# IB DP Maths Topic 9.4 The integral as a limit of a sum; lower and upper Riemann sums. HL Paper 3

## Question The diagram shows part of the graph of $$y = \frac{1}{{{x^3}}}$$ together with line segments parallel to the coordinate axes.

(a)     Using the diagram, show that $$\frac{1}{{{4^3}}} + \frac{1}{{{5^3}}} + \frac{1}{{{6^3}}} + … < \int_3^\infty {\frac{1}{{{x^3}}}{\text{d}}x < \frac{1}{{{3^3}}} + \frac{1}{{{4^3}}} + \frac{1}{{{5^3}}} + …}$$ .

(b)     Hence find upper and lower bounds for $$\sum\limits_{n = 1}^\infty {\frac{1}{{{n^3}}}}$$.

## Markscheme

(a)     The area under the curve is sandwiched between the sum of the areas of the lower rectangles and the upper rectangles.     M2

Therefore

$$1 \times \frac{1}{{{4^3}}} + 1 \times \frac{1}{{{5^3}}} + 1 \times \frac{1}{{{6^3}}} + … < \int_3^\infty {\frac{{{\text{d}}x}}{{{x^3}}} < 1 \times \frac{1}{{{3^3}}} + 1 \times \frac{1}{{{4^3}}} + 1 \times \frac{1}{{{5^3}}} + …}$$     A1

which leads to the printed result.

[3 marks]

(b)     We note first that

$$\int_3^\infty {\frac{{{\text{d}}x}}{{{x^3}}} = \left[ { – \frac{1}{{2{x^2}}}} \right]_3^\infty = \frac{1}{{18}}}$$     M1A1

Consider first

$$\sum\limits_{n = 1}^\infty {\frac{1}{{{n^3}}}} = 1 + \frac{1}{{{2^3}}} + \frac{1}{{{3^3}}} + \left( {\frac{1}{{{4^3}}} + \frac{1}{{{5^3}}} + \frac{1}{{{6^3}}} + …} \right)$$     M1A1

$$< 1 + \frac{1}{8} + \frac{1}{{27}} + \frac{1}{{18}}$$     M1A1

$$= \frac{{263}}{{216}}{\text{ (1.22)}}$$ (which is an upper bound)     A1

$$\sum\limits_{n = 1}^\infty {\frac{1}{{{n^3}}}} = 1 + \frac{1}{{{2^3}}} + \left( {\frac{1}{{{3^3}}} + \frac{1}{{{4^3}}} + \frac{1}{{{5^3}}} + …} \right)$$     M1A1

$$> 1 + \frac{1}{8} + \frac{1}{{18}}$$     M1A1

$$= \frac{{85}}{{72}}\left( {\frac{{255}}{{216}}} \right){\text{ (1.18)}}$$ (which is a lower bound)     A1

[12 marks]

Total [15 marks]

## Examiners report

Many candidates failed to give a convincing argument to establish the inequality. In (b), few candidates progressed beyond simply evaluating the integral.

## Question

Using the integral test, show that $$\sum\limits_{n = 1}^\infty {\frac{1}{{4{n^2} + 1}}}$$ is convergent.


a.

(i)     Show, by means of a diagram, that $$\sum\limits_{n = 1}^\infty {\frac{1}{{4{n^2} + 1}}} < \frac{1}{{4 \times {1^2} + 1}} + \int_1^\infty {\frac{1}{{4{x^2} + 1}}{\text{d}}x}$$.

(ii)     Hence find an upper bound for $$\sum\limits_{n = 1}^\infty {\frac{1}{{4{n^2} + 1}}}$$


b.

## Markscheme

$$\int {\frac{1}{{4{x^2} + 1}}{\text{d}}x = \frac{1}{2}\arctan 2x + k}$$     (M1)(A1)

Note: Do not penalize the absence of “+k”.

$$\int_1^\infty {\frac{1}{{4{x^2} + 1}}{\text{d}}x = \frac{1}{2}\mathop {\lim }\limits_{a \to \infty } } [\arctan 2x]_1^a$$     (M1)

Note: Accept $$\frac{1}{2}[\arctan 2x]_1^\infty$$.

$$= \frac{1}{2}\left( {\frac{\pi }{2} – \arctan 2} \right)\,\,\,\,\,( = 0.232)$$     A1

hence the series converges     AG

[4 marks]

a.

