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8.4

Understand and use integration as the limit of a sum.

Draft — not yet indexed

Integration as the limit of a sum

Worked answers and methods for 8.4 on Edexcel A-level Maths 9MA0.

Explanation

  • Partition an interval into strips of width Δx\Delta x and form ∑f(xr)Δx\sum f(x_r)\Delta x; its limit as the maximum strip width tends to zero is the definite integral.
  • To recognise a limit, identify the factor playing the role of Δx\Delta x and rewrite the sampled expression in terms of an endpoint such as xr=a+rΔxx_r=a+r\Delta x.
  • A sum with factor 1/n1/n usually samples an interval of length 11; a different interval width introduces a corresponding scale factor.
  • A common error is to identify the integrand but omit the width factor, which changes the value of the limiting integral.

Worked example

Evaluate lim⁡n→∞1n∑r=1n(1+3rn)2\lim_{n\to\infty}\frac1n\sum_{r=1}^n\left(1+\frac{3r}{n}\right)^2 by expressing it as a definite integral.

  1. 1.With x=r/nx=r/n, the limit is ∫01(1+3x)2 dx\int_0^1(1+3x)^2\,\mathrm dx.
  2. 2.Equivalently, with u=1+3xu=1+3x, it is 13∫14u2 du=19(43−13)=7\frac13\int_1^4u^2\,\mathrm du=\frac19(4^3-1^3)=7.

Answer: 77

Common mistakes

  • Don't read i/ni/n as the strip width rather than as the sample-point position within the interval.
  • Don't match a Riemann sum to the wrong interval by overlooking the index increment and endpoint expression.

Exam tip

Rewrite the summand as a function of the sample point and identify both the strip width and integration limits.

Worked practice

Q1
Tier 1 · Easy

1.

Express ∫02x2 dx\int_0^2x^2\,\mathrm dx as a limit of a right-endpoint sum and evaluate it.

(4)

(Total for Question 1 is 4 marks)

Mark scheme

Mark scheme for question 1
QuestionSchemeMarks
1
  • lim⁡n→∞8n3∑r=1nr2=∫02x2 dx=83\lim_{n\to\infty}\frac8{n^3}\sum_{r=1}^nr^2=\int_0^2x^2\,\mathrm dx=\frac83
4
Notes
Use Δx=2/n\Delta x=2/n and xr=2r/nx_r=2r/n. Then ∑f(xr)Δx=∑r=1n(2r/n)2(2/n)=8n3∑r=1nr2\sum f(x_r)\Delta x=\sum_{r=1}^n(2r/n)^2(2/n)=\frac8{n^3}\sum_{r=1}^nr^2. Its limit is ∫02x2 dx=[x3/3]02=83\int_0^2x^2\,\mathrm dx=[x^3/3]_0^2=\frac83.

(4 marks)

Q2
Tier 2 · Standard

2.

Express ∫251x dx\displaystyle\int_2^5\dfrac1x\,\mathrm dx as the limit of a sum using nn equal subintervals and right-hand endpoints, and hence evaluate the limit.

(4)

(Total for Question 2 is 4 marks)

Mark scheme

Mark scheme for question 2
QuestionSchemeMarks
2
  • lim⁡n→∞3n∑r=1n12+3r/n=ln⁡ ⁣(52)\displaystyle\lim_{n\to\infty}\frac3n\sum_{r=1}^n\frac{1}{2+3r/n}=\ln\!\left(\dfrac52\right)
4
Notes
The interval length is 33, so each subinterval has width Δx=3/n\Delta x=3/n. Using right endpoints xr=2+3r/nx_r=2+3r/n, the Riemann sum is 3n∑r=1n12+3r/n\frac3n\sum_{r=1}^n\frac{1}{2+3r/n}. Its limit is the given integral, which evaluates to [ln⁡x]25=ln⁡5−ln⁡2=ln⁡(5/2)[\ln x]_2^5=\ln5-\ln2=\ln(5/2).

