IB Maths AA HL Topic 5 — Calculus Paper 1 & 2 HL ~11 min read

Differentiating & Integrating Maclaurin Series

A Maclaurin series IS just a polynomial (with infinitely many terms). So you can differentiate it term-by-term and the result is the Maclaurin series of the function’s derivative. Same for integration: integrating term-by-term gives the Maclaurin series of an antiderivative, plus a constant of integration to determine. This builds new series quickly from booklet ones — without going through the general formula every time.

📘 What you need to know

Differentiation rule: if f(x) has Maclaurin series a₀ + a₁x + a₂x² + a₃x³ + …, then f′(x) has series a₁ + 2a₂x + 3a₃x² + … — just differentiate each term using d/dx(xⁿ) = nxⁿ⁻¹.
Integration rule: integrating term-by-term gives a₀x + (a₁/2)x² + (a₂/3)x³ + … + c. The constant c must be determined from a known value (usually the antiderivative’s value at 0).
Determine c from f(0): if you integrate the series for f′(x) to recover f(x), then evaluating at x = 0 collapses every x^k term to 0, so f(0) = c.
Power shifts: differentiation lowers each power by 1; integration raises each power by 1. The leading constant of f vanishes in f′; an “initial-condition” constant c appears when integrating.
Cross-check standard pairs: sin x ↔ cos x, e^x ↔ e^x (its own derivative!), ln(1+x) ↔ 1/(1+x), arctan x ↔ 1/(1+x²). All these pairs are visible at the series level.
Build new series from known ones: e.g., differentiate the geometric series 1/(1+x) to get the series of 1/(1+x)², or integrate it to get the series of ln(1+x).
Application to definite integrals: when an antiderivative is hard or impossible in closed form (like ∫ cos(x²) dx), integrate the Maclaurin series term-by-term and evaluate to approximate the answer.
Faster than first principles: avoid computing f⁽ⁿ⁾(0) when you can differentiate or integrate an existing series term-by-term.

The term-by-term differentiation rule

Term-by-term differentiation If f(x) = a₀ + a₁x + a₂x² + a₃x³ + …, then f′(x) = a₁ + 2a₂x + 3a₃x² + 4a₄x³ + …

Each xⁿ/n term in ln(1+x) differentiates to ±xⁿ⁻¹ (the n‘s cancel). Integration runs the same diagram in reverse: integrate each derivative term and add +c at the end.

The term-by-term integration rule

Term-by-term integration ∫(a₀ + a₁x + a₂x² + …) dx = c + a₀x + a₁2x² + a₂3x³ + …

The constant of integration c is determined by the known value at x = 0. If integrating g(x) recovers f(x) (so g = f′), evaluating the integrated series at x = 0 gives c = f(0) — every other term vanishes.

Two paired operations

Differentiate

a_nxⁿ → n·a_nxⁿ⁻¹

Power drops by 1, coefficient multiplied by old power. Constant term disappears. No +c.

Integrate

a_nxⁿ → a_nn+1xⁿ⁺¹

Power increases by 1, coefficient divided by new power. Always add a +c at the end and determine from f(0).

Cross-check pairs: (sin x, cos x), (e^x, e^x), (ln(1+x), 1/(1+x)), (arctan x, 1/(1+x²)). Differentiate the first to get the second; integrate the second (with appropriate c) to get the first.

🧭 Recipe — using differentiation or integration

Identify which technique applies: do you want the derivative or antiderivative of a function whose series you already know?
Write the known series from the formula booklet (or one you’ve derived earlier) with enough terms for the target accuracy.
For differentiation: apply d/dx(xⁿ) = nxⁿ⁻¹ to every term. The constant term vanishes.
For integration: apply ∫xⁿ dx = xⁿ⁺¹/(n+1) to every term. Add a single +c.
Determine c from the value of the antiderivative at x = 0 (which equals c). Simplify the resulting series.

Worked examples

WE 1

Differentiate ln(1+x) series to derive 1/(1+x) series

(a) Write down the derivative of ln(1+x). (b) Hence use the Maclaurin series for ln(1+x) to derive the Maclaurin series for 1/(1+x) up to and including the term in x⁴.