(i) A2

The shaded rectangles lie within the area below the graph so that $$\sum\limits_{n = 2}^\infty {\frac{1}{{4{n^2} + 1}}} < \int_1^\infty {\frac{1}{{4{x^2} + 1}}{\text{d}}x}$$. Adding the first term in the series, $$\frac{1}{{4 \times {1^2} + 1}}$$, gives $$\sum\limits_{n = 1}^\infty {\frac{1}{{4{n^2} + 1}}} < \frac{1}{{4 \times {1^2} + 1}} + \int_1^\infty {\frac{1}{{4{x^2} + 1}}{\text{d}}x}$$     R1AG

(ii)     upper bound $$= \frac{1}{5} + \frac{1}{2}\left( {\frac{\pi }{2} – \arctan 2} \right)\,\,\,\,\,( = 0.432)$$     A1

[4 marks]

b.

## Examiners report

This proved to be a hard question for most candidates. A number of fully correct answers to (a) were seen, but a significant number were unable to integrate $${\frac{1}{{4{x^2} + 1}}}$$ successfully. Part (b) was found the hardest by candidates with most candidates unable to draw a relevant diagram, without which the proof of the inequality was virtually impossible.

a.

This proved to be a hard question for most candidates. A number of fully correct answers to (a) were seen, but a significant number were unable to integrate $${\frac{1}{{4{x^2} + 1}}}$$ successfully. Part (b) was found the hardest by candidates with most candidates unable to draw a relevant diagram, without which the proof of the inequality was virtually impossible.

b.

## Question

Prove that $$\mathop {\lim }\limits_{H \to \infty } \int_a^H {\frac{1}{{{x^2}}}{\text{d}}x}$$ exists and find its value in terms of $$a{\text{ (where }}a \in {\mathbb{R}^ + })$$.


a.

Use the integral test to prove that $$\sum\limits_{n = 1}^\infty {\frac{1}{{{n^2}}}}$$ converges.


b.

Let $$\sum\limits_{n = 1}^\infty {\frac{1}{{{n^2}}}} = L$$ .

The diagram below shows the graph of $$y = \frac{1}{{{x^2}}}$$. (i)     Shade suitable regions on a copy of the diagram above and show that

$$\sum\limits_{n = 1}^k {\frac{1}{{{n^2}}}} + \int_{k + 1}^\infty {\frac{1}{{{x^2}}}} {\text{d}}x < L$$ .

(ii)     Similarly shade suitable regions on another copy of the diagram above and

show that $$L < \sum\limits_{n = 1}^k {\frac{1}{{{n^2}}}} + \int_k^\infty {\frac{1}{{{x^2}}}} {\text{d}}x$$ .


c.

Hence show that $$\sum\limits_{n = 1}^k {\frac{1}{{{n^2}}}} + \frac{1}{{k + 1}} < L < \sum\limits_{n = 1}^k {\frac{1}{{{n^2}}}} + \frac{1}{k}$$


d.

You are given that $$L = \frac{{{\pi ^2}}}{6}$$.

By taking k = 4 , use the upper bound and lower bound for L to find an upper bound and lower bound for $$\pi$$ . Give your bounds to three significant figures.


e.

## Markscheme

$$\mathop {\lim }\limits_{H \to \infty } \int_a^H {\frac{1}{{{x^2}}}{\text{d}}x} = \mathop {\lim }\limits_{H \to \infty } \left[ {\frac{{ – 1}}{x}} \right]_a^H$$     A1

$$\mathop {\lim }\limits_{H \to \infty } \left( {\frac{{ – 1}}{H} + \frac{1}{a}} \right)$$     A1

$$= \frac{1}{a}$$     A1

[3 marks]

a.

as $$\left\{ {\frac{1}{{{n^2}}}} \right\}$$ is a positive decreasing sequence we consider the function $$\frac{1}{{{x^2}}}$$

we look at $$\int_1^\infty {\frac{1}{{{x^2}}}} {\text{d}}x$$     M1

$$\int_1^\infty {\frac{1}{{{x^2}}}} {\text{d}}x = 1$$     A1

since this is finite (allow “limit exists” or equivalent statement)     R1

$$\sum\limits_{n = 1}^\infty {\frac{1}{{{n^2}}}}$$ converges     AG

[3 marks]

b.

