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2024 Paper 2 Q8
D: 1500.0 B: 1500.0
arithmetic-geometric mean sequences and convergence integration by substitution inequality bounded monotone convergence improper integral

In this question, the following theorem may be used without proof. Let \(u_1, u_2, \ldots\) be a sequence of real numbers. If the sequence is

  • bounded above, so \(u_n \leqslant b\) for all \(n\), where \(b\) is some fixed number
  • and increasing, so \(u_n \leqslant u_{n+1}\) for all \(n\)
then there is a number \(L \leqslant b\) such that \(u_n \to L\) as \(n \to \infty\). For positive real numbers \(x\) and \(y\), define \(\mathrm{a}(x,y) = \frac{1}{2}(x+y)\) and \(\mathrm{g}(x,y) = \sqrt{xy}\). Let \(x_0\) and \(y_0\) be two positive real numbers with \(y_0 < x_0\) and define, for \(n \geqslant 0\) \[ x_{n+1} = \mathrm{a}(x_n, y_n)\,, \] \[ y_{n+1} = \mathrm{g}(x_n, y_n)\,. \]
  1. By considering \((\sqrt{x_n} - \sqrt{y_n})^2\), show that \(y_{n+1} < x_{n+1}\), for \(n \geqslant 0\). Show further that, for \(n \geqslant 0\)
    • \(x_{n+1} < x_n\)
    • \(y_n < y_{n+1}\).
    Deduce that there is a value \(M\) such that \(y_n \to M\) as \(n \to \infty\). Show that \(0 < x_{n+1} - y_{n+1} < \frac{1}{2}(x_n - y_n)\) and hence that \(x_n - y_n \to 0\) as \(n \to \infty\). Explain why \(x_n\) also tends to \(M\) as \(n \to \infty\).
  2. Let \[ \mathrm{I}(p,q) = \int_0^{\infty} \frac{1}{\sqrt{(p^2 + x^2)(q^2 + x^2)}}\,\mathrm{d}x, \] where \(p\) and \(q\) are positive real numbers with \(q < p\). Show, using the substitution \(t = \frac{1}{2}\!\left(x - \dfrac{pq}{x}\right)\) in the integral \[ \int_{-\infty}^{\infty} \frac{1}{\sqrt{\left(\frac{1}{4}(p+q)^2 + t^2\right)(pq + t^2)}}\,\mathrm{d}t, \] that \[ \mathrm{I}(p,q) = \mathrm{I}\!\left(\mathrm{a}(p,q),\, \mathrm{g}(p,q)\right). \] Hence evaluate \(\mathrm{I}(x_0, y_0)\) in terms of \(M\).

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2020 Paper 2 Q1
D: 1500.0 B: 1500.0
integration substitution definite integral algebraic manipulation improper integral

  1. Use the substitution \(x = \dfrac{1}{1-u}\), where \(0 < u < 1\), to find in terms of \(x\) the integral \[\int \frac{1}{x^{\frac{3}{2}}(x-1)^{\frac{1}{2}}}\,\mathrm{d}x \quad \text{(where } x > 1\text{).}\]
  2. Find in terms of \(x\) the integral \[\int \frac{1}{(x-2)^{\frac{3}{2}}(x+1)^{\frac{1}{2}}}\,\mathrm{d}x \quad \text{(where } x > 2\text{).}\]
  3. Show that \[\int_2^{\infty} \frac{1}{(x-1)(x-2)^{\frac{1}{2}}(3x-2)^{\frac{1}{2}}}\,\mathrm{d}x = \tfrac{1}{3}\pi.\]

Solution:

