Evaluate simple improper integrals using endpoint limits
Replace singular or infinite bounds with valid finite bounds and decide whether the accumulated value approaches a finite limit.
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
- Evaluate definite rational-power integrals on valid finite intervals.
- Interpret a limit approached from within the function's domain.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
For ε > 0, what is ∫ε¹ x^(−1/2) dx?
What you will learn
- Evaluate simple integrals with a singular endpoint or an infinite bound using limits.
- Distinguish a finite convergent integral from a divergent one and state modelling limits.
A singular endpoint must be approached through allowed inputs
For ∫₀¹ x^(−1/2) dx, the integrand is undefined at zero. First integrate from ε to 1 with ε > 0: [2√x]ε¹ = 2 − 2√ε. As ε approaches zero from above, this tends to the finite value 2. The improper integral converges even though its integrand becomes unbounded at the endpoint. The procedure uses a limit, not substitution of zero into 1/√x.
Not every endpoint singularity gives a finite integral. For ∫₀¹ x^(−3/2) dx, the truncated result is [−2/√x]ε¹ = 2/√ε − 2, which grows without bound as ε approaches zero. This integral diverges; writing a finite answer by discarding the endpoint term would be incorrect. An improper integral is finite only when the relevant endpoint limit exists as a finite value.
An infinite bound is a limit, not an input value
For ∫₁∞ x^(−2) dx, integrate first to a finite R > 1: [−1/x]₁ᴿ = 1 − 1/R. As R tends to infinity, this approaches 1. In contrast ∫₁ᴿ x^(−1/2) dx = 2√R − 2 grows without bound, so that integral to infinity diverges. Do not write infinity as an ordinary number in endpoint arithmetic.
An unbounded rate near an initial instant can have a finite integral in an idealised model. For q(t) = 2/√t litres per minute on 0 < t ≤ 9, the limit of the volume added from ε to 9 is 4√9 − 4√ε → 12 litres. This is a mathematical accumulation in the supplied model; it does not establish a physically achievable infinite initial flow. State both the valid time interval and that limitation.
Pause and explain
What happens to ∫ε¹ x^(−3/2) dx as ε → 0⁺?
Worked example
Evaluate the improper integral ∫₀¹ x^(−1/2) dx with a valid endpoint argument.
Show the worked solution
- Replace zero by ε > 0 and integrate on [ε, 1].
- The result is [2√x]ε¹ = 2 − 2√ε.
- Taking ε → 0⁺ gives the finite limit 2; the integral converges although its integrand is undefined at zero.
Answer The improper integral converges to 2.
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.
Evaluate ∫₀⁴ x^(−1/2) dx as an improper integral. Also evaluate the truncated integral from 0.01 to 4.
Give me a hint
Use an ε > 0 lower bound before taking a one-sided limit.
Compare my reasoning
- For ε > 0, [2√x]ε⁴ = 4 − 2√ε.
- As ε → 0⁺, this tends to 4, so the improper integral converges.
- At ε = 0.01, the truncated value is 4 − 0.2 = 3.8, approaching 4 from below.
Improper integral 4; truncated value 3.8.
Look for these in your work
- Show a valid one-sided endpoint limit.
- Distinguish a finite truncation from the limiting value.
Evaluate ∫₂∞ 3x^(−2) dx. For an upper cutoff R = 20, find the accumulated value and the omitted tail.
Give me a hint
Integrate to a finite R first; the remaining tail is the difference from the limiting total.
Compare my reasoning
- The truncated integral is [−3/x]₂ᴿ = 3/2 − 3/R.
- Taking R → ∞ gives the convergent total 3/2.
- At R = 20, the accumulated value is 27/20 = 1.35 and the omitted tail is 3/20 = 0.15.
Total 3/2; cutoff value 27/20; omitted tail 3/20.
Look for these in your work
- Use a finite upper bound before the infinite limit.
- Account for both the observed accumulation and the tail.
A supplied ideal flow model is q(t) = 2/√t litres per minute for 0 < t ≤ 9. Find its limiting added volume from the initial instant to 9 minutes and explain one physical limitation.
Give me a hint
Compute the integral from ε to 9, then separate mathematical convergence from a physical interpretation.
Compare my reasoning
- An antiderivative is 4√t, so the added volume from ε to 9 is 12 − 4√ε litres.
- As ε → 0⁺, the limit is 12 litres, a finite mathematical accumulation.
- The rate becomes unbounded near zero and is undefined at zero; a real pump does not supply an infinite initial flow, so the model needs a physical starting-time qualification.
Limiting added volume 12 litres; the model's unbounded initial rate is a physical limitation.
Look for these in your work
- Use the endpoint limit rather than substitute into the undefined rate.
- State the valid time interval and distinguish model behaviour from physical feasibility.
Common mistakes
- Never treat an undefined endpoint value as an ordinary finite input.
- State a finite limit when claiming convergence; not every unbounded integrand is integrable.
- A mathematically finite accumulation does not validate an infinite physical rate.
What can you explain now?
Compare ∫₁∞ x^(−2) dx and ∫₁∞ x^(−1/2) dx.
Compare with the explanation
The first converges to 1; the second diverges.
Their truncated values are 1 − 1/R → 1 and 2√R − 2 → ∞ as R grows.
After trying it yourself, choose your next review. This is your self-assessment.
Your review choice appears on Today. Sign in to sync it across devices.
Make it stick
For a singular endpoint and an infinite bound, write the finite truncated integral before its limit. Explain what would prove divergence.
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.