IB Maths AA HL Topic 5 — Calculus Paper 1 & 2 ~6 min read

Introduction to Integration

Integration is the reverse of differentiation: if differentiation takes f to f′, integration takes f′ back to f. The output is called an antiderivative and is written F(x). Because constants disappear when differentiating, any antiderivative carries a “+ c” — the constant of integration. The notation ∫f(x) dx = F(x) + c is the standard form.

📘 What you need to know

Integration = reverse of differentiation. Also called antidifferentiation; the result is an antiderivative.
Notation: ∫f(x) dx = F(x) + c. The ∫ symbol means “integrate”; dx tells you the variable.
Integrand: the function f(x) being integrated.
Constant of integration “+ c“: every antiderivative includes an unknown constant, because d/dx(constant) = 0.
Family of antiderivatives: f(x) has infinitely many antiderivatives — they all differ only by a constant.
Verification by differentiation: if you suspect F is an antiderivative of f, check that d/dx(F) = f.
Indefinite integral: just another name for the antiderivative + c; written ∫f(x) dx.
Two equivalent notations: ∫f(x) dx and ∫(dy/dx) dx — the second integrates a derivative to recover y.

The integration symbol — anatomy of an integral

Indefinite integral ∫ f(x) dx = F(x) + c

Part of ∫f(x) dx	What it means
∫	the integral sign — “integrate the following”
f(x)	the integrand — the function being integrated
dx	the variable of integration (here, x)
F(x)	the antiderivative — its derivative gives f(x)
+ c	the constant of integration (mandatory in indefinite integrals)

The reverse relationship: d/dx takes F to f. ∫ … dx takes f back to F (+ c). They undo each other — almost. The “almost” is the lost constant: when you differentiate F, any constant in F disappears, so when you integrate back you must add “+ c“.

Why “+ c” — the family of antiderivatives

Consider f(x) = 2x. Every function of the form F(x) = x² + (any constant) differentiates back to 2x: the constant term vanishes. So x², x² + 5, x² − 17, x² + 100 are all antiderivatives of 2x. They form a family of curves, identical in shape, shifted vertically up or down.

x² shifted up or down x y y = x² + 0 y = x² + 5 y = x² − 3 y = x² + 10 all four curves have the same gradient function f(x) = 2x at every value of x

Each curve is y = x² + c for a different c. They all share the same shape and have the same gradient at every x — so all are antiderivatives of f(x) = 2x.

🧭 Recipe — verify an antiderivative

Differentiate F(x) using the standard rules (power, chain, product, etc).
Compare F′(x) with f(x): if they match, F is an antiderivative.
Constants don’t matter for verification: F(x) and F(x) + 7 both work.
For the indefinite integral, always write “+ c“ at the end of your final answer.
If something is missing, adjust the coefficient (the formula for ∫xⁿ dx comes in the next note).

Worked examples

WE 1

Verify an antiderivative by differentiating

Show that F(x) = 2x³ − x² + 4x is an antiderivative of f(x) = 6x² − 2x + 4.

Differentiate F(x) term by term d/dx [2x³] = 6x² d/dx [−x²] = −2x d/dx [4x] = 4 Add the derivatives F′(x) = 6x² − 2x + 4 Compare with f(x) F′(x) = f(x) ✓ F(x) is an antiderivative of f(x), as required F(x) + c is the general antiderivative — any constant added to F still differentiates to f

WE 2

Three antiderivatives of the same function — all valid

Show that each of F(x) = x⁴ + 7, G(x) = x⁴ − 3, and H(x) = x⁴ + 100 is an antiderivative of f(x) = 4x³.

Differentiate each function d/dx [x⁴ + 7] = 4x³ + 0 = 4x³ d/dx [x⁴ − 3] = 4x³ − 0 = 4x³ d/dx [x⁴ + 100] = 4x³ + 0 = 4x³ All three give the same derivative → all three are antiderivatives of f(x) = 4x³ F, G, H all differentiate to 4x³ — only differ by a constant this is the WHOLE reason for “+ c” — the constant disappears when you differentiate

WE 3

Trig antiderivative — verify

Show that F(x) = −cos(x) is an antiderivative of f(x) = sin(x).