  1. \(\,\) \begin{align*} && x &= \frac1{1-u} \\ \Rightarrow && \d x &= \frac{1}{(1-u)^2} \d u \\ && I &= \int \frac{1}{x^{\frac32}(x-1)^{\frac12} } \d x \\ &&&= \int \frac1{(1-u)^{-\frac32}u^{\frac12}(1-u)^{-\frac12}} (1-u)^{-2} \d u \\ &&&= \int u^{-\frac12} \d u \\ &&&= 2\sqrt{u} + C \\ &&&= 2\sqrt{1-\frac{1}{x}} + C \end{align*}
  2. \(\,\) \begin{align*} && J &= \int \frac{1}{(x-2)^{\frac32}(x+1)^{\frac12}} \d x \\ y = x+1: &&&= \int \frac{1}{(y-3)^{\frac32}y^{\frac12}} \d y \\ y = 9(3-u)^{-1}: &&&= \int \frac1{\left (9(3-u)^{-1}-3 \right)^{\frac32}3(3-u)^{-\frac12}} \frac{9}{(3-u)^2} \d u \\ &&&= \int \frac1{\left (3u(3-u)^{-1} \right)^{\frac32}3(3-u)^{-\frac12}} \frac{9}{(3-u)^2} \d u \\ &&&= \frac{1}{\sqrt3} \int u^{-\frac32} \d u \\ &&&= -\frac2{\sqrt3} u^{-\frac12} + C \\ &&&= -\frac2{\sqrt3} \sqrt{\frac{y}{3(y-3)}} + C \\ &&&= -\frac2{3} \sqrt{\frac{x+1}{x-2}} + C \\ \end{align*}
  3. \(\,\) \begin{align*} && K &= \int_2^{\infty} \frac{1}{(x-1)(x-2)^{\frac12}(3x-2)^{\frac12}} \d x \\ y = x - 1: &&&=\int_{y=1}^{\infty} \frac{1}{y(y-1)^{\frac12}(3y+1)^{\frac12}} \d y \\ y = (1-u)^{-1}: &&&= \int_{u=0}^{u=1} \frac{1}{(1-u)^{-1}(u(1-u)^{-1})^{\frac12}((4-u)(1-u)^{-1})^{\frac12}} \frac{1}{(1-u)^2} \d u \\ &&&= \int_0^1 \frac{1}{u^{\frac12}(4-u)^{\frac12}} \d u \\ &&&= \int_0^1 \frac{1}{\sqrt{4-(u-2)^2}} \d u \\ &&&= \left [-\sin^{-1} \left ( \frac{2-u}{2} \right) \right]_0^1 \\ &&&= \sin^{-1} 1 - \sin^{-1} \tfrac12 \\ &&&= \frac{\pi}{2} - \frac{\pi}{6} = \frac{\pi}{3} \end{align*}

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2020 Paper 3 Q5
D: 1500.0 B: 1500.0
partial fractions integration polynomial identity differentiation improper integral limit evaluation

Show that for positive integer \(n\), \(x^n - y^n = (x-y)\displaystyle\sum_{r=1}^{n} x^{n-r} y^{r-1}\).

  1. Let \(\mathrm{F}\) be defined by \[ \mathrm{F}(x) = \frac{1}{x^n(x-k)} \quad \text{for } x \neq 0,\, k \] where \(n\) is a positive integer and \(k \neq 0\).
    1. Given that \[ \mathrm{F}(x) = \frac{A}{x-k} + \frac{\mathrm{f}(x)}{x^n}, \] where \(A\) is a constant and \(\mathrm{f}(x)\) is a polynomial, show that \[ \mathrm{f}(x) = \frac{1}{x-k}\left(1 - \left(\frac{x}{k}\right)^n\right). \] Deduce that \[ \mathrm{F}(x) = \frac{1}{k^n(x-k)} - \frac{1}{k}\sum_{r=1}^{n} \frac{1}{k^{n-r}x^r}. \]
    2. By differentiating \(x^n \mathrm{F}(x)\), prove that \[ \frac{1}{x^n(x-k)^2} = \frac{1}{k^n(x-k)^2} - \frac{n}{xk^n(x-k)} + \sum_{r=1}^{n} \frac{n-r}{k^{n+1-r}x^{r+1}}. \]
  2. Hence evaluate the limit of \[ \int_2^N \frac{1}{x^3(x-1)^2} \; \mathrm{d}x \] as \(N \to \infty\), justifying your answer.

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2016 Paper 3 Q1
D: 1700.0 B: 1500.0
integration reduction formula proof by induction trigonometric substitution improper integral completing the square binomial coefficient integration by parts

Let \[ \displaystyle I_n= \int_{-\infty}^\infty \frac 1 {(x^2+2ax+b)^n} \, \d x \] where \(a\) and \(b\) are constants with \(b > a^2\), and \(n\) is a positive integer.

