Use an initial condition to determine the integration constant
Turn a family of antiderivatives into the particular curve or accumulated quantity that satisfies the supplied information.
Section 1.8 · Integration
Pure Mathematics 1 (Paper 1) · AS & A Level Mathematics 9709, 2026–2027. Section 1.8 includes rational and linear-composite powers except exponent −1, simple improper integrals, areas and volumes about either coordinate axis, including regions away from the rotation axis. Written practice is self-checked.
Before you begin
- Find polynomial and linear-composite antiderivatives.
- Substitute a coordinate into an equation and solve for a constant.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
If y = x² + x + C passes through (1, −2), what equation determines C?
What you will learn
- Use a point or initial quantity to find C after integrating.
- Check both the derivative and the original condition on a valid interval.
A derivative alone does not determine the height
Curves y = F(x) + C for different C have the same derivative and differ by a vertical translation. A supplied point (a, b) selects the member satisfying b = F(a) + C. Integrate first; then use the original y value, not a gradient value, in the condition. A single valid point determines one additive constant on its connected interval.
If dy/dx = 2x + 1 and the curve passes through (1, −2), integration gives y = x² + x + C. Substitute −2 = 1 + 1 + C to obtain C = −4, so y = x² + x − 4. Differentiating reproduces 2x + 1, and substituting x = 1 reproduces −2. Those checks test two different parts of the solution.
Interpret initial information in its own units and domain
If a volume rate is given in litres per minute, its integral is a volume in litres and C is an initial volume, not a rate. For V′(t) = 6t + 4 and V(0) = 20, the result is V(t) = 3t² + 4t + 20. The total at t = 3 is 59 litres while the increase is 39 litres; the initial stock is included only in the total.
An initial point must lie in the domain of the derivative. For dy/dx = (2x + 1)^(−2), the point (0, 3) selects a curve on x > −1/2. It does not specify the constant on the separate interval x < −1/2, and the curve cannot be continued through the singular input by pretending the same rate remains defined. State the relevant interval when a condition chooses one branch.
Pause and explain
A volume has rate V′(t) = 6t + 4 and V(0) = 20. What is V(3)?
Worked example
A curve has dy/dx = 2x + 1 and passes through (1, −2). Find its equation.
Show the worked solution
- Integrate to get y = x² + x + C.
- Use the point: −2 = 1² + 1 + C, so C = −4.
- Check the result has derivative 2x + 1 and value −2 at x = 1.
Answer y = x² + x − 4.
Try a topic test
Choose Easy, Normal or Difficult. Save your answers and open worked review after finishing.
Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Find the curve with dy/dx = 3x² − 4x passing through (1, 2).
Give me a hint
Integrate both terms before substituting the point.
Compare my reasoning
- The curve has form y = x³ − 2x² + C.
- Substitute (1, 2): 2 = 1 − 2 + C, giving C = 3.
- The equation y = x³ − 2x² + 3 satisfies both the derivative and the point.
y = x³ − 2x² + 3.
Look for these in your work
- Find the full antiderivative before applying the condition.
- Check the point and derivative separately.
For x > −1/2, find y with dy/dx = (2x + 1)^(−2) and y(0) = 3. Then find y(1).
Give me a hint
The antiderivative contains the divisor 2 from the inner expression.
Compare my reasoning
- Integrate to y = −1/[2(2x + 1)] + C.
- At x = 0, 3 = −1/2 + C, so C = 7/2.
- At x = 1, y = −1/6 + 7/2 = 10/3; this branch remains on x > −1/2.
y = −1/[2(2x + 1)] + 7/2; y(1) = 10/3, for x > −1/2.
Look for these in your work
- Reverse the inner scale and the negative exponent correctly.
- Keep the interval containing the initial point.
A tank initially holds 20 litres. The supplied inflow is 6t + 4 litres per minute for 0 ≤ t ≤ 3, with no outflow. Find V(t), the volume after 3 minutes and the increase.
Give me a hint
Integrate the rate and use the initial stock to choose C.
Compare my reasoning
- Integration gives V(t) = 3t² + 4t + C.
- Since V(0) = 20, C = 20, and V(3) = 27 + 12 + 20 = 59 litres.
- The increase is V(3) − V(0) = 39 litres, distinct from the final total.
V(t) = 3t² + 4t + 20 litres; final volume 59 litres; increase 39 litres.
Look for these in your work
- Use the initial stock once in the total.
- Distinguish the rate, total and accumulated change with appropriate units.
Common mistakes
- Use a point's y coordinate as a function value, not as a derivative.
- Do not drop the initial quantity when a total is requested.
- One condition does not connect two intervals separated by an undefined input.
What can you explain now?
Find y if y′ = 4x and y(2) = 7.
Compare with the explanation
y = 2x² − 1.
Integrate to 2x² + C and use 7 = 8 + C, giving C = −1.
After trying it yourself, choose your next review. This is your self-assessment.
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Make it stick
Explain why a derivative and a point provide different information. Verify your particular solution against both.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.