The Constant of Integration — Pure Mathematics 1 (9709) — Mathematics

Introduction

When you reverse the process of differentiation to find an indefinite integral, you cannot recover the original constant term — it vanishes upon differentiation. The result is a family of curves, all differing by a vertical shift, and the constant of integration $c$ represents this unknown shift. In the 9709 exam, questions regularly supply an additional piece of information — typically a point that lies on the curve — so that you can calculate the exact value of $c$ and state the unique curve. This skill sits at the heart of the syllabus objective for ref 1.8 and appears in nearly every integration question on Paper 1.

Core Concept

Suppose you are told that $\dfrac{\mathrm{d}y}{\mathrm{d}x} = f(x)$ . Integrating gives:

y = \int f(x)\, \mathrm{d}x = F(x) + c

where $F(x)$ is any antiderivative of $f(x)$ and $c$ is an arbitrary constant. Without further information, infinitely many curves satisfy this equation — each value of $c$ produces a distinct curve, all with the same gradient function.

To pin down the unique curve, you need one point $(x_0,\, y_0)$ that lies on it. Substituting $x = x_0$ and $y = y_0$ into $y = F(x) + c$ gives a single equation in the single unknown $c$ , which you solve directly. The resulting expression is the particular integral (or particular solution) for that specific curve.

In summary:

Integrate $\dfrac{\mathrm{d}y}{\mathrm{d}x}$ to obtain $y = F(x) + c$ .
Substitute the given point $(x_0, y_0)$ .
Solve for $c$ .
Write the complete equation of the curve.

Key Formulae & Definitions

General power rule for integration:

\int x^n \, \mathrm{d}x = \frac{x^{n+1}}{n+1} + c, \qquad n \neq -1

Linearity of integration:

\int \bigl[af(x) + bg(x)\bigr] \mathrm{d}x = a\int f(x)\,\mathrm{d}x + b\int g(x)\,\mathrm{d}x

Evaluating the constant — the key step:

y_0 = F(x_0) + c \implies c = y_0 - F(x_0)

Symbol	Meaning
$\dfrac{\mathrm{d}y}{\mathrm{d}x}$	Given gradient function
$F(x)$	Antiderivative (without constant)
$c$	Constant of integration (to be found)
$(x_0, y_0)$	Given point on the curve

Worked Examples

Example 1 — Polynomial gradient function

A curve has gradient function $\dfrac{\mathrm{d}y}{\mathrm{d}x} = 3x^2 - 4x + 1$ and passes through the point $(2,\, 5)$ . Find the equation of the curve.

Step 1 — Integrate the gradient function.

y = \int \left(3x^2 - 4x + 1\right)\mathrm{d}x

Apply the power rule term by term:

y = \frac{3x^3}{3} - \frac{4x^2}{2} + x + c = x^3 - 2x^2 + x + c

Step 2 — Substitute the known point $(2, 5)$ .

5 = (2)^3 - 2(2)^2 + 2 + c

5 = 8 - 8 + 2 + c

5 = 2 + c

Step 3 — Solve for $c$ .

c = 3

Step 4 — Write the equation of the curve.

\boxed{y = x^3 - 2x^2 + x + 3}

Example 2 — Gradient function involving a negative power

A curve satisfies $\dfrac{\mathrm{d}y}{\mathrm{d}x} = 6x^2 - \dfrac{4}{x^3}$ and passes through the point $\left(1,\, 7\right)$ . Find the equation of the curve.

Step 1 — Rewrite using index notation and integrate.

\frac{\mathrm{d}y}{\mathrm{d}x} = 6x^2 - 4x^{-3}

y = \int \left(6x^2 - 4x^{-3}\right)\mathrm{d}x

y = \frac{6x^3}{3} - \frac{4x^{-2}}{-2} + c

y = 2x^3 + 2x^{-2} + c

Step 2 — Substitute the known point $(1,\, 7)$ .

