IB DP Maths Topic 9.4 Improper integrals of the type HL Paper 3

 

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Question

(a)     Using l’Hopital’s Rule, show that \(\mathop {\lim }\limits_{x \to \infty } x{{\text{e}}^{ – x}} = 0\) .

(b)     Determine \(\int_0^a {x{{\text{e}}^{ – x}}{\text{d}}x} \) .

(c)     Show that the integral \(\int_0^\infty  {x{{\text{e}}^{ – x}}{\text{d}}x} \) is convergent and find its value.

Answer/Explanation

Markscheme

(a)     \(\mathop {\lim }\limits_{x \to \infty } \frac{x}{{{{\text{e}}^x}}} = \mathop {\lim }\limits_{x \to \infty } \frac{1}{{{{\text{e}}^x}}}\)     M1A1

= 0     AG

[2 marks]

 

(b)     Using integration by parts     M1

\(\int_0^a {x{{\text{e}}^{ – x}}{\text{d}}x}  = \left[ { – x{{\text{e}}^{ – x}}} \right]_0^a + \int_0^a {{{\text{e}}^{ – x}}{\text{d}}x} \)     A1A1

\( = – a{{\text{e}}^{ – a}} – \left[ {{e^{ – x}}} \right]_0^a\)     A1

\( = 1 – a{{\text{e}}^{ – a}} – {{\text{e}}^{ – a}}\)     A1

[5 marks]

 

(c)     Since \({{\text{e}}^{ – a}}\) and \(a{{\text{e}}^{ – a}}\) are both convergent (to zero), the integral is convergent.     R1

Its value is 1.     A1

[2 marks]

Total [9 marks]

Examiners report

Most candidates made a reasonable attempt at (a). In (b), however, it was disappointing to note that some candidates were unable to use integration by parts to perform the integration. In (c), while many candidates obtained the correct value of the integral, proof of its convergence was often unconvincing.

Question

Find the exact value of \(\int_0^\infty  {\frac{{{\text{d}}x}}{{(x + 2)(2x + 1)}}} \).

Answer/Explanation

Markscheme

Let \(\frac{1}{{(x + 2)(2x + 1)}} = \frac{A}{{x + 2}} + \frac{B}{{2x + 1}} = \frac{{A(2x + 1) + B(x + 2)}}{{(x + 2)(2x + 1)}}\)     M1A1

\(x = – 2 \to A = – \frac{1}{3}\)     A1

\(x = – \frac{1}{2} \to B = \frac{2}{3}\)     A1     N3

\(I = \frac{1}{3}\int_0^h {\left[ {\frac{2}{{(2x + 1)}} – \frac{1}{{(x + 2)}}} \right]{\text{d}}x} \)     M1

\( = \frac{1}{3}\left[ {\ln (2x + 1) – \ln (x + 2)} \right]_0^h\)     A1

\( = \frac{1}{3}\left[ {\mathop {\lim }\limits_{h \to \infty } \left( {\ln \left( {\frac{{2h + 1}}{{h + 2}}} \right)} \right) – \ln \frac{1}{2}} \right]\)     A1

\( = \frac{1}{3}\left( {\ln 2 – \ln \frac{1}{2}} \right)\)     A1

\( = \frac{2}{3}\ln 2\)     A1

Note: If the logarithms are not combined in the third from last line the last three A1 marks cannot be awarded.

 

Total [9 marks]

Examiners report

Not a difficult question but combination of the logarithms obtained by integration was often replaced by a spurious argument with infinities to get an answer. \(\log (\infty + 1)\) was often seen.

Question

Find the set of values of k for which the improper integral \(\int_2^\infty {\frac{{{\text{d}}x}}{{x{{(\ln x)}^k}}}} \) converges.

[6]
a.

Show that the series \(\sum\limits_{r = 2}^\infty {\frac{{{{( – 1)}^r}}}{{r\ln r}}} \) is convergent but not absolutely convergent.