(i) correct start and finish points for rectangles     A1

since the area shaded is less that the area of the required staircase we have     R1

$$\sum\limits_{n = 1}^k {\frac{1}{{{n^2}}}} + \int_{k + 1}^\infty {\frac{1}{{{x^2}}}} {\text{d}}x < L$$     AG

(ii) correct start and finish points for rectangles     A1

since the area shaded is greater that the area of the required staircase we have     R1

$$L < \sum\limits_{n = 1}^k {\frac{1}{{{n^2}}}} + \int_k^\infty {\frac{1}{{{x^2}}}} {\text{d}}x$$     AG

Note: Alternative shading and rearranging of the inequality is acceptable.

[6 marks]

c.

$$\int_{k + 1}^\infty {\frac{1}{{{x^2}}}} {\text{d}}x = \frac{1}{{k + 1}},{\text{ }}\int_k^\infty {\frac{1}{{{x^2}}}} {\text{d}}x = \frac{1}{k}$$     A1A1

$$\sum\limits_{n = 1}^k {\frac{1}{{{n^2}}}} + \frac{1}{{k + 1}} < L < \sum\limits_{n = 1}^k {\frac{1}{{{n^2}}}} + \frac{1}{k}$$     AG

[2 marks]

d.

$$\frac{{205}}{{144}} + \frac{1}{5} < \frac{{{\pi ^2}}}{6} < \frac{{205}}{{144}} + \frac{1}{4}{\text{ }}\left( {1.6236… < \frac{{{\pi ^2}}}{6} < 1.6736…} \right)$$     A1

$$\sqrt {6\left( {\frac{{205}}{{144}} + \frac{1}{5}} \right)} < \pi < \sqrt {6\left( {\frac{{205}}{{144}} + \frac{1}{4}} \right)}$$     (M1)

$$3.12 < \pi < 3.17$$     A1     N2

[3 marks]

e.

## Examiners report

Most candidates correctly obtained the result in part (a). Many then failed to realise that having obtained this result once it could then simply be stated when doing parts (b) and (d)

a.

Most candidates correctly obtained the result in part (a). Many then failed to realise that having obtained this result once it could then simply be stated when doing parts (b) and (d)

In part (b) the calculation of the integral as equal to 1 only scored 2 of the 3 marks. The final mark was for stating that ‘because the value of the integral is finite (or ‘the limit exists’ or an equivalent statement) then the series converges. Quite a few candidates left out this phrase.

b.

Most candidates correctly obtained the result in part (a). Many then failed to realise that having obtained this result once it could then simply be stated when doing parts (b) and (d)

Candidates found part (c) difficult. Very few drew the correct series of rectangles and some clearly had no idea of what was expected of them.

c.

Most candidates correctly obtained the result in part (a). Many then failed to realise that having obtained this result once it could then simply be stated when doing parts (b) and (d)

d.

Though part (e) could be done without doing any of the previous parts of the question many students were probably put off by the notation because only a minority attempted it.

e.

## Question Figure 1 Figure 2

Figure 1 shows part of the graph of $$y = \frac{1}{x}$$ together with line segments parallel to the coordinate axes.

(i)     By considering the areas of appropriate rectangles, show that

$\frac{{2a + 1}}{{a(a + 1)}} < \ln \left( {\frac{{a + 1}}{{a – 1}}} \right) < \frac{{2a – 1}}{{a(a – 1)}}.$

(ii)     Hence find lower and upper bounds for $$\ln (1.2)$$.


a.

An improved upper bound can be found by considering Figure 2 which again shows part of the graph of $$y = \frac{1}{x}$$.

(i)     By considering the areas of appropriate regions, show that

$\ln \left( {\frac{a}{{a – 1}}} \right) < \frac{{2a – 1}}{{2a(a – 1)}}.$

(ii)     Hence find an upper bound for $$\ln (1.2)$$.


b.