7 = 2(1)^3 + 2(1)^{-2} + c

7 = 2 + 2 + c

7 = 4 + c

Step 3 — Solve for $c$ .

c = 3

Step 4 — Write the equation of the curve.

\boxed{y = 2x^3 + \frac{2}{x^2} + 3}

Example 3 — Second derivative given

The second derivative of a curve is $\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} = 6x - 2$ . The curve passes through $(0,\, 4)$ and the gradient at $x = 1$ is $3$ . Find the equation of the curve.

Step 1 — Integrate once to find $\dfrac{\mathrm{d}y}{\mathrm{d}x}$ .

\frac{\mathrm{d}y}{\mathrm{d}x} = \int (6x - 2)\,\mathrm{d}x = 3x^2 - 2x + A

Step 2 — Use the gradient condition: $\dfrac{\mathrm{d}y}{\mathrm{d}x} = 3$ when $x = 1$ .

3 = 3(1)^2 - 2(1) + A \implies 3 = 1 + A \implies A = 2

So $\dfrac{\mathrm{d}y}{\mathrm{d}x} = 3x^2 - 2x + 2$ .

Step 3 — Integrate again to find $y$ .

y = \int (3x^2 - 2x + 2)\,\mathrm{d}x = x^3 - x^2 + 2x + B

Step 4 — Use the point $(0,\, 4)$ to find $B$ .

4 = 0 - 0 + 0 + B \implies B = 4

Step 5 — Write the equation of the curve.

\boxed{y = x^3 - x^2 + 2x + 4}

Common Mistakes & Examiner Pitfalls

Forgetting $c$ entirely. Omitting the constant of integration and then wondering why the point condition cannot be used — this earns no method marks for the constant.
Integrating but not evaluating $c$ . Writing $y = F(x) + c$ and stopping — the question asks you to find $c$ , so always complete the substitution step.
Substituting into the derivative instead of the integral. A very common slip: using $\dfrac{\mathrm{d}y}{\mathrm{d}x} = f(x)$ to substitute the point rather than $y = F(x) + c$ . Remember, the given point lies on the curve $y$ , not on the gradient function.
Index errors when integrating negative or fractional powers. For example, $\displaystyle\int x^{-3}\,\mathrm{d}x = \dfrac{x^{-2}}{-2}$ , not $\dfrac{x^{-2}}{2}$ . Always add 1 to the power and divide by the new power, carefully tracking signs.
Misreading the given point. Ensure you correctly identify which coordinate is $x$ and which is $y$ before substituting.
For second-derivative problems: each integration introduces a new constant ( $A$ , then $B$ ). Both must be evaluated using the two separate conditions provided. Missing either constant loses marks.

Practice Questions

Q1. A curve has gradient function $\dfrac{\mathrm{d}y}{\mathrm{d}x} = 4x^3 - 6x + 2$ and passes through the point $(1,\, 0)$ . Find the equation of the curve.

<details><summary>Show answer</summary>

Integrate:

y = \int (4x^3 - 6x + 2)\,\mathrm{d}x = x^4 - 3x^2 + 2x + c

Substitute $(1,\, 0)$ :

0 = 1 - 3 + 2 + c = 0 + c \implies c = 0

Equation of curve:

y = x^4 - 3x^2 + 2x

</details>

Q2. A curve satisfies $\dfrac{\mathrm{d}y}{\mathrm{d}x} = \sqrt{x} + \dfrac{1}{\sqrt{x}}$ and passes through $(4,\, 10)$ . Find $y$ in terms of $x$ .