  1. By using the substitution \(x + a = \sqrt{b- a^2} \, \tan u\,\), or otherwise, show that \[ I_1 = \dfrac \pi {\sqrt{b-a^2}}\, . \]
  2. Show that \(2n(b - a^2)\, I_{n+1} =(2n - 1) \, I_n\,\).
  3. Hence prove by induction that \[ I_n =\frac{\pi}{2^{2n-2}( b - a^2)^{n-\frac12}} \, \binom {2n-2}{n-1} \]

Solution:

  1. \(\,\) \begin{align*} && I_1 &= \int_{-\infty}^{\infty} \frac{1}{x^2+2ax+b} \d x \\ &&&= \int_{-\infty}^{\infty} \frac{1}{b-a^2 +(x+a)^2} \d x \\ &&&= \left [ \frac{1}{\sqrt{b-a^2}} \tan^{-1} \frac{x+a}{\sqrt{b-a^2}} \right]_{-\infty}^{\infty} \\ &&&= \frac{\pi}{\sqrt{b-a^2}} \end{align*}
  2. \(\,\) Here is the corrected LaTeX code for the second part, maintaining your exact styling and notation.
  3. \(\,\) \begin{align*} && I_{n} &= \int_{-\infty}^{\infty} \frac{1}{(x^2+2ax+b)^{n}} \d x \\ &&&= \left[ \frac{x}{(x^2+2ax+b)^n} \right]_{-\infty}^{\infty} - \int_{-\infty}^\infty x \cdot \frac{-n(2x+2a)}{(x^2+2ax+b)^{n+1}} \d x \\ &&&= 0 + n \int_{-\infty}^\infty \frac{2x^2+2ax}{(x^2+2ax+b)^{n+1}} \d x \\ &&&= n \int_{-\infty}^\infty \frac{2(x^2+2ax+b) - (2ax+2b)}{(x^2+2ax+b)^{n+1}} \d x \\ &&&= 2n I_n - n \int_{-\infty}^\infty \frac{2ax+2b}{(x^2+2ax+b)^{n+1}} \d x \\ &&&= 2n I_n - n \int_{-\infty}^\infty \frac{a(2x+2a) + 2(b-a^2)}{(x^2+2ax+b)^{n+1}} \d x \\ &&&= 2n I_n - n \int_{-\infty}^\infty \frac{a(2x+2a)}{(x^2+2ax+b)^{n+1}} \d x - 2n(b-a^2) I_{n+1} \\ &&&= 2n I_n - n \left[ \frac{-a}{n(x^2+2ax+b)^n} \right]_{-\infty}^\infty - 2n(b-a^2) I{n+1} \\ &&&= 2n I_n - 0 - 2n(b-a^2) I_{n+1} \\ \Rightarrow && 2n(b-a^2)I_{n+1} &= (2n-1)I_n \end{align*}
  4. \(\,\) \begin{align*} && I_{n+1} &= \frac{2n-1}{2n(b-a^2)} I_n \\ &&&= \frac{2n-1}{2n(b-a^2)} \cdot \frac{2n-3}{2(n-1)(b-a^2)} I_{n-1} \\ &&&= \frac{2n-1}{2n(b-a^2)} \cdot \frac{2n-3}{2(n-1)(b-a^2)} \cdots I_{1} \\ &&&= \frac{(2n-1)(2n-3) \cdots 1}{2n \cdot 2(n-1) \cdots 2 (b-a^2)^n} \frac{\pi}{\sqrt{b-a^2}} \\ &&&= \frac{(2n-1)(2n-3) \cdots 1}{2^n n!} \frac{\pi}{(b-a^2)^{n+\frac12}} \\ &&&= \frac{(2n-1)(2n-3) \cdots 1 \cdot 2n \cdot 2(n-1) \cdots 2}{2^{2n} n!n!} \frac{\pi}{(b-a^2)^{n+\frac12}} \\ &&&= \frac{(2n)!}{2^{2n}n!n!}\frac{\pi}{(b-a^2)^{n+\frac12}} \\ &&&= \frac{\pi}{2^{2n}(b-a^2)^{n+\frac12}} \binom{2n}{n} \\ \Rightarrow && I_n &= \frac{\pi}{2^{2n-2}(b-a^2)^{n-\frac12}} \binom{2n-2}{n-1} \\ \end{align*}

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2015 Paper 3 Q1
D: 1700.0 B: 1500.0
integration reduction formula substitution proof rational function improper integral factorial recurrence relation

