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Date	May 2018	Marks available	4	Reference code	18M.3.AHL.TZ0.Hca_3
Level	Additional Higher Level	Paper	Paper 3	Time zone	Time zone 0
Command term	Show and Show that	Question number	Hca_3	Adapted from	N/A

Question

Find the value of $\int\limits_4^\infty {\frac{1}{{{x^3}}}{\text{d}}x}$ .

[3]

Illustrate graphically the inequality $\sum\limits_{n = 5}^\infty {\frac{1}{{{n^3}}}} < \int\limits_4^\infty {\frac{1}{{{x^3}}}{\text{d}}x} < \sum\limits_{n = 4}^\infty {\frac{1}{{{n^3}}}}$ .

[4]

Hence write down a lower bound for $\sum\limits_{n = 4}^\infty {\frac{1}{{{n^3}}}}$ .

[1]

Find an upper bound for $\sum\limits_{n = 4}^\infty {\frac{1}{{{n^3}}}}$ .

[3]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

$\int\limits_4^\infty {\frac{1}{{{x^3}}}{\text{d}}x} = \mathop {{\text{lim}}}\limits_{R \to \infty } \int\limits_4^R {\frac{1}{{{x^3}}}{\text{d}}x}$ (A1)

Note: The above A1 for using a limit can be awarded at any stage.
Condone the use of $\mathop {{\text{lim}}}\limits_{x \to \infty }$ .

Do not award this mark to candidates who use $\infty$ as the upper limit throughout.

= $\mathop {{\text{lim}}}\limits_{R \to \infty } \left[ { - \frac{1}{2}{x^{ - 2}}} \right]_4^R\left( { = \left[ { - \frac{1}{2}{x^{ - 2}}} \right]_4^\infty } \right)$ M1

$= \mathop {{\text{lim}}}\limits_{R \to \infty } \left( { - \frac{1}{2}\left( {{R^{ - 2}} - {4^{ - 2}}} \right)} \right)$

$= \frac{1}{{32}}$ A1

[3 marks]

A1A1A1A1

A1 for the curve
A1 for rectangles starting at $x = 4$
A1 for at least three upper rectangles
A1 for at least three lower rectangles

Note: Award A0A1 for two upper rectangles and two lower rectangles.

sum of areas of the lower rectangles < the area under the curve < the sum of the areas of the upper rectangles so

$\sum\limits_{n = 5}^\infty {\frac{1}{{{n^3}}}} < \int\limits_4^\infty {\frac{1}{{{x^3}}}{\text{d}}x} < \sum\limits_{n = 4}^\infty {\frac{1}{{{n^3}}}}$ AG

[4 marks]

a lower bound is $\frac{1}{{32}}$ A1

Note: Allow FT from part (a).

[1 mark]

METHOD 1

$\sum\limits_{n = 5}^\infty {\frac{1}{{{n^3}}}} < \frac{1}{{32}}$ (M1)

$\frac{1}{{64}} + \sum\limits_{n = 5}^\infty {\frac{1}{{{n^3}}}} = \frac{1}{{32}} + \frac{1}{{64}}$ (M1)

$\sum\limits_{n = 4}^\infty {\frac{1}{{{n^3}}}} < \frac{3}{{64}}$ , an upper bound A1

Note: Allow FT from part (a).

METHOD 2

changing the lower limit in the inequality in part (b) gives

$\sum\limits_{n = 4}^\infty {\frac{1}{{{n^3}}}} < \int\limits_3^\infty {\frac{1}{{{x^3}}}{\text{d}}x} \left( { < \sum\limits_{n = 3}^\infty {\frac{1}{{{n^3}}}} } \right)$ (A1)

$\sum\limits_{n = 4}^\infty {\frac{1}{{{n^3}}}} < \mathop {{\text{lim}}}\limits_{R \to \infty } \left[ { - \frac{1}{2}{x^{ - 2}}} \right]_3^R$ (M1)

$\sum\limits_{n = 4}^\infty {\frac{1}{{{n^3}}}} < \frac{1}{{18}}$ , an upper bound A1

Note: Condone candidates who do not use a limit.

[3 marks]

Examiners report

[N/A]

Syllabus sections

Topic 5—Calculus » SL 5.5—Integration introduction, areas between curve and x axis

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