## Markscheme

(i)     the area under the curve between a – 1 and  a + 1

$$= \int_{a – 1}^{a + 1} {\frac{{{\text{d}}x}}{x}}$$     M1

$$= [\ln x]_{a – 1}^{a + 1}$$     A1

$$= \ln \left( {\frac{{a + 1}}{{a – 1}}} \right)$$     A1

lower sum $$= \frac{1}{a} + \frac{1}{{a + 1}}$$     M1A1

$$= \frac{{2a + 1}}{{a(a + 1)}}$$     AG

upper sum $$= \frac{1}{{a – 1}} + \frac{1}{a}$$     A1

$$= \frac{{2a – 1}}{{a(a – 1)}}$$     AG

it follows that

$$\frac{{2a + 1}}{{a(a + 1)}} < \ln \left( {\frac{{a + 1}}{{a – 1}}} \right) < \frac{{2a – 1}}{{a(a – 1)}}$$

because the area of the region under the curve lies between the areas of the regions defined by the lower and upper sums     R1

(ii)     putting

$$\left( {\frac{{a + 1}}{{a – 1}} = 1.2} \right) \Rightarrow a = 11$$     A1

therefore, $${\text{UB}} = \frac{{21}}{{110}}( = 0.191),{\text{ LB}} = \frac{{23}}{{132}}( = 0.174)$$     A1

[9 marks]

a.

(i)     the area under the curve between a – 1 and a

$$= \int_{a – 1}^a {\frac{{{\text{d}}x}}{x}}$$     A1

$$= [\ln x]_{a – 1}^a = \ln \left( {\frac{a}{{a – 1}}} \right)$$

attempt to find area of trapezium     M1

area of trapezoidal “upper sum” $$= \frac{1}{2}\left( {\frac{1}{{a – 1}} + \frac{1}{a}} \right)$$ or equivalent     A1

$$= \frac{{2a – 1}}{{2a(a – 1)}}$$

it follows that $$\ln \left( {\frac{a}{{a – 1}}} \right) < \frac{{2a – 1}}{{2a(a – 1)}}$$     AG

(ii)     putting

$$\left( {\frac{a}{{a – 1}} = 1.2} \right) \Rightarrow a = 6$$     A1

therefore, $${\text{UB}} = \frac{{11}}{{60}}( = 0.183)$$     A1

[5 marks]

b.

## Examiners report

Many candidates made progress with this problem. This was pleasing since whilst being relatively straightforward it was not a standard problem. There were still some candidates who did not use the definite integral correctly to find the area under the curve in part (a) and part (b). Also candidates should take care to show all the required working in a “show that” question, even when demonstrating familiar results. The ability to find upper and lower bounds was often well done in parts (a) (ii) and (b) (ii).

a.

Many candidates made progress with this problem. This was pleasing since whilst being relatively straightforward it was not a standard problem. There were still some candidates who did not use the definite integral correctly to find the area under the curve in part (a) and part (b). Also candidates should take care to show all the required working in a “show that” question, even when demonstrating familiar results. The ability to find upper and lower bounds was often well done in parts (a) (ii) and (b) (ii).

b.

## Question

Consider the function $$f(x) = \frac{1}{{1 + {x^2}}},{\text{ }}x \in \mathbb{R}$$.

Illustrate graphically the inequality, $$\frac{1}{5}\sum\limits_{r = 1}^5 {f\left( {\frac{r}{5}} \right) < \int_0^1 {f(x){\text{d}}x < \frac{1}{5}\sum\limits_{r = 0}^4 {f\left( {\frac{r}{5}} \right)} } }$$.


a.

Use the inequality in part (a) to find a lower and upper bound for $$\pi$$.


b.

Show that $$\sum\limits_{r = 0}^{n – 1} {{{( – 1)}^r}{x^{2r}} = \frac{{1 + {{( – 1)}^{n – 1}}{x^{2n}}}}{{1 + {x^2}}}}$$.


c.

Hence show that $$\pi = 4\left( {\sum\limits_{r = 0}^{n – 1} {\frac{{{{( – 1)}^r}}}{{2r + 1}} – {{( – 1)}^{n – 1}}\int_0^1 {\frac{{{x^{2n}}}}{{1 + {x^2}}}{\text{d}}x} } } \right)$$.


d.