(4 marks)

Q3
Tier 3 · Hard

3.

Evaluate exactly lim⁡n→∞2n∑r=1nln⁡(1+2rn)\lim_{n\to\infty}\frac2n\sum_{r=1}^n\ln\left(1+\frac{2r}{n}\right).

(6)

(Total for Question 3 is 6 marks)

Mark scheme

Mark scheme for question 3
QuestionSchemeMarks
3
  • 3ln⁡3−23\ln3-2
6
Notes
Here Δx=2/n\Delta x=2/n and the right endpoints are xr=2r/nx_r=2r/n, so the limit is ∫02ln⁡(1+x) dx\int_0^2\ln(1+x)\,\mathrm dx. By substitution followed by integration by parts, an antiderivative is (1+x)ln⁡(1+x)−(1+x)(1+x)\ln(1+x)-(1+x). Evaluation from 00 to 22 gives (3ln⁡3−3)−(−1)=3ln⁡3−2(3\ln3-3)-(-1)=3\ln3-2.

(6 marks)

Q4
Tier 1 · Easy

4.

Evaluate lim⁡n→∞1n∑r=1n(2+rn)\displaystyle\lim_{n\to\infty}\frac1n\sum_{r=1}^n\left(2+\frac rn\right) by expressing it as a definite integral.

(2)

(Total for Question 4 is 2 marks)

Mark scheme

Mark scheme for question 4
QuestionSchemeMarks
4
  • 52\dfrac52
2
Notes
Here Δx=1/n\Delta x=1/n and the right endpoint is xr=r/nx_r=r/n, so the limit is ∫01(2+x) dx=[2x+x2/2]01=5/2\int_0^1(2+x)\,\mathrm dx=[2x+x^2/2]_0^1=5/2.

(2 marks)

Q5
Tier 2 · Standard

5.

Evaluate exactly lim⁡n→∞4n∑r=1n11+4r/n\displaystyle\lim_{n\to\infty}\frac4n\sum_{r=1}^n\dfrac{1}{1+4r/n}.

(4)

(Total for Question 5 is 4 marks)

Mark scheme

Mark scheme for question 5
QuestionSchemeMarks
5
  • ln⁡5\ln5
4
Notes
The strip width is 4/n4/n and the right endpoints on [0,4][0,4] are xr=4r/nx_r=4r/n. Hence the limit is ∫0411+x dx=[ln⁡(1+x)]04=ln⁡5\int_0^4\frac{1}{1+x}\,\mathrm dx=[\ln(1+x)]_0^4=\ln5.

(4 marks)

Q6
Tier 3 · Hard

6.

Evaluate exactly lim⁡n→∞2n∑r=1n3+2rn\displaystyle\lim_{n\to\infty}\frac2n\sum_{r=1}^n\sqrt{3+\frac{2r}{n}}.

(6)

(Total for Question 6 is 6 marks)

Mark scheme

Mark scheme for question 6
QuestionSchemeMarks
6
  • 23(55−33)\dfrac23(5\sqrt5-3\sqrt3)
6
Notes
Divide [1,3][1,3] into nn strips of width Δx=2/n\Delta x=2/n. The right endpoints are xr=1+2r/nx_r=1+2r/n, and 3+2r/n=2+xr\sqrt{3+2r/n}=\sqrt{2+x_r}. Hence the limit is ∫13x+2 dx=[23(x+2)3/2]13=23(55−33)\int_1^3\sqrt{x+2}\,\mathrm dx=[\frac23(x+2)^{3/2}]_1^3=\frac23(5\sqrt5-3\sqrt3).

(6 marks)

Q7
Tier 2 · Standard

7.

Evaluate lim⁡n→∞1n∑r=1n(2+rn)3\displaystyle\lim_{n\to\infty}\frac1n\sum_{r=1}^n\left(2+\frac{r}{n}\right)^3 by expressing it as a definite integral.