(a) Derivative of ln(1+x) d/dx [ln(1+x)] = 1/(1+x) (b) Booklet series for ln(1+x) ln(1+x) = x – x²/2 + x³/3 – x⁴/4 + x⁵/5 – … Differentiate each term using d/dx (xⁿ) = n xⁿ⁻¹ d/dx (x) = 1 d/dx (-x²/2) = -x d/dx (x³/3) = x² d/dx (-x⁴/4) = -x³ d/dx (x⁵/5) = x⁴ Collect — by part (a), this is the series for 1/(1+x) 11 + x ≈ 1 − x + x² − x³ + x⁴ notice the n’s cancel beautifully: xⁿ/n → xⁿ⁻¹ via differentiation. The result is the geometric series 1/(1+x) = 1 − x + x² − x³ + …, also obtainable from the binomial expansion of (1+x)⁻¹.

WE 2

Integrate cos x series to derive sin x series

Use the Maclaurin series for cos x from the formula booklet to derive the Maclaurin series for sin x by integration, up to and including the term in x⁵.

Step 1 — note sin x is an antiderivative of cos x d/dx (sin x) = cos x, so ∫ cos x dx = sin x + c Step 2 — booklet series for cos x cos x = 1 – x²/2! + x⁴/4! – x⁶/6! + … Step 3 — integrate term by term ∫ 1 dx = x ∫ -x²/2! dx = -x³/(3 · 2!) = -x³/6 = -x³/3! ∫ x⁴/4! dx = x⁵/(5 · 4!) = x⁵/120 = x⁵/5! Step 4 — apply IC: sin(0) = 0, so c = 0 sin x = x – x³/3! + x⁵/5! + c At x = 0: sin(0) = 0 + 0 + 0 + c = c, so c = 0. sin x ≈ x − x³3! + x⁵5! elegant pattern: integrating x^(2k)/(2k)! gives x^(2k+1)/(2k+1)!. The factorials in the denominators grow correctly. Matches the booklet series for sin x ✓.

WE 3

Integrate 1/(1−x) series to derive ln(1−x) series

Given that 1/(1 − x) = 1 + x + x² + x³ + … (the geometric series), use integration to derive the Maclaurin series for ln(1 − x) up to and including the term in x⁵.

Step 1 — find an antiderivative of 1/(1-x) ∫ 1/(1-x) dx = -ln|1-x| + c So if we integrate the series, we’ll get -ln(1-x) + c (for 1-x > 0). Step 2 — integrate the geometric series term by term ∫ (1 + x + x² + x³ + x⁴) dx = x + x²/2 + x³/3 + x⁴/4 + x⁵/5 + c Step 3 — match to -ln(1-x) -ln(1-x) = x + x²/2 + x³/3 + x⁴/4 + x⁵/5 + c At x = 0: -ln(1) = 0, so c = 0. ⟹ -ln(1-x) = x + x²/2 + x³/3 + x⁴/4 + x⁵/5 Step 4 — multiply by -1 ln(1 − x) ≈ −x − x²2 − x³3 − x⁴4 − x⁵5 cross-check: substitute -x into the booklet ln(1+x) series. That gives the same result. Either method works — pick whichever is faster.

WE 4

Differentiate 1/(1+x) to derive 1/(1+x)² series

Use the geometric series 1/(1+x) = 1 − x + x² − x³ + x⁴ − x⁵ + … and term-by-term differentiation to derive the Maclaurin series for 1/(1+x)² up to and including the term in x⁴.

Step 1 — differentiate the function d/dx [1/(1+x)] = d/dx [(1+x)⁻¹] = -(1+x)⁻² = -1/(1+x)² Step 2 — differentiate the series term by term d/dx [1 – x + x² – x³ + x⁴ – x⁵] = 0 – 1 + 2x – 3x² + 4x³ – 5x⁴ = -1 + 2x – 3x² + 4x³ – 5x⁴ Step 3 — equate and solve -1/(1+x)² = -1 + 2x – 3x² + 4x³ – 5x⁴ Multiply both sides by -1: 1(1 + x)² ≈ 1 − 2x + 3x² − 4x³ + 5x⁴ verify by binomial: (1+x)⁻² has coefficients C(-2, n)·1^(-2-n). C(-2, 0)=1, C(-2, 1)=-2, C(-2, 2)= (-2)(-3)/2=3, … → 1 – 2x + 3x² – 4x³ + … ✓ matches.

