IB Maths AA SL Topic 5 — Calculus Paper 1 & 2 🎯 Skill ~3 min practice

AA SL Definite Integrals skills

Definite integrals come with limits — the numbers above and below the ∫ symbol. The result is just F(upper) − F(lower). No C needed (it cancels), and on Paper 2 the GDC handles the whole thing in one go.

The Method

∫_a^b f(x) dx = F(b) − F(a) F is the antiderivative of f · upper minus lower · ✓ in formula booklet

Find the antiderivative F(x) using the integration rules. Skip the + C — it cancels.
Substitute the upper limit b into F(x) → F(b).
Subtract F(a) — the antiderivative evaluated at the lower limit.

The square-bracket notation

When you write up a definite integral, this is the standard layout:

∫₁³ 2x dx = [x²]₁³ = (3)² − (1)² = 8

Square brackets show the antiderivative with the limits attached. Then evaluate top minus bottom. The IB markers want this layout — show all three steps for full method marks.

The recipe — every definite integral

Antidifferentiate using the power rule (or special-function rules). Skip + C.
Write the antiderivative inside square brackets with the limits: [F(x)]_a^b.
Substitute the upper limit b into F(x) → F(b).
Subtract F(a) — keep brackets around a if it’s negative to avoid sign errors.
Simplify — your answer is a number, not an expression in x.

GDC shortcut for Paper 2

TI-84 Plus

Press MATH → 9: fnInt(
Enter: function, X, lower, upper
e.g. fnInt(2X, X, 1, 3) = 8
Press ENTER → answer in one go

Casio fx-CG50

From RUN-MAT, press OPTN → CALC (F4)
Choose ∫dx (F4)
Enter: function, lower, upper
e.g. ∫(2X, 1, 3) = 8

Use this on Paper 2 as a quick check or to skip the algebra entirely. Paper 1 still needs full working — show the antiderivative and the substitution.

Worked examples

WE 1 EASY

Evaluate ∫₀² (3x² + 2) dx.

step 1 — antidifferentiate [x³ + 2x]₀² step 2 — substitute upper F(2) = 8 + 4 = 12 step 3 — subtract lower F(0) = 0 12 − 0 = 12 ∫₀² (3x² + 2) dx = 12 when the lower limit is 0, F(0) often vanishes — but always write it out!

WE 2 MEDIUM

Evaluate ∫₋₁² (4x³ − x) dx.

step 1 — antidifferentiate [x⁴ − x²/2]₋₁² step 2 — substitute upper F(2) = 16 − 4/2 = 16 − 2 = 14 step 3 — substitute lower (careful with signs!) F(−1) = (−1)⁴ − (−1)²/2 = 1 − 1/2 = 1/2 step 4 — subtract 14 − 1/2 = 27/2 ∫ = 27/2 (or 13.5) brackets around negative limits — (−1)⁴ = 1, not −1!

WE 3 HARD

Evaluate ∫₁⁴ (2/√x − x) dx, giving the answer in exact form.

step 1 — rewrite as powers 2/√x = 2x^−1/2 step 2 — antidifferentiate 2x^−1/2 → 2 × x^1/2/(1/2) = 4x^1/2 = 4√x −x → −x²/2 [4√x − x²/2]₁⁴ step 3 — evaluate F(4) = 4(2) − 16/2 = 8 − 8 = 0 F(1) = 4(1) − 1/2 = 7/2 step 4 — subtract 0 − 7/2 = −7/2 ∫ = −7/2 a negative answer just means the curve was below the x-axis on net — that’s fine for a definite integral!

Practice questions

Try each one yourself first, then click the question to reveal the worked answer. Always evaluate upper minus lower.

Q1 EASY Evaluate ∫₀³ 2x dx. Show answer ▼Hide answer ▲

[x²]₀³ = 9 − 0 = 9

Q2 EASY Evaluate ∫₁² (3x² − 1) dx. Show answer ▼Hide answer ▲

[x³ − x]₁² = (8 − 2) − (1 − 1) = 6

Q3 MEDIUM Evaluate ∫₋₂¹ (x² + 2x) dx. Show answer ▼Hide answer ▲

[x³/3 + x²]₋₂¹ F(1) = 1/3 + 1 = 4/3 F(−2) = −8/3 + 4 = 4/3 4/3 − 4/3 = 0 = 0 a result of 0 means the positive and negative areas cancelled — check by sketching!

Q4 MEDIUM Evaluate ∫₁⁹ √x dx. Show answer ▼Hide answer ▲

rewrite √x = x^1/2 [2x^3/2/3]₁⁹ F(9) = 2(27)/3 = 18 F(1) = 2/3 18 − 2/3 = 52/3 = 52/3

Q5 HARD Find k > 0 such that ∫₀^k (2x + 1) dx = 12. Show answer ▼Hide answer ▲

step 1 — antidifferentiate and evaluate [x² + x]₀^k = k² + k step 2 — set equal to 12 k² + k − 12 = 0 (k + 4)(k − 3) = 0 k = 3 (since k > 0) k = 3 “find the unknown limit” → set up the integral as an equation, solve for the unknown!

⚠ Common mistakes

Adding + C on a definite integral. The C cancels (C − C = 0), so don’t include it. Examiners can deduct method marks.
Subtracting in the wrong order. It’s upper minus lower: F(b) − F(a). Reversing the limits flips the sign.
Sign errors with negative limits. Always put brackets around: F(−2) means substitute (−2), not −2 just dropped in.
Treating a negative answer as a mistake. A definite integral can legitimately be negative if the curve is below the x-axis. That’s not the same as “area” (which is always positive).
Stopping at the antiderivative. The answer needs to be a number, not an expression in x.
Using GDC on Paper 1. Paper 1 is non-calculator — show the algebra. The GDC shortcut is for Paper 2 only.

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Read the full Definite Integrals notes for the link to area under a curve, the fundamental theorem of calculus, and how negative integrals signal regions below the x-axis.

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AA SL Definite Integrals skills

The Method

The square-bracket notation

The recipe — every definite integral

GDC shortcut for Paper 2

Worked examples

Practice questions

⚠ Common mistakes

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