IB Maths AI SL Integration Paper 1 & 2 the constant c ~6 min read

Finding the Constant of Integration

An indefinite integral comes with a “+ c” — and on its own, c could be anything. But give one point that the curve passes through, and c is fixed. This note shows how a single known point turns a whole family of antiderivatives into one definite function.

📘 What you need to know

Integrating gives a family of curves y = F(x) + c — all the same shape, shifted vertically.
A single known point on the curve picks out exactly one member of the family — it fixes the value of c.
To find c: integrate (remembering + c), then substitute the point’s coordinates into the result.
This gives an equation in c — solve it for the value of c.
The same process recovers f(x) from f′(x), or y from dy/dx, whenever a point is given.
If a question gives a point, you are expected to find c — don’t leave the answer as “+ c“.

Choosing a curve from the family

Integrating a function produces not one answer but a whole family of curves, y = F(x) + c — every value of c giving a different curve. All the curves have the same shape; they are simply shifted up or down.

The family of antiderivatives ∫ f(x) dx = F(x) + c one curve for each value of c — all the same shape, shifted vertically

All four curves are antiderivatives of the same function — same shape, shifted by c. The known point (x₁, y₁) lies on just one of them; that curve (bold) is the answer.

So which curve is the right one? A single known point settles it. Only one member of the family passes through that point, and finding which one means finding the value of c.

Finding the value of c

Once you know a point on the curve, the constant is quick to find. Integrate as usual — keeping the + c — then substitute the coordinates of the known point.

Finding the constant at a known point (x₁, y₁): y₁ = F(x₁) + c substitute the point’s coordinates into F(x) + c, then solve for c

Putting the point’s x– and y-values into F(x) + c gives an equation with only c unknown. Solve it, and write the final function with that value in place — never leave a “+ c” once the question has given you enough to find it.

Functions from f′(x) or dy/dx

In practice these questions hand you a gradient function — written f′(x) or dy/dx — together with one point, and ask for the original function.

The process is the same either way: integrating f′(x) recovers f(x), and integrating dy/dx recovers y — in both cases the given point fixes the constant. If the gradient function is not yet a sum of powers, rewrite it first: expand brackets, and turn fractions into negative powers.

🧭 Recipe — finding the constant of integration

Rewrite the function into integrable form — every term a power of x or a constant.
Integrate term by term, including the constant of integration + c.
Substitute the x– and y-coordinates of the known point into the result.
Solve the resulting equation for c.
Write the final function with the value of c filled in — no “+ c” left.

Worked examples

WE 1

A gradient function and a point

A curve has gradient function f′(x) = 2x + 3 and passes through the point (1, 5). Find f(x).

integrate f′(x), keeping + c f(x) = x² + 3x + c substitute the point (1, 5): f(1) = 5 (1)² + 3(1) + c = 5 ⇒ 4 + c = 5 c = 1, so f(x) = x² + 3x + 1 integrate first, then use the point — the + c is essential until the point pins it down.

WE 2

Working with dy/dx

A curve has dy/dx = 6x² − 4, and y = 7 when x = 2. Find y.

integrate dy/dx, keeping + c y = 2x³ − 4x + c substitute x = 2, y = 7 2(8) − 4(2) + c = 7 ⇒ 8 + c = 7 c = −1, so y = 2x³ − 4x − 1 the same method works with dy/dx — integrate to recover y, then use the point.

WE 3

A point on the y-axis

The gradient of a curve is f′(x) = 3x² − 10x, and the curve passes through (0, 4). Find f(x).

integrate, keeping + c f(x) = x³ − 5x² + c substitute (0, 4) (0)³ − 5(0)² + c = 4 ⇒ c = 4 f(x) = x³ − 5x² + 4 when the point is on the y-axis (x = 0) the constant c is simply the y-value.

WE 4

Rewriting before integrating

A curve has dy/dx = 4x − 6/x² and passes through (1, 9). Find y.

rewrite: dy/dx = 4x − 6x⁻² y = 2x² + 6x⁻¹ + c = 2x² + 6/x + c substitute (1, 9) 2(1)² + 6/1 + c = 9 ⇒ 8 + c = 9 c = 1, so y = 2x² + 6/x + 1 rewrite the fraction as a negative power before integrating; ∫(−6x⁻²) dx = 6x⁻¹.

WE 5

Expanding before integrating

The gradient function of f(x) is f′(x) = x(x − 6), and f(3) = −10. Find f(x).

expand the product: x(x − 6) = x² − 6x f(x) = x³/3 − 3x² + c substitute (3, −10): f(3) = −10 (3)³/3 − 3(3)² + c = −10 ⇒ −18 + c = −10 c = 8, so f(x) = x³/3 − 3x² + 8 expand the product before integrating — you cannot integrate brackets directly.

WE 6

Full question: a profit model

A company finds that the rate of change of its profit P (thousand dollars) with respect to the number of units x (in hundreds) sold is dP/dx = 9 − 2x. With no units sold (x = 0) the profit is −5 thousand dollars, a fixed cost. (a) Find an expression for P in terms of x. (b) Hence find the profit when 600 units are sold.

(a) integrate the rate, keeping + c P = ∫(9 − 2x) dx = 9x − x² + c at x = 0, P = −5: 0 − 0 + c = −5 ⇒ c = −5 P = 9x − x² − 5 (b) 600 units means x = 6 P = 9(6) − (6)² − 5 = 54 − 36 − 5 (a) P = 9x − x² − 5 · (b) profit = 13 thousand dollars a “rate of change” is a derivative — integrate it to recover the quantity, and the given starting value fixes c.

💡 Top tips

Always integrate first, with the + c, before using the point.
A point gives one equation — exactly enough to solve for one unknown, c.
When the point lies on the y-axis (x = 0), c is just the y-coordinate.
Rewrite fractions and expand products before integrating, as always.
A “rate of change” in a worded question is a derivative — integrate it, then use the starting value to find c.

⚠ Common mistakes

Leaving the answer as “+ c“ when a point was given to find it.
Substituting the point into the gradient function instead of into the integrated function.
Forgetting the + c entirely, so there is nothing to solve for.
Sign slips when solving the equation for c.
Not rewriting fractions or expanding products before integrating.

Next up: Finding Areas Using a GDC — definite integrals, where the “+ c” quietly cancels and integration finally delivers an exact area. The habit to keep: whenever a question gives a point alongside a gradient function, that point is there to be used — integrate, substitute, solve for c.

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