  1. Let \[ I_n= \int_0^\infty \frac 1 {(1+u^2)^n}\, \d u \,, \] where \(n\) is a positive integer. Show that \[ I_n - I_{n+1} = \frac 1 {2n} I_n \] and deduce that \[ I_{n+1} = \frac{(2n)!\, \pi}{2^{2n+1}(n!)^2} \,. \]
  2. Let \[ J = \int_0^\infty \f\big( (x- x^{-1})^2\big ) \, \d x \,, \] where \(\f\) is any function for which the integral exists. Show that \[ J = \int_0^\infty x^{-2} \f\big( (x- x^{-1})^2\big) \, \d x \, = \frac12 \int_0^\infty (1 + x^{-2}) \f\big( (x- x^{-1})^2\big ) \, \d x \, = \int_0^\infty \f\big(u^2\big) \,\d u \,. \]
  3. Hence evaluate \[ \int_0^\infty \frac {x^{2n-2}}{(x^4-x^2+1)^n} \, \d x \,, \] where \(n\) is a positive integer.

Solution: \begin{align*} I_n - I_{n+1} &= \int_0^\infty \frac 1 {(1+u^2)^n}\, \d u - \int_0^\infty \frac 1 {(1+u^2)^{n+1}}\, \d u \\ &= \int_0^\infty \l \frac 1 {(1+u^2)^n}- \frac 1 {(1+u^2)^{n+1}} \r\, \d u \\ &= \int_0^\infty \frac {u^2} {(1+u^2)^{n+1}} \, \d u \\ &= \left [ u \frac{u}{(1+u^2)^{n+1}} \right]_0^{\infty} - \frac{-1}{2n}\int_0^{\infty} \frac{1}{(1+u^2)^n} \d u \tag{\(IBP: u = u, v' = \frac{u}{(1+u^2)^{n+1}}\)}\\ &= \frac{1}{2n} I_n \end{align*} \(\displaystyle I_1 = \int_0^{\infty} \frac{1}{1+u^2} \d u = \left [ \tan^{-1} u \right]_0^\infty = \frac{\pi}{2}\) as expected. We also have, \(I_{n+1} = \frac{2n(2n-1)}{2n \cdot 2n} I_n \), by rearranging the recurrence relation. Therefore, when we multiply out the top we will have \(2n!\) and the bottom we will have two factors of \(n!\) and two factors of \(2^n\) combined with the original \(\frac{\pi}{2}\) we get \[ I_{n+1} = \frac{(2n)! \pi}{2^{2n+1} (n!)^2} \] \begin{align*} J = \int_0^\infty f\big( (x- x^{-1})^2\big ) \, \d x &= \int_{u = \infty}^{u = 0} f((u^{-1}-u)^2)(-u^{-2} )\d u \tag{\(u = x^{-1}, \d u = -x^{-2} \d x\)} \\ &= \int^{u = \infty}_{u = 0} f((u^{-1}-u)^2)u^{-2} \d u \\ &= \int^{\infty}_{0} u^{-2}f((u-u^{-1})^2) \d u \\ \end{align*} Therefore adding the two forms for \(J\) we have \begin{align*} 2 J &= \int_0^\infty f\big( (x- x^{-1})^2\big ) \, \d x + \int_0^\infty x^{-2} f\big( (x- x^{-1})^2\big ) \, \d x \\ &= \int_0^\infty (1+x^{-2}) f\big( (x- x^{-1})^2\big ) \, \d x \end{align*} And letting \(u = x - x^{-1}\), we have \(\d u = (1 + x^{-2}) \d x\), and \(u\) runs from \(-\infty\) to \(\infty\) so we have: \begin{align*} \int_0^\infty (1+x^{-2}) f\big( (x- x^{-1})^2\big ) \, \d x &= \int_{-\infty}^\infty f(u^2) \, \d u \\ &=2 \int_{0}^\infty f(u^2) \, \d u \end{align*} Since both of these are \(2J\) we have the result we are after. Finally, \begin{align*} \int_0^\infty \frac {x^{2n-2}}{(x^4-x^2+1)^n} \, \d x &= \int_0^{\infty} \frac{x^{2n-2}}{x^{2n}(x^2-1+x^{-2})^n} \d x \\ &= \int_0^{\infty} \frac{x^{-2}}{((x-x^{-1})^2+1)^n} \d x \\ &= \int_0^{\infty} \frac{1}{(x^2+1)^n} \d x \tag{Where \(f(x) = (1+x^2)^{-n}\) in \(J\) integral} \\ &= I_n = \frac{(2n-2)! \pi}{2^{2n-1} ((n-1)!)^2} \end{align*}

