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

Finding the Constant of Integration

Without extra info, integration leaves you with an unknown c. But if the question gives you one point on the curve, you can pin it down. Just substitute the coordinates and solve for c. Quick, three steps, easy marks.

📘 What you need to know

Without info → indefinite integral has “+ c” (a whole family of curves).
With one point → you can find the exact value of c and pin down a single curve.
Method: integrate, substitute, solve.
Don’t leave your answer with “+ c” if extra info was given — that’s a lost mark.

How one point pins down the curve

Before vs after using a known point

Before — many curves

→ use the point

After — exact curve

one point is all you need to find c

Every “+ c” in the answer means there’s a whole family of parallel curves. The point is what locks in the right one.

The 3-step method

How to find c

Integrate the gradient function (or rate). Don’t forget + c.
Substitute the given point’s x and y into the integrated expression.
Solve the resulting equation for c, then write the final answer.

🧠

“Integrate, sub, solve”

Three words. Three steps. That’s the whole method.

Worked examples

WE 1

Basic case

The curve y = f(x) passes through (1, 5) and dy/dx = 4x. Find y in terms of x.

step 1 — integrate y = ∫ 4x dx = 4x²/2 + c = 2x² + cstep 2 — substitute (1, 5) 5 = 2(1)² + c 5 = 2 + c → c = 3step 3 — write final y = 2x² + 3 simple — integrate, sub, solve!

WE 2

Cubic gradient through a given point

The graph of y = f(x) passes through (3, −4). The gradient function is f′(x) = 3x² − 4x − 4. Find f(x).

step 1 — integrate f(x) = 3x³/3 − 4x²/2 − 4x + c f(x) = x³ − 2x² − 4x + cstep 2 — substitute (3, −4) f(3) = (3)³ − 2(3)² − 4(3) + c = −4 27 − 18 − 12 + c = −4 −3 + c = −4 → c = −1step 3 — write final f(x) = x³ − 2x² − 4x − 1 don’t forget to write the FULL function in your final answer!

WE 3

With fractional powers — rewrite first

The curve y = f(x) passes through (4, 12) and f′(x) = 6√x. Find f(x).

step 1 — rewrite & integrate f′(x) = 6x^(1/2) f(x) = 6x^(3/2)/(3/2) + c = 4x^(3/2) + cstep 2 — substitute (4, 12) 12 = 4(4)^(3/2) + c (4)^(3/2) = √4³ = 8 12 = 4(8) + c = 32 + c → c = −20step 3 — final f(x) = 4x^(3/2) − 20 (or 4√x³ − 20) divide by 3/2 = multiply by 2/3 → 6 × 2/3 = 4!

WE 4

Find a specific value once c is known

The graph of y = f(x) passes through (2, 7) and f′(x) = 6x − 4. Find the value of f(5).

step 1 — integrate f(x) = 3x² − 4x + cstep 2 — find c using (2, 7) 7 = 3(4) − 4(2) + c = 12 − 8 + c → c = 3 f(x) = 3x² − 4x + 3step 3 — find f(5) f(5) = 3(25) − 4(5) + 3 = 75 − 20 + 3 f(5) = 58 find c first, THEN evaluate at the new x-value!

WE 5

When the point IS the y-intercept

The curve y = f(x) has y-intercept −5 and dy/dx = 9x² − 2x. Find y in terms of x.

step 1 — integrate y = 9x³/3 − 2x²/2 + c = 3x³ − x² + cstep 2 — substitute (0, −5) y-intercept means x = 0, y = −5: −5 = 3(0)³ − (0)² + c → c = −5step 3 — final y = 3x³ − x² − 5 when the point is the y-intercept, c is just the y-value directly!

💡 Top tips

Integrate, sub, solve — three quick steps, easy marks.
Always include “+ c” when integrating, even if you’ll find it later.
Read the question carefully to spot the given point — sometimes it’s “y-intercept”, sometimes “passes through (a, b)”.
If the point is the y-intercept (x = 0), then c = y-value directly — no algebra needed.
Find c first, then write the full function before answering any follow-up questions.
Check your answer by substituting the given point — you should get the right y-value.

⚠ Common mistakes

Leaving “+ c” in your final answer when extra info was given. Find the value!
Sign errors when solving for c, especially with negative coordinates.
Substituting into f′(x) instead of f(x). The point lies on the original curve, not the gradient.
Forgetting to write the full function in the final answer — just stating “c = 3” isn’t enough.
Mixing up which is x and which is y in the given point — careful with order.

Now you can find the exact antiderivative when given a point. The next note moves to definite integrals — using your GDC to compute the area under a curve between two limits.

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