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

Definite Integrals

∫_a^b f(x) dx = F(b) − F(a) — that’s the Fundamental Theorem of Calculus. Beyond evaluating by hand or GDC, the SIX standard properties (constant factor, term-by-term, equal limits, swapping limits, splitting intervals, horizontal translation) let you answer “given ∫f = 12, find ∫6f(x+5) =” questions without ever computing an antiderivative.

📘 What you need to know

Fundamental Theorem of Calculus: ∫_a^b f(x) dx = [F(x)]_a^b = F(b) − F(a) — provided f is continuous on [a, b].
No “+ c“ in definite integrals — the constant cancels in F(b) − F(a).
Constant factor: ∫_a^b k f(x) dx = k ∫_a^b f(x) dx — pull constants outside.
Term by term: ∫_a^b[f(x) ± g(x)] dx = ∫_a^b f dx ± ∫_a^b g dx.
Equal limits: ∫_a^a f(x) dx = 0. Always.
Swap limits: ∫_b^a f(x) dx = −∫_a^b f(x) dx — reversing the limits flips the sign.
Splitting: ∫_a^b f(x) dx = ∫_a^c f(x) dx + ∫_c^b f(x) dx for any c in [a, b].
Horizontal translation: ∫_a^b f(x) dx = ∫_a−k^b−k f(x + k) dx — shifting the function shifts the limits.

Evaluating definite integrals — by hand and by GDC

Fundamental Theorem of Calculus ∫_a^b f(x) dx = [F(x)]_a^b = F(b) − F(a)

For polynomials and standard integrands, evaluate by hand using the antiderivative. For “ugly” integrands without an elementary antiderivative (e.g. e^{sin x}, sin(x²), ln(x² + 1)), use your GDC’s numerical integration. For Paper 1 the integral will always be analytical (no GDC); Paper 2 lets you use GDC.

Always check with GDC when allowed — even if you’ve computed by hand, the GDC gives a decimal you can compare. Catches sign errors and arithmetic slips instantly.

The six properties — summary

Property	Formula	When useful
Constant factor	∫_a^b k f dx = k ∫_a^b f dx	pull out a constant multiplier
Term by term	∫(f ± g) dx = ∫f dx ± ∫g dx	break sum/diff into separate integrals
Equal limits	∫_a^a f dx = 0	quick recognition — answer is just 0
Swap limits	∫_b^a f dx = −∫_a^b f dx	given a “wrong way” integral
Split interval	∫_a^b = ∫_a^c + ∫_c^b	combining or breaking ranges
Horizontal translation	∫_a^b f(x) dx = ∫_a−k^b−k f(x + k) dx	shift function ↔ shift limits

Translation intuition: if you replace x by (x + k) in the function, the graph slides LEFT by k. To keep the same area, the limits must also shift left by k. So ∫₀⁶ f(x + 2) dx = ∫₂⁸ f(x) dx.

🧭 Recipe — evaluate a definite integral

Rewrite the integrand if needed (expand products, simplify fractions, rewrite roots/reciprocals as powers).
Find the antiderivative F(x) — no “+ c” needed.
Write it with bracket notation: I = [F(x)]_a^b.
Substitute the upper limit, then the lower: F(b) − F(a). Subtract — careful with negatives.
For “ugly” integrands: use GDC numerical integration. For “given ∫f = …, find …” questions: use the SIX properties — never compute an antiderivative.

Worked examples

WE 1

“Show that” — by hand evaluation

Show that ∫₂³ (4x³ + 6x) dx = 80.

Step 1 — find the antiderivative ∫(4x³ + 6x) dx = x⁴ + 3x² Step 2 — write with bracket notation I = [x⁴ + 3x²]₂³ Step 3 — evaluate at upper limit then lower F(3) = 3⁴ + 3·3² = 81 + 27 = 108 F(2) = 2⁴ + 3·2² = 16 + 12 = 28 Step 4 — subtract I = 108 − 28 = 80 ✓ ∫₂³ (4x³ + 6x) dx = 80 (as required) “show that” means you must give full working — quoting just “80” doesn’t earn marks

WE 2

Use a GDC for a non-elementary integrand

Use your GDC to evaluate ∫₀¹ x² sin(x²) dx, giving your answer to 3 significant figures.

Why use GDC? — x² sin(x²) has no elementary antiderivative (no antiderivative in terms of standard functions exists) On GDC: enter integrand, lower limit 0, upper limit 1 GDC displays: 0.1821109660… Round to 3 s.f. 0.1821… → 0.182 ∫₀¹ x² sin(x²) dx = 0.182 (3 s.f.) 3 s.f. for 0.1821… → start counting from the first non-zero digit: 1, 8, 2 → 0.182

WE 3

Properties — equal limits and swapping limits

h is continuous on [2, 12]. Given ∫₂⁸ h(x) dx = 9 and ∫₈¹² h(x) dx = 14, write down the values of:

