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
a.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]
b.Show that the series \(\sum\limits_{r = 2}^\infty {\frac{{{{( – 1)}^r}}}{{r\ln r}}} \) is convergent but not absolutely convergent.[5]
▶️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]
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]
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 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.
Question
A function \(f\) is given by \(f(x) = \int_0^x {\ln (2 + \sin t){\text{d}}t} \).
a.Write down \(f'(x)\).[1]
b.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]
c.Hence obtain an expression for the derivative of \(\int_x^{{x^2}} {\ln (2 + \sin t){\text{d}}t} \) with respect to \(x\).[3]
▶️Answer/Explanation
Markscheme
\(\ln (2 + \sin x)\) A1
Note: Do not accept \(\ln (2 + \sin t)\).
[1 mark]
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]
\(\int_x^{{x^2}} {\ln (2 + \sin t){\text{d}}t = \int_0^{{x^2}} {\ln (2 + \sin t){\text{d}}t – \int_0^x {\ln (2 + \sin t){\text{d}}t} } } \) (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]
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.
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.
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.