Evaluate definite integrals and interpret signed accumulation
Use upper minus lower antiderivative values, track orientation and distinguish net change from total area or distance.
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
- Integrate polynomial and rational-power expressions.
- Substitute and subtract signed endpoint values accurately.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
An antiderivative has F(3) = 0 and F(1) = −2. What is ∫₁³ F′(x) dx?
What you will learn
- Evaluate definite integrals using an antiderivative at both bounds.
- Explain reversed bounds, cancellation of C and signed accumulation.
Subtract the lower endpoint value from the upper
For a continuous integrand f on [a, b] with antiderivative F, ∫ₐᵇ f(x) dx = F(b) − F(a). Substitute each complete bound into the complete antiderivative and subtract in that order. A constant C cancels, so a definite integral is a value rather than another family of functions. When a bound is singular or infinite, a limit is required instead; that is an improper integral, covered next.
For ∫₁³(2x − 3) dx, take F(x) = x² − 3x. Then F(3) = 0 and F(1) = −2, so the integral is 0 − (−2) = 2. Reversing the bounds gives −2. If both bounds are the same allowed input, the integral is zero. Keep exact fractions and roots until the final requested rounding.
The value measures net accumulation
Positive integrand values contribute positively and negative values contribute negatively. Therefore a definite integral can be negative or zero despite a graph enclosing a non-zero total area. For 2x − 3 from 1 to 3, the graph crosses at x = 3/2: the negative contribution is −1/4 and the positive contribution is 9/4, giving net 2. The total geometric area is their magnitudes added, 5/2, a different request.
For a signed velocity v(t), ∫v(t) dt is displacement. Total distance requires the speed |v(t)| and usually a split where velocity changes sign. Similarly a net flow model gives a change in stored volume; it does not by itself give the final volume unless the initial stock is known. Explain what the integrand means before attaching units to the result.
Pause and explain
A definite integral of signed velocity over a time interval gives which quantity?
Worked example
Evaluate ∫₁³(2x − 3) dx and the integral with its bounds reversed.
Show the worked solution
- An antiderivative is F(x) = x² − 3x.
- F(3) − F(1) = 0 − (−2) = 2.
- Reversing the endpoints changes the sign, giving ∫₃¹(2x − 3) dx = −2.
Answer Forward integral 2; reversed integral −2.
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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.
Evaluate ∫₀²(3x² − 2x + 1) dx, then reverse the bounds.
Give me a hint
Use x³ − x² + x as the antiderivative and evaluate both ends.
Compare my reasoning
- The antiderivative is F(x) = x³ − x² + x.
- F(2) − F(0) = 8 − 4 + 2 − 0 = 6.
- The reversed integral is −6 because endpoint subtraction is reversed.
∫₀² = 6; ∫₂⁰ = −6 for the stated integrand.
Look for these in your work
- Evaluate the whole antiderivative at each bound.
- Change the sign when reversing the bounds.
Evaluate ∫₁⁴ 1/√x dx exactly and explain why C is absent from the result.
Give me a hint
Use the power −1/2 and subtract the same antiderivative family at the two bounds.
Compare my reasoning
- An antiderivative is 2√x + C.
- Endpoint subtraction is (4 + C) − (2 + C) = 2.
- The common constant cancels; the definite integral is the number 2.
2; the common integration constant cancels.
Look for these in your work
- Use exponent −1/2 and its correct antiderivative.
- Explain cancellation rather than leave a final + C.
A particle has signed velocity v(t) = 2t − 4 m/s for 0 ≤ t ≤ 3. Find its displacement and interpret the sign.
Give me a hint
Integrate the signed velocity, not its absolute value, for net position change.
Compare my reasoning
- An antiderivative of velocity is t² − 4t.
- The displacement is [t² − 4t]₀³ = 9 − 12 = −3 m.
- The final position is 3 m below the initial position in the chosen positive direction; this result is not total distance.
Displacement −3 m; net motion 3 m in the negative direction.
Look for these in your work
- Retain negative contributions from signed velocity.
- Interpret a position change without claiming total distance.
Common mistakes
- Use F(upper) − F(lower), including parentheses around negative values.
- A definite integral has no remaining arbitrary constant.
- A signed integral is not automatically total geometric area or distance.
What can you explain now?
Evaluate ∫₁⁴ x^(−1/2) dx.
Compare with the explanation
2.
The antiderivative 2√x gives 2√4 − 2√1 = 4 − 2 = 2; neither endpoint is singular.
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Make it stick
Describe upper minus lower endpoint subtraction, then give an example where net accumulation differs from total magnitude.
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.