(4)

(Total for Question 7 is 4 marks)

Mark scheme

Mark scheme for question 7
QuestionSchemeMarks
7
  • 654\dfrac{65}{4}
4
Notes
Divide [0,1][0,1] into nn equal subintervals of width Δx=1/n\Delta x=1/n. Their right endpoints are xr=0+rΔx=r/nx_r=0+r\Delta x=r/n, so the sum is a right-endpoint Riemann sum for ∫01(2+x)3 dx\int_0^1(2+x)^3\,\mathrm dx. Therefore the limit is [(2+x)4/4]01=(81−16)/4=65/4[(2+x)^4/4]_0^1=(81-16)/4=65/4.

(4 marks)

Q8
Tier 3 · Hard

8.

Evaluate exactly lim⁡n→∞∑r=1n13n+2r\displaystyle\lim_{n\to\infty}\sum_{r=1}^n\frac{1}{3n+2r} by expressing it as a definite integral.

(5)

(Total for Question 8 is 5 marks)

Mark scheme

Mark scheme for question 8
QuestionSchemeMarks
8
  • 12ln⁡(53)\dfrac12\ln\left(\dfrac53\right)
5
Notes
Rewrite each term as 1/(3n+2r)=1n⋅13+2r/n1/(3n+2r)=\frac1n\cdot\frac{1}{3+2r/n}. The sum is therefore a right-endpoint sum of width 1/n1/n on [0,1][0,1], so its limit is ∫0113+2x dx=[12ln⁡(3+2x)]01=12ln⁡(5/3)\int_0^1\frac{1}{3+2x}\,\mathrm dx=[\frac12\ln(3+2x)]_0^1=\frac12\ln(5/3).

(5 marks)

Q9
Tier 3 · Hard

9.

Evaluate exactly lim⁡n→∞3n∑r=1ne 2+3r/n\displaystyle\lim_{n\to\infty}\frac3n\sum_{r=1}^n e^{\,2+3r/n} by expressing it as a definite integral.

(5)

(Total for Question 9 is 5 marks)

Mark scheme

Mark scheme for question 9
QuestionSchemeMarks
9
  • ∫25ex dx=e5−e2\displaystyle\int_2^5 e^x\,\mathrm dx=e^5-e^2
5
Notes
Here Δx=3/n\Delta x=3/n and the sample point is the right endpoint xr=2+rΔx=2+3r/nx_r=2+r\Delta x=2+3r/n. The given limit is therefore ∫25ex dx\int_2^5e^x\,\mathrm dx. Evaluating gives [ex]25=e5−e2[e^x]_2^5=e^5-e^2.

(5 marks)

Q10
Tier 3 · Hard

10.

Evaluate exactly lim⁡n→∞2π3n∑r=1nsin⁡(π6+2πr3n)\displaystyle\lim_{n\to\infty}\frac{2\pi}{3n}\sum_{r=1}^n\sin\left(\frac{\pi}{6}+\frac{2\pi r}{3n}\right) by expressing it as a definite integral.

(5)

(Total for Question 10 is 5 marks)

Mark scheme

Mark scheme for question 10
QuestionSchemeMarks
10
  • ∫π/65π/6sin⁡x dx=3\displaystyle\int_{\pi/6}^{5\pi/6}\sin x\,\mathrm dx=\sqrt3
5
Notes
The strip width is Δx=2π/(3n)\Delta x=2\pi/(3n) and the sample point is the right endpoint xr=π/6+rΔxx_r=\pi/6+r\Delta x. The upper limit is π/6+2π/3=5π/6\pi/6+2\pi/3=5\pi/6, so the given limit is ∫π/65π/6sin⁡x dx\int_{\pi/6}^{5\pi/6}\sin x\,\mathrm dx. Hence its value is [−cos⁡x]π/65π/6=3[-\cos x]_{\pi/6}^{5\pi/6}=\sqrt3.

(5 marks)

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