area is the magnitude.
$\text{Area} = |{-18}| = 18$

Final answer:

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.
$4x - x^2 = x \;\Rightarrow\; 3x - x^2 = 0 \;\Rightarrow\; x = 0,\, 3$
On (0, 3) the parabola is on top. Integrate (top − bottom) = 3x − x².
$\int_0^3 (3x - x^2)\,dx = \left[\tfrac{3x^2}{2} - \tfrac{x^3}{3}\right]_0^3 = \tfrac{27}{2} - 9 = \tfrac{9}{2}$

Final answer:

Area = 9/2 = 4.5.

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.

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.

IB Math AA SL — Definite integrals & areas

Key concepts in Definite integrals & areas

∫ Evaluating a definite integral

📐 Finding areas

✏️ IB-style worked examples

IB-style question — evaluate with a property (Paper 1)

IB-style question — area where the curve dips below the axis (Paper 1)

IB-style question — area enclosed between two curves (Paper 1)

Exam Tips

What you'll learn in Topic 5.11

Study resources — 5.11 Definite integrals & areas

Definite integrals

Area between curves

Ready to study Definite integrals & areas?

Ready to practice?

IB Math AA SL — Definite integrals & areas

Key concepts in Definite integrals & areas

∫ Evaluating a definite integral

📐 Finding areas

✏️ IB-style worked examples

IB-style question — evaluate with a property (Paper 1)

IB-style question — area where the curve dips below the axis (Paper 1)

IB-style question — area enclosed between two curves (Paper 1)

Exam Tips

What you'll learn in Topic 5.11

Study resources — 5.11 Definite integrals & areas

Definite integrals

Area between curves

Ready to study Definite integrals & areas?

Ready to practice?