[5]
b.
Answer/Explanation

Markscheme

consider the limit as \(R \to \infty \) of the (proper) integral

\(\int_2^R {\frac{{{\text{d}}x}}{{x{{(\ln x)}^k}}}} \)     (M1)

substitute \(u = \ln x,{\text{ d}}u = \frac{1}{x}{\text{d}}x\)     (M1)

obtain \(\int_{\ln 2}^{\ln R} {\frac{1}{{{u^k}}}{\text{d}}u = \left[ { – \frac{1}{{k – 1}}\frac{1}{{{u^{k – 1}}}}} \right]_{\ln 2}^{\ln R}} \)     A1 

Note: Ignore incorrect limits or omission of limits at this stage.

or \([\ln u]_{\ln 2}^{\ln R}\) if k = 1     A1 

Note: Ignore incorrect limits or omission of limits at this stage.

because \(\ln R{\text{ }}({\text{and }}\ln \ln R) \to \infty {\text{ as }}R \to \infty \)     (M1)

converges in the limit if k > 1     A1

[6 marks]

a.

C: \({\text{terms}} \to 0{\text{ as }}r \to \infty \)     A1

\(\left| {{u_{r + 1}}} \right| < \left| {{u_r}} \right|\) for all r     A1

convergence by alternating series test     R1

AC: \({(x\ln x)^{ – 1}}\) is positive and decreasing on \([2,\,\infty )\)     A1

not absolutely convergent by integral test using part (a) for k = 1     R1

[5 marks]

b.

Examiners report

A good number of candidates were able to find the integral in part (a) although the vast majority did not consider separately the integral when k = 1. Many candidates did not explicitly set a limit for the integral to let this limit go to infinity in the anti – derivative and it seemed that some candidates were “substituting for infinity”. This did not always prevent candidates finding a correct final answer but the lack of good technique is a concern. In part (b) many candidates seemed to have some knowledge of the relevant test for convergence but this test was not always rigorously applied. In showing that the series was not absolutely convergent candidates were often not clear in showing that the function being tested had to meet a number of criteria and in so doing lost marks.

a.

A good number of candidates were able to find the integral in part (a) although the vast majority did not consider separately the integral when k = 1. Many candidates did not explicitly set a limit for the integral to let this limit go to infinity in the anti – derivative and it seemed that some candidates were “substituting for infinity”. This did not always prevent candidates finding a correct final answer but the lack of good technique is a concern. In part (b) many candidates seemed to have some knowledge of the relevant test for convergence but this test was not always rigorously applied. In showing that the series was not absolutely convergent candidates were often not clear in showing that the function being tested had to meet a number of criteria and in so doing lost marks.

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 \(a{\text{ (where }}a \in {\mathbb{R}^ + })\).

[3]
a.

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

[3]
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 \(L < \sum\limits_{n = 1}^k {\frac{1}{{{n^2}}}}  + \int_k^\infty  {\frac{1}{{{x^2}}}} {\text{d}}x\) .

[6]
c.

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

[2]
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.

[3]
e.
Answer/Explanation

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)

attempt to shade rectangles     M1

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)

attempt to shade rectangles     M1

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

Consider the functions \(f\) and \(g\) given by \(f(x) = \frac{{{{\text{e}}^x} + {{\text{e}}^{ – x}}}}{2}{\text{ and }}g(x) = \frac{{{{\text{e}}^x} – {{\text{e}}^{ – x}}}}{2}\).

Show that \(f'(x) = g(x)\) and \(g'(x) = f(x)\).

[2]
a.

Find the first three non-zero terms in the Maclaurin expansion of \(f(x)\).

[5]
b.

Hence find the value of \(\mathop {{\text{lim}}}\limits_{x \to 0} \frac{{1 – f(x)}}{{{x^2}}}\).

[3]
c.

Find the value of the improper integral \(\int_0^\infty  {\frac{{g(x)}}{{{{\left[ {f(x)} \right]}^2}}}{\text{d}}x} \).

