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

AA SL Integrating Polynomials skills

Integration is differentiation in reverse. For powers of x, the rule flips: add 1 to the power, then divide by the new power. Apply it term by term and never forget the + C on indefinite integrals.

The Method

∫ xⁿ dx = xⁿ⁺¹n + 1 + C add 1 to the power, divide by the new power · ✓ in formula booklet

Add 1 to the power — xⁿ becomes xⁿ⁺¹.
Divide by the new power — write xⁿ⁺¹ over (n + 1).
Add + C at the end (indefinite integrals only — never on definite integrals).

The reverse power move

Example: integrate x³

Step 1 — power +1 x³ → x⁴

→

Step 2 — divide by new x⁴4

Step 3 — add C C

Final answer: ∫ x³ dx = x⁴/4 + C. Sanity check: differentiating x⁴/4 gives 4x³/4 = x³ ✓

⚠️

Don’t forget + C on indefinite integrals

Any constant differentiates to zero, so when integrating you don’t know what was there before. Adding + C covers all possibilities. Indefinite integrals always need + C; definite integrals never do — the C cancels when you subtract.

Worked examples

WE 1 EASY

Find ∫ (3x² − 4x + 5) dx.

step 1 — integrate term by term 3x² → 3 × x³/3 = x³ −4x → −4 × x²/2 = −2x² 5 → 5xstep 2 — combine + Cx³ − 2x² + 5x + C a constant integrates to a constant times x. always include + C!

WE 2 MEDIUM

Find ∫ (6√x − 1/x²) dx.

step 1 — rewrite as powers 6√x = 6x^1/2 −1/x² = −x⁻²step 2 — integrate each term 6x^1/2 → 6 × x^3/2/(3/2) = 4x^3/2 −x⁻² → −x⁻¹/(−1) = 1/x4x^3/2 + 1/x + C always rewrite roots and fractions as powers FIRST — then the rule applies cleanly!

WE 3 HARD

Find ∫ x(2x − 3)² dx.

step 1 — expand the bracket (2x − 3)² = 4x² − 12x + 9step 2 — multiply by x x(4x² − 12x + 9) = 4x³ − 12x² + 9xstep 3 — integrate term by term 4x³ → x⁴ −12x² → −4x³ 9x → 9x²/2x⁴ − 4x³ + 9x²/2 + C products and powers of brackets — always expand BEFORE integrating!

Practice questions

Try each one yourself first, then click the question to reveal the worked answer. Always check + C is there!

Q1 EASY Find ∫ (4x³ + 2x) dx. Show answer ▼Hide answer ▲

4x³ → 4 × x⁴/4 = x⁴ 2x → 2 × x²/2 = x² x⁴ + x² + C

Q2 EASY Find ∫ (5x⁴ − 6x² + 7) dx. Show answer ▼Hide answer ▲

5x⁴ → x⁵ −6x² → −2x³ 7 → 7x x⁵ − 2x³ + 7x + C

Q3 MEDIUM Find ∫ (√x + 2/x³) dx. Show answer ▼Hide answer ▲

rewrite as powers √x = x^1/2 2/x³ = 2x⁻³ x^1/2 → x^3/2/(3/2) = 2x^3/2/3 2x⁻³ → 2x⁻²/(−2) = −x⁻² = −1/x² 2x^3/2/3 − 1/x² + C

Q4 MEDIUM Find ∫ (3x² − 1)(x + 2) dx. Show answer ▼Hide answer ▲

expand first (3x² − 1)(x + 2) = 3x³ + 6x² − x − 2 integrate term by term 3x³ → 3x⁴/4 6x² → 2x³ −x → −x²/2 −2 → −2x 3x⁴/4 + 2x³ − x²/2 − 2x + C

Q5 HARD Find ∫ (x² + 3x)/√x dx. Show answer ▼Hide answer ▲

split the fraction first (x² + 3x)/x^1/2 = x^3/2 + 3x^1/2 integrate x^3/2 → x^5/2/(5/2) = 2x^5/2/5 3x^1/2 → 3 × x^3/2/(3/2) = 2x^3/2 2x^5/2/5 + 2x^3/2 + C when there’s a fraction, split the numerator over the denominator first — never integrate as one piece!

⚠ Common mistakes

Forgetting + C. The most common integration error. Always include it on indefinite integrals (no limits).
Subtracting 1 from the power. That’s differentiation. Integration adds 1.
Forgetting to divide by the new power. ∫x⁴ dx = x⁵/5 + C — the /5 matters.
Not rewriting roots and fractions as powers. √x = x^1/2, 1/x² = x⁻² — much easier to apply the rule from this form.
Trying to integrate a product directly. ∫x(x+1) dx ≠ x × ∫(x+1) dx — expand first, then integrate term by term.
Special case: ∫1/x dx. The power rule fails when n = −1 (would divide by 0). Use ∫1/x dx = ln|x| + C instead.

📖

Want the theory?

Read the full Integrating Powers of x notes for why integration reverses differentiation, the link to area under curves, and how integration scales linearly across sums and constants.

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