WE 5

Verify e^x is its own derivative at the series level

Use term-by-term differentiation of the Maclaurin series for e^x to confirm that e^x is its own derivative.

Step 1 — booklet series for e^x e^x = 1 + x + x²/2! + x³/3! + x⁴/4! + … Step 2 — differentiate each term d/dx (1) = 0 (constant disappears) d/dx (x) = 1 d/dx (x²/2!) = 2x/2! = x/1! = x d/dx (x³/3!) = 3x²/3! = x²/2! d/dx (x⁴/4!) = 4x³/4! = x³/3! Step 3 — collect d/dx (e^x) = 1 + x + x²/2! + x³/3! + … = e^x ✓ d/dx(e^x) = e^x ✓ (every term reproduces the term before it) the trick is n/n! = 1/(n-1)!. Differentiating xⁿ/n! gives n·xⁿ⁻¹/n! = xⁿ⁻¹/(n-1)!. So term-by-term, the series shifts down by one and the n-th coefficient becomes the (n-1)-th coefficient — leaving the series unchanged.

WE 6

Approximate a definite integral with no closed form

The integral ∫₀^0.5 cos(x²) dx has no elementary antiderivative. Use the Maclaurin series for cos(x²) and term-by-term integration to estimate its value to 4 decimal places.

Step 1 — get the Maclaurin series for cos(x²) cos x = 1 – x²/2! + x⁴/4! – x⁶/6! + … Substitute x²: cos(x²) = 1 – x⁴/2 + x⁸/24 – x¹²/720 + … Step 2 — integrate term by term ∫ cos(x²) dx ≈ x – x⁵/(5·2) + x⁹/(9·24) – … + c = x – x⁵/10 + x⁹/216 – … + c Step 3 — evaluate from 0 to 0.5 ∫₀^0.5 cos(x²) dx ≈ [x – x⁵/10 + x⁹/216]₀^0.5 = 0.5 – (0.5)⁵/10 + (0.5)⁹/216 – 0 = 0.5 – 0.003125 + 0.00000904 ≈ 0.4969 (to 4 d.p.) ∫₀^0.5 cos(x²) dx ≈ 0.4969 the actual value (computed numerically) is 0.49688403, so the three-term approximation matches to 7+ decimal places. This technique works for many “non-elementary” integrals — error functions, Fresnel integrals, sine/cosine integrals, etc.

💡 Top tips

Use what you already have: if a question asks for the series of f′(x) or ∫f(x) dx and you have the series for f(x), don’t compute derivatives at 0 — just differentiate or integrate term-by-term.
The constant of integration matters: every integration introduces a +c. Substitute x = 0 into both sides to find c from the known antiderivative value at 0.
Cancel cleverly: when differentiating xⁿ/n, the n‘s cancel. When integrating xⁿ, the denominator becomes (n+1) automatically.
Definite integrals of non-elementary functions: term-by-term integration of the series is a standard technique. Each extra term added to the truncation improves accuracy by roughly a power of x.
Justify your method in exam answers: state that you’re “differentiating the Maclaurin series term-by-term” or “integrating the Maclaurin series term-by-term, with constant of integration determined from f(0)”.

⚠ Common mistakes

Forgetting the constant of integration: integration always adds a +c. Don’t skip determining it — at x = 0, c equals the antiderivative’s known value.
Forgetting that the constant term vanishes when differentiating: if f(0) = 1, the leading “1” in the series for f goes to 0 in f′. Don’t carry it through.
Off-by-one power errors: differentiation lowers each power by 1; integration raises by 1. If you forget this, the entire resulting series is misaligned.
Wrong coefficient after integrating: ∫xⁿdx = xⁿ⁺¹/(n+1). It’s easy to write the wrong denominator (n instead of n+1).
Not enough terms for accuracy: when approximating a definite integral, keep enough terms so the next discarded term is smaller than the required accuracy. Adding one more term and checking it’s negligible confirms you have enough.

Up next (final note of the chapter): Maclaurin Series from Differential Equations. When you have dy/dx = g(x, y) and an initial condition y(0), you can build the Maclaurin series of the solution y = f(x) term by term — using implicit differentiation to find each successive derivative y′(0), y″(0), y‴(0), …. This works even when the DE has no closed-form solution.

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