Date | May 2018 | Marks available | 3 | Reference code | 18M.3ca.hl.TZ0.3 |
Level | HL only | Paper | Paper 3 Calculus | Time zone | TZ0 |
Command term | Find | Question number | 3 | Adapted from | N/A |
Question
Find the value of \(\int\limits_4^\infty {\frac{1}{{{x^3}}}{\text{d}}x} \).
Illustrate graphically the inequality \(\sum\limits_{n = 5}^\infty {\frac{1}{{{n^3}}}} < \int\limits_4^\infty {\frac{1}{{{x^3}}}{\text{d}}x} < \sum\limits_{n = 4}^\infty {\frac{1}{{{n^3}}}} \).
Hence write down a lower bound for \(\sum\limits_{n = 4}^\infty {\frac{1}{{{n^3}}}} \).
Find an upper bound for \(\sum\limits_{n = 4}^\infty {\frac{1}{{{n^3}}}} \).
Markscheme
\(\int\limits_4^\infty {\frac{1}{{{x^3}}}{\text{d}}x} = \mathop {{\text{lim}}}\limits_{R \to \infty } \int\limits_4^R {\frac{1}{{{x^3}}}{\text{d}}x} \) (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]