(a) ∫₆⁶ h(x) dx (b) ∫₁₂⁸ h(x) dx

(a) Equal limits — the integral is just 0 property: ∫ₐᵃ f dx = 0 (because F(a) − F(a) = 0) ∫₆⁶ h(x) dx = 0 (b) Swap limits — flips the sign property: ∫ᵇᵃ f dx = −∫ₐᵇ f dx ∫₁₂⁸ h(x) dx = − ∫₈¹² h(x) dx = − 14 (a) 0 | (b) −14 no antiderivative needed — these are pure property questions, instant marks once you spot the pattern

WE 4

Properties — splitting interval and constant factor

Same h as in WE 3. Given ∫₂⁸ h(x) dx = 9 and ∫₈¹² h(x) dx = 14, find:

(a) ∫₂¹² h(x) dx (b) ∫₂⁸ 5h(x) dx

(a) Split at x = 8 — combine the two given pieces property: ∫ₐᵇ f dx = ∫ₐᶜ f dx + ∫ᶜᵇ f dx ∫₂¹² h(x) dx = ∫₂⁸ h(x) dx + ∫₈¹² h(x) dx = 9 + 14 = 23 (b) Constant factor — pull 5 out property: ∫ₐᵇ k f dx = k · ∫ₐᵇ f dx ∫₂⁸ 5 h(x) dx = 5 · ∫₂⁸ h(x) dx = 5 · 9 = 45 (a) 23 | (b) 45 splitting works ONLY at a point inside [a, b] — you can’t split ∫₂⁸ at x = 10 because 10 lies outside

WE 5

Horizontal translation property

Same h. Given ∫₂⁸ h(x) dx = 9, find ∫₀⁶ 3 h(x + 2) dx.

Pull out the constant 3 first ∫₀⁶ 3 h(x + 2) dx = 3 · ∫₀⁶ h(x + 2) dx Apply horizontal translation: ∫ₐᵇ f(x) dx = ∫(a-k)^(b-k) f(x+k) dx read backwards: ∫(a-k)^(b-k) f(x+k) dx = ∫ₐᵇ f(x) dx here a − k = 0 and b − k = 6 with k = 2, so a = 2 and b = 8 ∫₀⁶ h(x + 2) dx = ∫₂⁸ h(x) dx = 9 Multiply by the 3 from the start ∫₀⁶ 3 h(x + 2) dx = 3 · 9 = 27 ∫₀⁶ 3 h(x + 2) dx = 27 replacing x by (x + 2) shifts the graph LEFT by 2, so to capture the same area the limits also shift left by 2: (0, 6) ↔ (2, 8)

WE 6

Combining ALL the properties

f is continuous on [1, 7] with ∫₁⁴ f(x) dx = 6 and ∫₄⁷ f(x) dx = −2. Find ∫₀⁶ 4 f(x + 1) dx.

Step 1 — pull out the constant 4 (constant factor property) ∫₀⁶ 4 f(x + 1) dx = 4 · ∫₀⁶ f(x + 1) dx Step 2 — apply horizontal translation with k = 1 a − k = 0 and b − k = 6 with k = 1, so a = 1 and b = 7 ∫₀⁶ f(x + 1) dx = ∫₁⁷ f(x) dx Step 3 — split at x = 4 (since both given pieces meet there) ∫₁⁷ f(x) dx = ∫₁⁴ f(x) dx + ∫₄⁷ f(x) dx = 6 + (−2) = 4 Step 4 — multiply by the 4 from the start ∫₀⁶ 4 f(x + 1) dx = 4 · 4 = 16 ∫₀⁶ 4 f(x + 1) dx = 16 three properties chained together: constant factor → translation → splitting. Each step is a one-liner if you know the rules

💡 Top tips

“+ c” is NOT needed in definite integrals — it always cancels. Don’t write it; examiners may penalise the redundancy.
Property questions are FREE marks if you know the six rules. No antiderivative, no algebra — just pattern recognition.
Spot equal limits FIRST — ∫_a^a f dx = 0 is the easiest property to use. Always answer 0 immediately on seeing identical limits.
For horizontal translation: think “if the function shifts LEFT by k, the limits also shift LEFT by k to capture the same area”.
GDC for Paper 2: use numerical integration to verify any by-hand work. Saves marks when you make arithmetic errors.

⚠ Common mistakes

Writing “+ c“ in a definite integral — wrong notation. Definite integrals give a NUMBER, not an expression.
Swapping upper and lower limits when evaluating: it’s F(upper) − F(lower), not the other way round. Reversing flips the sign.
Wrong direction on horizontal translation: ∫₀⁶ f(x + 2) dx = ∫₂⁸ f(x) dx (limits shift left by 2 when reading from the f(x) side — be careful with direction).
Trying to split ∫₂⁸ at a point outside [2, 8] — splitting only works at a point INSIDE the interval.
Sign error when subtracting F(a) — especially when F(a) is negative. F(b) − F(a) where F(a) = −5 means F(b) + 5, not F(b) − 5. Watch parentheses.

Up next: Negative Integrals. So far we’ve assumed the curve stays above the x-axis. But if part of the curve dips below, the integral over that part is NEGATIVE. To find the actual area (which must be positive), you take the modulus — ∫|y| dx — or split the integral into “above-axis” and “below-axis” pieces and add the absolute values.

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