Differentiate F(x) = −cos(x) d/dx [cos(x)] = −sin(x) d/dx [−cos(x)] = −(−sin(x)) = sin(x) Compare with f(x) F′(x) = sin(x) = f(x) ✓ ∫ sin(x) dx = −cos(x) + c the minus on cos derivative gets cancelled by the minus on F — that’s why ∫ sin(x) dx has a minus

WE 4

Chain-rule reverse — verify a more complex antiderivative

Show that F(x) = (3x + 1)⁴ / 12 is an antiderivative of f(x) = (3x + 1)³.

Differentiate F using chain rule F(x) = (1/12) · (3x + 1)⁴ d/dx [(3x + 1)⁴] = 4(3x + 1)³ · 3 = 12(3x + 1)³ Apply the 1/12 factor F′(x) = (1/12) · 12(3x + 1)³ = (3x + 1)³ Compare with f(x) F′(x) = (3x + 1)³ = f(x) ✓ ∫ (3x + 1)³ dx = (3x + 1)⁴ / 12 + c the 1/12 is the “reverse chain rule” — power goes up by 1 (→ 4), then divide by 4·3 = 12

WE 5

Find an antiderivative — e^x

Find F(x) such that F′(x) = e^x. State the general antiderivative.

Recall: d/dx [e^x] = e^x → F(x) = e^x is a candidate Verify by differentiating F′(x) = e^x ✓ General antiderivative — add the constant F(x) = e^x + c ∫ e^x dx = e^x + c e^x is special — it’s the only function (up to a constant multiple) that’s its own derivative AND its own antiderivative

WE 6

Read the notation; state alternative antiderivatives

Consider the indefinite integral ∫(3x² + 5x − 2) dx.

(a) State the integrand and the variable of integration. (b) Explain why the answer must include “+ c“. (c) Given that F(x) = x³ + (5/2)x² − 2x is ONE valid antiderivative, write down two more.

(a) Read off from the notation integrand: 3x² + 5x − 2 variable of integration: x (from “dx”) (b) Why “+ c”? d/dx [constant] = 0, so any constant in F vanishes on differentiating → infinitely many antiderivatives exist, differing by a constant (c) Two more antiderivatives — just shift by any constant G(x) = x³ + (5/2)x² − 2x + 1 H(x) = x³ + (5/2)x² − 2x − 7 Verify (sanity check) G′(x) = 3x² + 5x − 2 ✓ H′(x) = 3x² + 5x − 2 ✓ (a) 3x² + 5x − 2; variable x | (b) constants vanish on differentiating | (c) e.g. F(x) + 1 and F(x) − 7 any number of valid alternative antiderivatives — there are infinitely many. + c covers them all in one expression

💡 Top tips

“Differentiate to check” is the fastest way to verify any antiderivative — if d/dx(F) = f, you’re done.
Always write “+ c“ at the end of an indefinite integral. Markers deduct for forgetting it.
Read “dx“ as “with respect to x“. If a question uses dt instead, integrate with respect to t.
Constants don’t need + c in each step — just on the final answer. Adding constants together gives one constant.
If you forget the integral of something, guess a candidate F and differentiate it; adjust the coefficient until F′ matches f.

⚠ Common mistakes

Forgetting “+ c“ on indefinite integrals — the most common Paper 1 slip in integration.
Confusing notation: ∫f(x) dx integrates f. The dx at the end isn’t a multiplication — it tells you the variable.
Mixing up f and F: lowercase f is what you START with; uppercase F is the antiderivative (what you END with).
Thinking the antiderivative is unique — it’s not. There are infinitely many, differing by a constant.
Differentiating to “verify” and not noticing missing factors — sometimes F′ has the right shape but wrong coefficient. Always match coefficients carefully.

Up next: Integrating Powers of x. The formula ∫xⁿ dx = xⁿ⁺¹/(n+1) + c — raise the power by 1, then divide by the new power. Works for ANY rational n except n = −1 (because that would mean dividing by 0). After that comes “finding the + c” using an initial condition, and then GDC area calculations.

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