IB Maths AI HL
Integration
Paper 1 & 2
~5 min read
Finding the Constant of Integration
An indefinite integral gives a whole family of curves — all the same shape, shifted up or down by the unknown “+ c“. One known point on the curve is enough to single out the right one: substitute it in, solve for c, and the family collapses to a single function.
📘 What you need to know
- Family of curves: every value of c gives a curve of the same shape, shifted vertically.
- One point pins it down: a known point on y = F(x) + c fixes c.
- Method: integrate (keep “+ c“), substitute the point, solve for c.
- Rewrite first if needed — each term a power of x before integrating.
- Don’t stop at “+ c“: if the question gives a point, you’re expected to find c.
- Final answer: a specific function with c replaced by its value.
The family of curves
🤔 Why does one point fix c?
All the antiderivatives differ only by a vertical shift, so at any given x they each pass through a different height. Naming one actual point — say the curve goes through (3, −4) — picks out exactly the member of the family at that height, which fixes c to a single value.
🧠 “Integrate, substitute, solve”
Integrate to get F(x) + c, substitute the known point to make an equation in c, then solve. Three steps, every time.
Finding c
🧭 Recipe — the constant of integration
- Rewrite f(x) into integrable form (each term a power of x) if needed.
- Integrate each term, remembering “+ c“.
- Substitute the x and y of the given point to form an equation in c.
- Solve for c and write the final function.
Worked examples
The graph of y = f(x) passes through (3, −4), with f′(x) = 3x2 − 4x − 4.
Already in integrable form — integrate term by term, keep “+ c“.
f(x) = 3x³3 − 4x²2 − 4x + c
f(x) = x³ − 2x² − 4x + c
WE 2Substitute the point (3, −4)
Put x = 3 and f(x) = −4 to form an equation in c.
f(3) = −4
(3)³ − 2(3)² − 4(3) + c = −4
27 − 18 − 12 + c = −4
−3 + c = −4
WE 3Solve for c and state f(x)
Solve the equation, then write the specific function.
c = −4 + 3 = −1
f(x) = x³ − 2x² − 4x − 1
WE 4A curve has dydx = 6x − 2 and passes through (1, 5). Find y.
Integrate, substitute, solve.
y = 3x² − 2x + c
at (1, 5): 3 − 2 + c = 5 → 1 + c = 5
c = 4
y = 3x² − 2x + 4
WE 5A curve has dydx = 4√x and passes through (4, 10). Find y.
Rewrite the root first, then integrate and solve.
dydx = 4x^(1/2) → y = 4x^(3/2)3/2 + c = 83x^(3/2) + c
at (4, 10): 83(8) + c = 10 → 643 + c = 10
c = 10 − 643 = −343
y = 83√x³ − 343
💡 Top tips
- Keep “+ c“ right through the integration — you need it to solve for.
- Substitute carefully — match the x and y of the given point.
- Always finish the job — a given point means you must find c.
- Rewrite roots/fractions before integrating.
- State the specific function at the end, with c replaced.
- Check by substituting the point back into your final answer.
⚠ Common mistakes
- Leaving the answer with “+ c“ when a point was given.
- Dropping “+ c” during integration — then there’s nothing to solve.
- Swapping x and y when substituting the point.
- Arithmetic slips with negatives when solving for c.
- Forgetting to rewrite roots/fractions before integrating.
Next up — Finding Areas Using a GDC. You’ve now mastered the indefinite integral and how to pin down its constant. The final topic of the unit turns to the definite integral — where limits replace the “+ c” entirely — to find the exact area under a curve, with your GDC doing the heavy lifting.
Need help with Integration?
Get 1-on-1 help from an IB examiner who knows exactly what Paper 1 & 2 are looking for.
Book Free Session →