## Markscheme A1A1A1

A1 for upper rectangles, A1 for lower rectangles, A1 for curve in between with $$0 \le x \le 1$$

hence $$\frac{1}{5}\sum\limits_{r = 1}^5 {f\left( {\frac{r}{5}} \right) < \int_0^1 {f(x){\text{d}}x < \frac{1}{5}\sum\limits_{r = 0}^4 {f\left( {\frac{r}{5}} \right)} } }$$     AG

[3 marks]

a.

attempting to integrate from $$0$$ to $$1$$     (M1)

$$\int_0^1 {f(x){\text{d}}x = [\arctan x]_0^1}$$

$$= \frac{\pi }{4}$$     A1

attempt to evaluate either summation     (M1)

$$\frac{1}{5}\sum\limits_{r = 1}^5 {f\left( {\frac{r}{5}} \right) < \frac{\pi }{4} < \frac{1}{5}\sum\limits_{r = 0}^4 {f\left( {\frac{r}{5}} \right)} }$$

hence $$\frac{4}{5}\sum\limits_{r = 1}^5 {f\left( {\frac{r}{5}} \right) < \pi < \frac{4}{5}\sum\limits_{r = 0}^4 {f\left( {\frac{r}{5}} \right)} }$$

so $$2.93 < \pi < 3.33$$     A1A1

Note:     Accept any answers that round to $$2.9$$ and $$3.3$$.

[5 marks]

b.

EITHER

recognise $$\sum\limits_{r = 0}^{n – 1} {{{( – 1)}^r}{x^{2r}}}$$ as a geometric series with $$r = – {x^2}$$     M1

sum of $$n$$ terms is $$\frac{{1 – {{( – {x^2})}^n}}}{{1 – – {x^2}}} = \frac{{1 + {{( – 1)}^{n – 1}}{x^{2n}}}}{{1 + {x^2}}}$$     M1AG

OR

$$\sum\limits_{r = 0}^{n – 1} {{{( – 1)}^r}(1 + {x^2}){x^{2r}} = (1 + {x^2}){x^0} – (1 + {x^2}){x^2} + (1 + {x^2}){x^4} + \ldots }$$

$$+ {( – 1)^{n – 1}}(1 + {x^2}){x^{2n – 2}}$$     M1

cancelling out middle terms     M1

$$= 1 + {( – 1)^{n – 1}}{x^{2n}}$$     AG

[2 marks]

c.

$$\sum\limits_{r = 0}^{n – 1} {{{( – 1)}^r}{x^{2r}} = \frac{1}{{1 + {x^2}}} + {{( – 1)}^{n – 1}}\frac{{{x^{2n}}}}{{1 + {x^2}}}}$$

integrating from $$0$$ to $$1$$     M1

$$\left[ {\sum\limits_{r = 0}^{n – 1} {{{( – 1)}^r}\frac{{{x^{2r + 1}}}}{{2r + 1}}} } \right]_0^1 = \int_0^1 {f(x){\text{d}}x + {{( – 1)}^{n – 1}}\int_0^1 {\frac{{{x^{2n}}}}{{1 + {x^2}}}{\text{d}}x} }$$     A1A1

$$\int_0^1 {f(x){\text{d}}x = \frac{\pi }{4}}$$     A1

so $$\pi = 4\left( {\sum\limits_{r = 0}^{n – 1} {\frac{{{{( – 1)}^r}}}{{2r + 1}} – {{( – 1)}^{n – 1}}\int_0^1 {\frac{{{x^{2n}}}}{{1 + {x^2}}}{\text{d}}x} } } \right)$$     AG

[4 marks]

Total [14 marks]

d.

[N/A]

a.

[N/A]

b.

[N/A]

c.

[N/A]

d.

## Question

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


a.

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


b.

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


c.

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


d.

## 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]

a. 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]

b.

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

Note: Allow FT from part (a).

[1 mark]

c.

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]

d.

[N/A]

a.

[N/A]

b.

[N/A]

c.

[N/A]

d.