IB Maths AA HL Topic 5 — Calculus Paper 1 & 2 ~7 min read

Finding the Constant of Integration

An indefinite integral leaves “+ c” as an unknown. To pin down the exact antiderivative, you need an extra piece of information — usually a known point on the curve, or an initial condition like f(0) = 2. Substitute the values, solve for c, and you’ve turned an antiderivative into the antiderivative.

📘 What you need to know

Why c matters: integrating gives the family F(x) + c. Each value of c shifts the curve vertically; you need one piece of information to lock down which curve in the family is the right one.
What “more information” looks like: a point (x₀, y₀) on the curve, or a stated value like f(2) = 7 or v(0) = 3.
Plug-and-solve: integrate to get F(x) + c; substitute the known x-value to get F(x₀) + c; set this equal to the known y-value; solve for c.
Multiple constants for multiple integrations: integrating f″ to get f gives TWO constants (one per integration), needing TWO conditions.
Common kinematics setup: ∫v(t) dt = s(t) + c; the initial displacement s(0) gives c directly.
For Paper 1 (no GDC): be especially careful with arithmetic when substituting; for Paper 2 you can use a GDC to verify.
Don’t leave “+ c” in the final answer when c has been found — substitute the numerical value of c.

The setup — picking the right curve from a family

Every antiderivative of a function f produces a family of vertically shifted curves y = F(x) + c. The given point selects one specific curve from the family — the one that actually passes through that point.

The family of antiderivatives all have the same shape. The given point determines which specific curve passes through that location — that’s the curve whose constant c matches.

🧭 Recipe — find the constant of integration

Rewrite the integrand if needed (roots → fractional powers, reciprocals → negative powers).
Integrate term by term, remembering “+ c” at the end of the antiderivative.
Substitute the known point (x₀, y₀) into the antiderivative: F(x₀) + c = y₀.
Solve for c — a single linear equation.
Write out the full antiderivative with c substituted in (NO “+ c” left over).

Worked examples

WE 1

Cubic from a quadratic derivative

The graph of y = f(x) passes through (1, 8). Given f′(x) = 6x² − 4x + 3, find f(x).

Step 1 — integrate f′(x) f(x) = ∫(6x² − 4x + 3) dx = 6x³/3 − 4x²/2 + 3x + c = 2x³ − 2x² + 3x + c Step 2 — substitute (1, 8) and solve for c f(1) = 2 − 2 + 3 + c = 3 + c = 8 → c = 5 Write out the answer f(x) = 2x³ − 2x² + 3x + 5 f(x) = 2x³ − 2x² + 3x + 5 verify: f(1) = 2 − 2 + 3 + 5 = 8 ✓

WE 2

Negative powers — reciprocals + initial condition

The function f satisfies f′(x) = 4x − 6/x² and f(2) = 7. Find f(x).

Step 1 — rewrite and integrate f′(x) = 4x − 6x⁻² f(x) = ∫(4x − 6x⁻²) dx = 2x² − 6x⁻¹/(−1) + c = 2x² + 6/x + c Step 2 — substitute f(2) = 7 f(2) = 2(4) + 6/2 + c = 8 + 3 + c = 11 + c = 7 → c = −4 Write out the answer f(x) = 2x² + 6/x − 4 f(x) = 2x² + 6/x − 4 careful with double negatives: −6x⁻¹/(−1) = +6/x

WE 3

Root + constant; given f(4)

Given f′(x) = 3√x + 2 and f(4) = 12, find f(x).

Step 1 — rewrite and integrate f′(x) = 3x^(1/2) + 2 f(x) = ∫(3x^(1/2) + 2) dx = 3 · x^(3/2)/(3/2) + 2x + c = 2x^(3/2) + 2x + c Step 2 — substitute f(4) = 12 f(4) = 2(4)^(3/2) + 2(4) + c = 2·8 + 8 + c = 24 + c = 12 → c = −12 Write out the answer f(x) = 2x^(3/2) + 2x − 12 f(x) = 2x^(3/2) + 2x − 12 (also 2x√x + 2x − 12) 4^(3/2) = (√4)³ = 2³ = 8 — knowing standard powers of 4 saves time

WE 4

Curve through (1, 0) — dy/dx notation

A curve has gradient function dy/dx = 5 − 4x³ and passes through the point (1, 0). Find the equation of the curve.