[6]
d.
Answer/Explanation

Markscheme

any correct step before the given answer     A1AG

eg, \(f'(x) = \frac{{{{\left( {{{\text{e}}^x}} \right)}^\prime } + {{\left( {{{\text{e}}^{ – x}}} \right)}^\prime }}}{2} = \frac{{{{\text{e}}^x} – {{\text{e}}^{ – x}}}}{2} = g(x)\)

any correct step before the given answer     A1AG

eg, \(g'(x) = \frac{{{{\left( {{{\text{e}}^x}} \right)}^\prime } – {{\left( {{{\text{e}}^{ – x}}} \right)}^\prime }}}{2} = \frac{{{{\text{e}}^x} + {{\text{e}}^{ – x}}}}{2} = f(x)\)

[2 marks]

a.

METHOD 1

statement and attempted use of the general Maclaurin expansion formula     (M1)

\(f(0) = 1;{\text{ }}g(0) = 0\) (or equivalent in terms of derivative values)   A1A1

\(f(x) = 1 + \frac{{{x^2}}}{2} + \frac{{{x^4}}}{{24}}\) or \(f(x) = 1 + \frac{{{x^2}}}{{2!}} + \frac{{{x^4}}}{{4!}}\)     A1A1

METHOD 2

\({{\text{e}}^x} = 1 + x + \frac{{{x^2}}}{{2!}} + \frac{{{x^3}}}{{3!}} + \frac{{{x^4}}}{{4!}} +  \ldots \)     A1

\({{\text{e}}^{ – x}} = 1 – x + \frac{{{x^2}}}{{2!}} – \frac{{{x^3}}}{{3!}} + \frac{{{x^4}}}{{4!}} +  \ldots \)     A1

adding and dividing by 2     M1

\(f(x) = 1 + \frac{{{x^2}}}{2} + \frac{{{x^4}}}{{24}}\) or \(f(x) = 1 + \frac{{{x^2}}}{{2!}} + \frac{{{x^4}}}{{4!}}\)     A1A1

 

Notes: Accept 1, \(\frac{{{x^2}}}{2}\) and \(\frac{{{x^4}}}{{24}}\) or 1, \(\frac{{{x^2}}}{{2!}}\) and \(\frac{{{x^4}}}{{4!}}\).

     Award A1 if two correct terms are seen.

 

[5 marks]

b.

METHOD 1

attempted use of the Maclaurin expansion from (b)     M1

\(\mathop {{\text{lim}}}\limits_{x \to 0} \frac{{1 – f(x)}}{{{x^2}}} = \mathop {{\text{lim}}}\limits_{x \to 0} \frac{{1 – \left( {1 + \frac{{{x^2}}}{2} + \frac{{{x^4}}}{{24}} +  \ldots } \right)}}{{{x^2}}}\)

\(\mathop {{\text{lim}}}\limits_{x \to 0} \left( { – \frac{1}{2} – \frac{{{x^2}}}{{24}} –  \ldots } \right)\)     A1

\( =  – \frac{1}{2}\)     A1

METHOD 2

attempted use of L’Hôpital and result from (a)     M1

\(\mathop {{\text{lim}}}\limits_{x \to 0} \frac{{1 – f(x)}}{{{x^2}}} = \mathop {{\text{lim}}}\limits_{x \to 0} \frac{{ – g(x)}}{{2x}}\)

\(\mathop {{\text{lim}}}\limits_{x \to 0} \frac{{ – f(x)}}{2}\)     A1

\( =  – \frac{1}{2}\)     A1

[3 marks]

c.