<details><summary>Show answer</summary>

Rewrite in index form:

\frac{\mathrm{d}y}{\mathrm{d}x} = x^{1/2} + x^{-1/2}

Integrate:

y = \frac{x^{3/2}}{\tfrac{3}{2}} + \frac{x^{1/2}}{\tfrac{1}{2}} + c = \frac{2}{3}x^{3/2} + 2x^{1/2} + c

Substitute $(4,\, 10)$ :

10 = \tfrac{2}{3}(8) + 2(2) + c = \tfrac{16}{3} + 4 + c = \tfrac{28}{3} + c

c = 10 - \frac{28}{3} = \frac{30 - 28}{3} = \frac{2}{3}

Equation of curve:

y = \frac{2}{3}x^{3/2} + 2x^{1/2} + \frac{2}{3}

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Q3. The gradient of a curve at any point is $\dfrac{\mathrm{d}y}{\mathrm{d}x} = 2x - \dfrac{3}{x^2}$ . Given that the curve passes through $(3,\, 8)$ , find the equation of the curve.

<details><summary>Show answer</summary>

Rewrite and integrate:

y = \int \left(2x - 3x^{-2}\right)\mathrm{d}x = x^2 + 3x^{-1} + c = x^2 + \frac{3}{x} + c

Substitute $(3,\, 8)$ :

8 = 9 + 1 + c = 10 + c \implies c = -2

Equation of curve:

y = x^2 + \frac{3}{x} - 2

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Q4. A curve has $\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} = 12x - 6$ . The curve passes through $(2,\, 5)$ and has gradient $4$ at $x = 0$ . Find the equation of the curve.

<details><summary>Show answer</summary>

Integrate once:

\frac{\mathrm{d}y}{\mathrm{d}x} = 6x^2 - 6x + A

Use gradient condition at $x = 0$ :

4 = 0 - 0 + A \implies A = 4

So $\dfrac{\mathrm{d}y}{\mathrm{d}x} = 6x^2 - 6x + 4$ .

Integrate again:

y = 2x^3 - 3x^2 + 4x + B

Substitute $(2,\, 5)$ :

5 = 16 - 12 + 8 + B = 12 + B \implies B = -7

Equation of curve:

y = 2x^3 - 3x^2 + 4x - 7

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Q5. The gradient function of a curve is $\dfrac{\mathrm{d}y}{\mathrm{d}x} = kx^2 + 2$ , where $k$ is a constant. The curve passes through both $(0,\, 1)$ and $(3,\, 16)$ . Find the value of $k$ and hence the equation of the curve.

<details><summary>Show answer</summary>

Integrate:

y = \frac{k}{3}x^3 + 2x + c

Substitute $(0,\, 1)$ :

1 = 0 + 0 + c \implies c = 1

So $y = \dfrac{k}{3}x^3 + 2x + 1$ .

Substitute $(3,\, 16)$ :

16 = \frac{k}{3}(27) + 6 + 1 = 9k + 7

9k = 9 \implies k = 1

Equation of curve:

y = \frac{1}{3}x^3 + 2x + 1

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Connections

Prerequisite — Integration as the Reverse of Differentiation: The power rule $\displaystyle\int x^n\,\mathrm{d}x = \dfrac{x^{n+1}}{n+1} + c$ is used at every step here; ensure fluency with it, including negative and fractional indices.

Next subtopic — Definite Integration: Once limits are introduced, the constant of integration cancels between the bounds, but understanding why it cancels relies on knowing what $c$ represents here.

Connected topic — Kinematics (displacement, velocity, acceleration): In applied contexts, integrating $a(t)$ gives $v(t) + c$ , and an initial condition such as $v = u$ at $t = 0$ is used to evaluate $c$ — a direct application of exactly this skill.

Connected topic — Finding the Equation of a Curve from a Tangent/Normal: In coordinate geometry, you may be given a tangent line gradient and a point; combining this with integration to recover $y = F(x) + c$ links directly to this objective.

The Constant of Integration — Pure Mathematics 1 (9709)

Introduction

Core Concept

Key Formulae & Definitions

Worked Examples

Example 1 — Polynomial gradient function

Example 2 — Gradient function involving a negative power

Example 3 — Second derivative given

Common Mistakes & Examiner Pitfalls

Practice Questions

Connections

Figures

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