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How do you evaluate an integral with an infinite limit or an unbounded integrand, and how do you decide whether it converges?

Improper integrals with an infinite limit of integration or an integrand that is unbounded at an endpoint, evaluated as a limit, and deciding whether such an integral converges or diverges.

A focused answer to the OCR A-Level Further Mathematics A content on improper integrals, covering integrals with an infinite limit of integration and integrals whose integrand is unbounded at an endpoint, evaluating each as a limit of a proper integral, and deciding whether the integral converges to a finite value or diverges.

Generated by Claude Opus 4.811 min answer

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  1. What this dot point is asking
  2. What makes an integral improper
  3. Infinite limits of integration
  4. Unbounded integrands
  5. Convergence and divergence
  6. Try this

What this dot point is asking

OCR wants you to recognise an improper integral, one with an infinite limit of integration or an integrand that becomes unbounded at an endpoint, rewrite it as the limit of a proper integral over a finite interval, evaluate that limit, and state clearly whether the integral converges (the limit is finite) or diverges (the limit is infinite or does not exist).

What makes an integral improper

A definite integral ∫abf(x) dx\int_a^b f(x)\,dx is "proper" when the interval is finite and ff is bounded on it. It is improper in two situations: when one or both limits are infinite, or when ff shoots off to infinity at an endpoint (for example 1x\dfrac{1}{\sqrt{x}} near x=0x = 0). In both cases the ordinary definition does not apply directly, so you reach for a limit.

Infinite limits of integration

For an infinite upper limit, replace ∞\infty by a finite variable tt, integrate normally, and let tβ†’βˆžt \to \infty. The integral converges if and only if that limit is finite.

Unbounded integrands

When the integrand is unbounded at an endpoint, replace that endpoint by a variable approaching it and take the limit. The procedure is identical, but the limit is now towards the awkward endpoint rather than infinity.

Convergence and divergence

The whole question is whether the limit is finite. A clean benchmark to memorise is the family ∫1∞xβˆ’p dx\int_1^{\infty} x^{-p}\,dx: it converges for p>1p > 1 and diverges for p≀1p \le 1. Near zero the behaviour flips: ∫01xβˆ’p dx\int_0^{1} x^{-p}\,dx converges for p<1p < 1 and diverges for pβ‰₯1p \ge 1. These let you predict the outcome before computing.

Improper integrals support the further-calculus topics: an infinite volume of revolution and the mean value of a function over an infinite range both rely on the same limiting idea.

Try this

Q1. Evaluate ∫2∞1x3 dx\displaystyle\int_2^{\infty}\dfrac{1}{x^3}\,dx. [3 marks]

  • Cue. [βˆ’12xβˆ’2]2t=βˆ’12t2+18β†’18\left[-\tfrac{1}{2}x^{-2}\right]_2^{t} = -\tfrac{1}{2t^2} + \tfrac{1}{8} \to \tfrac{1}{8} as tβ†’βˆžt \to \infty.

Q2. State, with a reason, whether ∫1∞1x dx\displaystyle\int_1^{\infty}\dfrac{1}{x}\,dx converges. [2 marks]

  • Cue. [ln⁑x]1t=ln⁑tβ†’βˆž\left[\ln x\right]_1^{t} = \ln t \to \infty, so it diverges (this is the p=1p = 1 borderline case).

Exam-style practice questions

Practice questions written in the style of OCR exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

OCR 20185 marksEvaluate ∫1∞1x2 dx\displaystyle\int_1^{\infty} \dfrac{1}{x^2}\,dx, or show that it does not converge.
Show worked answer β†’

Replace the infinite limit by tt and take a limit (M1): ∫1∞1x2 dx=lim⁑tβ†’βˆžβˆ«1txβˆ’2 dx\displaystyle\int_1^{\infty}\dfrac{1}{x^2}\,dx = \lim_{t \to \infty}\int_1^{t} x^{-2}\,dx.

Integrate (M1, A1): ∫1txβˆ’2 dx=[βˆ’xβˆ’1]1t=βˆ’1t+1\int_1^{t} x^{-2}\,dx = \left[-x^{-1}\right]_1^{t} = -\dfrac{1}{t} + 1.

Take the limit (M1): as tβ†’βˆžt \to \infty, βˆ’1tβ†’0-\dfrac{1}{t} \to 0, so the value is 11 (A1).

The integral converges to 11. Markers reward replacing ∞\infty with tt, integrating, and evaluating the limit.

OCR 20226 marksDetermine whether ∫011x dx\displaystyle\int_0^{1} \dfrac{1}{\sqrt{x}}\,dx converges, and if so find its value. Then state, with a reason, whether ∫1∞1x dx\displaystyle\int_1^{\infty}\dfrac{1}{\sqrt{x}}\,dx converges.
Show worked answer β†’

The integrand 1x\dfrac{1}{\sqrt{x}} is unbounded as xβ†’0+x \to 0^+, so the first integral is improper at the lower limit (M1). Replace 00 by aa and take a limit: lim⁑aβ†’0+∫a1xβˆ’1/2 dx=lim⁑aβ†’0+[2x1/2]a1=lim⁑aβ†’0+(2βˆ’2a)\displaystyle\lim_{a \to 0^+}\int_a^{1} x^{-1/2}\,dx = \lim_{a \to 0^+}\left[2x^{1/2}\right]_a^{1} = \lim_{a \to 0^+}(2 - 2\sqrt{a}) (M1, A1).

As a→0+a \to 0^+, 2a→02\sqrt{a} \to 0, so the value is 22: the integral converges to 22 (A1).

For the second integral, ∫1txβˆ’1/2 dx=[2x1/2]1t=2tβˆ’2\displaystyle\int_1^{t} x^{-1/2}\,dx = \left[2x^{1/2}\right]_1^{t} = 2\sqrt{t} - 2, which tends to infinity as tβ†’βˆžt \to \infty (M1). So this integral diverges (A1).

Markers reward identifying the source of impropriety, the limit setup, the value 22, and the reasoned divergence of the second.

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