An indefinite integral reverses differentiation: ∫ f(x) dx = F(x) + c, where F′(x) = f(x). Because c can be any constant, one gradient function describes a whole family of curves. Given dy/dx and one point the curve passes through, integrate, then substitute the point to find c and the exact equation of the curve.
Indefinite integrals
- ∫ f(x) dx = F(x) + c, where F′(x) = f(x)
- To find a curve: integrate dy/dx, then substitute a known point to find c
- In context: integrate the rate, then use the initial condition
What does + c mean?
Integrating a function gives an antiderivative plus a constant. The constant appears because differentiating any constant gives zero, so infinitely many curves share the same gradient function. They are vertical translations of each other, and together they form a family of curves.
An indefinite integral has no limits and always carries + c. A definite integral, with limits, does not need it.
Finding a curve from its gradient function
Given dy/dx and a point on the curve, you can find the exact equation in two steps. Integrate dy/dx to get y = … + c, then substitute the coordinates of the point to work out c. Keep + c in your working right up to that substitution step; it is the whole reason the point is given.
Gradients with indices and surds
The same two steps work when the gradient contains negative or fractional powers. Rewrite roots and reciprocals as powers of x before integrating, then substitute the point using exact values. For example, at x = 9, x^(3/2) = 27.
Rewrite products before integrating
If the gradient is given as a product, such as x(x − 2), expand it first. There is no reverse product rule at this stage, and an unexpanded bracket cannot be integrated term by term.
Using an initial condition in context
Integration recovers a quantity from its rate of change. If you are given dP/dt and the value of P at a starting time, integrate and use that value to find the constant. The initial condition, often at t = 0, plays exactly the same role as the point on a curve.
Worked examples
Example 1. Given f′(x) = 3x² − 2, find the general form of f(x).
- Integrate term by term: 3x² gives x³, and −2 gives −2x.
- Add the constant.
Answer: f(x) = x³ − 2x + c.
Marks: M1 for integrating each term, A1 for the answer.
Example 2. A curve has dy/dx = 4x − 3 and passes through (1, 2). Find its equation.
- Integrate: y = 2x² − 3x + c.
- Substitute (1, 2): 2 = 2 − 3 + c, so c = 3.
Answer: y = 2x² − 3x + 3.
Marks: M1 for integrating, M1 for substituting the point, A1 for the equation.
Example 3. A curve has dy/dx = √x and passes through (9, 20). Find y.
- Integrate x^(1/2): y = ⅔x^(3/2) + c.
- At (9, 20): ⅔ × 27 + c = 20, so 18 + c = 20 and c = 2.
Answer: y = ⅔x^(3/2) + 2.
Marks: M1 for integrating x^(1/2), M1 for substituting, A1 for the equation.
Example 4. A curve has dy/dx = x(x − 2) and passes through (3, 4). Find y.
- Expand: x² − 2x.
- Integrate: y = x³/3 − x² + c.
- At (3, 4): 9 − 9 + c = 4, so c = 4.
Answer: y = x³/3 − x² + 4.
Marks: M1 for expanding, M1 for integrating and substituting, A1 for the equation.
Example 5. A population P changes at a rate dP/dt = 6t² − 2, and P = 5 when t = 0. Find P in terms of t.
- Integrate: P = 2t³ − 2t + c.
- At t = 0, P = 5, so c = 5.
Answer: P = 2t³ − 2t + 5.
Marks: M1 for integrating, A1 for the answer.
Common mistakes
- Dropping + c before substituting. The constant is what the point lets you find; keep it until then.
- Substituting the point into dy/dx. Substitute into the integrated equation for y, not into the gradient.
- Not expanding a product. Multiply out brackets before integrating.
- Using surds inaccurately. Evaluate powers such as 9^(3/2) = 27 exactly.
- Giving the family instead of the curve. If a point is given, the answer must have a numerical value of c.
How indefinite integrals are examined
Finding a curve from its gradient is a staple of the Pure 1 paper (WMA11) in Edexcel International A-Level, and of first-year Pure in Edexcel A-Level (9MA0). Questions often:
- give dy/dx and a point, and ask for the equation of the curve
- use fractional or negative indices in the gradient
- ask you to show another point lies on the curve
- use a rate of change and an initial value in context
You will find these multi-part questions, with mark schemes, in the Pure 1 integration exam questions. The integration worksheet and the integration revision engine give more practice.
This uses the integration power rule. Next come definite integrals, which use limits instead of + c.
Frequently asked questions
What is an indefinite integral?
An integral without limits. Its answer is a function plus an arbitrary constant, + c.
How do you find the equation of a curve from dy/dx?
Integrate dy/dx to get y in terms of x plus c, then substitute a known point on the curve to find c.
What is the difference between indefinite and definite integrals?
An indefinite integral gives a family of functions with + c. A definite integral has limits and gives a number.