METHOD 1

use of the substitution \(u = f(x)\) and \(\left( {{\text{d}}u = g(x){\text{d}}x} \right)\)     (M1)(A1)

attempt to integrate \(\int_1^\infty  {\frac{{{\text{d}}u}}{{{u^2}}}} \)     (M1)

obtain \(\left[ { – \frac{1}{u}} \right]_1^\infty \) or \(\left[ { – \frac{1}{{f(x)}}} \right]_0^\infty \)     A1

recognition of an improper integral by use of a limit or statement saying the integral converges     R1

obtain 1     A1     N0

METHOD 2

\(\int_0^\infty  {\frac{{\frac{{{{\text{e}}^x} – {{\text{e}}^{ – x}}}}{2}}}{{{{\left( {\frac{{{{\text{e}}^x} + {{\text{e}}^{ – x}}}}{2}} \right)}^2}}}{\text{d}}x = \int_0^\infty  {\frac{{2\left( {{{\text{e}}^x} – {{\text{e}}^{ – x}}} \right)}}{{{{\left( {{{\text{e}}^x} + {{\text{e}}^{ – x}}} \right)}^2}}}{\text{d}}x} } \)     (M1)

use of the substitution \(u = {{\text{e}}^x} + {{\text{e}}^{ – x}}\) and \(\left( {{\text{d}}u = {{\text{e}}^x} – {{\text{e}}^{ – x}}{\text{d}}x} \right)\)     (M1)

attempt to integrate \(\int_2^\infty  {\frac{{2{\text{d}}u}}{{{u^2}}}} \)     (M1)

obtain \(\left[ { – \frac{2}{u}} \right]_2^\infty \)     A1

recognition of an improper integral by use of a limit or statement saying the integral converges     R1

obtain 1     A1     N0

[6 marks]

d.

Examiners report

[N/A]

a.

[N/A]

b.

[N/A]

c.

[N/A]

d.

Question

A function \(f\) is given by \(f(x) = \int_0^x {\ln (2 + \sin t){\text{d}}t} \).

Write down \(f'(x)\).

[1]
a.

By differentiating \(f({x^2})\), obtain an expression for the derivative of \(\int_0^{{x^2}} {\ln (2 + \sin t){\text{d}}t} \) with respect to \(x\).

[3]
b.

Hence obtain an expression for the derivative of \(\int_x^{{x^2}} {\ln (2 + \sin t){\text{d}}t} \) with respect to \(x\).

[3]
c.
Answer/Explanation

Markscheme

\(\ln (2 + \sin x)\)    A1

Note:     Do not accept \(\ln (2 + \sin t)\).

[1 mark]

a.

attempt to use chain rule     (M1)

\(\frac{{\text{d}}}{{{\text{d}}x}}\left( {f({x^2})} \right) = 2xf'({x^2})\)    (A1)

\( = 2x\ln \left( {2 + \sin ({x^2})} \right)\)    A1

[3 marks]

b.

\(\int_x^{{x^2}} {\ln (2 + \sin t){\text{d}}t = \int_0^{{x^2}} {\ln (2 + \sin t){\text{d}}t – \int_0^x {\ln (2 + \sin t){\text{d}}t} } } \)    (M1)(A1)

\(\frac{{\text{d}}}{{{\text{d}}x}}\left( {\int_x^{{x^2}} {\ln (2 + \sin t){\text{d}}t} } \right) = 2x\ln \left( {2 + \sin ({x^2})} \right) – \ln (2 + \sin x)\)    A1

[3 marks]

c.

Examiners report

Many candidates answered this question well. Many others showed no knowledge of this part of the option; candidates that recognized the Fundamental Theorem of Calculus answered this question well. In general the scores were either very low or full marks.

a.

Many candidates answered this question well. Many others showed no knowledge of this part of the option; candidates that recognized the Fundamental Theorem of Calculus answered this question well. In general the scores were either very low or full marks.

b.

Many candidates answered this question well. Many others showed no knowledge of this part of the option; candidates that recognized the Fundamental Theorem of Calculus answered this question well. In general the scores were either very low or full marks.

c.

Question

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

[3]
a.

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

[4]
b.

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

[1]
c.

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

[3]
d.
Answer/Explanation

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.

Examiners report

[N/A]

a.

[N/A]

b.

[N/A]

c.

[N/A]

d.

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