Step 1 — integrate dy/dx y = ∫(5 − 4x³) dx = 5x − x⁴ + c Step 2 — substitute (1, 0) 0 = 5(1) − (1)⁴ + c = 5 − 1 + c = 4 + c → c = −4 Write out the answer y = 5x − x⁴ − 4 y = −x⁴ + 5x − 4 “passes through (a, b)” means f(a) = b — same thing as a known point

WE 5

Kinematics — initial displacement

A particle moves in a straight line with velocity v(t) = 3t² − 6t + 4 m/s, where t is time in seconds, t ≥ 0. The particle’s displacement from a fixed origin at t = 0 is 2 metres. Find an expression for the displacement s(t).

Step 1 — displacement = integral of velocity s(t) = ∫v(t) dt = ∫(3t² − 6t + 4) dt = t³ − 3t² + 4t + c Step 2 — use initial condition s(0) = 2 s(0) = 0 − 0 + 0 + c = c = 2 → c = 2 Write out the answer s(t) = t³ − 3t² + 4t + 2 s(t) = t³ − 3t² + 4t + 2 metres at t = 0, all polynomial terms with t vanish — c equals the y-intercept directly. Always quickest to use t = 0 if it’s available

WE 6

Two-stage integration — TWO constants needed

Given f″(x) = 12x − 6, f′(1) = 0, and f(2) = 5, find f(x).

Step 1 — integrate f″(x) to find f′(x) f′(x) = ∫(12x − 6) dx = 6x² − 6x + c₁ Use f′(1) = 0 to find c₁ f′(1) = 6 − 6 + c₁ = 0 + c₁ = 0 → c₁ = 0 → f′(x) = 6x² − 6x Step 2 — integrate f′(x) to find f(x) f(x) = ∫(6x² − 6x) dx = 2x³ − 3x² + c₂ Use f(2) = 5 to find c₂ f(2) = 16 − 12 + c₂ = 4 + c₂ = 5 → c₂ = 1 Write out the answer f(x) = 2x³ − 3x² + 1 f(x) = 2x³ − 3x² + 1 two integrations need two conditions — typically one for f′ and one for f. Order matters: use the f’ condition to find c₁ BEFORE integrating again

💡 Top tips

The given point pins down c — without it, the antiderivative is just a family. With it, you have the specific curve.
If a condition uses x = 0, every x-term vanishes and c = the y-value directly. The cleanest case.
Two integrations need two conditions — for f″ → f, you get c₁ from an f′ condition and c₂ from an f condition. Don’t mix them up.
Verify by differentiating — your final f(x) should give back f′(x) when differentiated.
Substitute the NUMERIC value of c in the final answer; leaving “+ c” still in is the most common mark loss on these questions.

⚠ Common mistakes

Leaving “+ c” in the final answer — once you’ve solved for c, substitute its value in.
Forgetting “+ c” during integration — then you have nothing to solve for. Always include it on every indefinite integral step.
Substituting into f′(x) instead of f(x) — read the condition carefully. f(2) = 7 means substitute into f, NOT f′.
For two-stage problems, using only ONE condition — f″ → f needs two integrations and two conditions; one of each won’t determine f uniquely.
Sign slips when substituting — especially with negative coordinates or negative powers of x. Take it slow.

Up next: Finding Areas Using a GDC. Indefinite integrals become DEFINITE integrals when you add limits a and b — and definite integrals compute the area under a curve. ∫ₐᵇ f(x) dx = F(b) − F(a). The constant of integration cancels out, so “+ c” is no longer needed. Your GDC can evaluate these directly — but knowing how to set them up by hand is essential for Paper 1.

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