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2009 Paper 3 Q4
D: 1700.0 B: 1500.0
integration Laplace transform improper integral substitution integration by parts exponential function trigonometric function transformation

For any given (suitable) function \(\f\), the Laplace transform of \(\f\) is the function \(\F\) defined by \[ \F(s) = \int_0^\infty \e^{-st}\f(t)\d t \quad \quad \, (s>0) \,. \]

  1. Show that the Laplace transform of \(\e^{-bt}\f(t)\), where \(b>0\), is \(\F(s+b)\).
  2. Show that the Laplace transform of \(\f(at)\), where \(a>0\), is \(a^{-1}\F(\frac s a)\,\).
  3. Show that the Laplace transform of \(\f'(t)\) is \(s\F(s) -\f(0)\,\).
  4. In the case \(\f(t)=\sin t\), show that \(\F(s)= \dfrac 1 {s^2+1}\,\).
Using only these four results, find the Laplace transform of \(\e^{-pt}\cos{qt}\,\), where \(p>0\) and \(q>0\).

Solution:

  1. \begin{align*} \mathcal{L}\{e^{-bt}f(t)\}(s) &= \int_0^{\infty}e^{-st}\{ e^{-bt}f(t) \} \d t \\ &= \int_0^{\infty} e^{-(s+b)t}f(t) \d t \\ &= F(s+b) \end{align*}
  2. \begin{align*} \mathcal{L}\{f(at)\}(s) &= \int_0^{\infty} e^{-st}f(at) \d t \\ &= \int_{u=0}^{\infty}e^{-s \frac{u}{a}} f\left(a \tfrac{u}{a}\right)\frac{1}{a} \d u \\ &= \int_0^{\infty}e^{-su/a}f(u) a^{-1} \d u \\ &= a^{-1} \int_0^{\infty} e^{-(s/a)u}f(u) \d u \\ &= a^{-1} F\left (\frac{s}{a} \right) \end{align*}
  3. \begin{align*} \mathcal{L}\{f'(t)\}(s) &= \int_0^{\infty} e^{-st}f'(t) \d t \\ &= \left [e^{-st} f(t) \right]_0^{\infty} - \int_0^{\infty} -s e^{-st} f(t) \d t\\ &= -f(0)+sF(s) \\ &= sF(s) - f(0) \end{align*}
  4. Since \(f''(t) = -f(t)\) we must have: \begin{align*} && -\mathcal{L}(f)&= \mathcal{L}(f'') \\ &&&= s\mathcal{L}(f') -f'(0) \\ &&&= s(s\mathcal{L}(f)-f(0)) - f'(0) \\ &&&= s^2\mathcal{L}(f) - 1 \\ \Rightarrow && (1+s^2) \mathcal{L}(f) &= 1 \\ \Rightarrow && F(s) &= \frac{1}{1+s^2} \end{align*}
\begin{align*} \mathcal{L}\{e^{-pt}\cos qt\}(s) &= \mathcal{L}\{\cos qt\}(s+p) \\ &= q^{-1}\mathcal{L}\{\cos t\}\left (\frac{s+p}{q} \right) \\ &= q^{-1}\mathcal{L}\{\sin'\}\left (\frac{s+p}{q} \right) \\ &= q^{-1} \left (\frac{s+p}{q} \right) \mathcal{L}\{\sin\} \left (\frac{s+p}{q} \right) - q^{-1}\sin \left (0\right) \\ &= \frac{s+p}{q^2} \frac{1}{1+\left (\frac{s+p}{q} \right)^2 } \\ &= \frac{s+p}{q^2+(s+p)^2} \end{align*}

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2004 Paper 3 Q1
D: 1700.0 B: 1603.9
integration hyperbolic functions inverse trigonometric functions partial fractions substitution improper integral logarithmic result definite integral

Show that \[ \int_0^a \frac{\sinh x}{2\cosh^2 x -1} \, \mathrm{d} x = \frac{1}{2 \sqrt{2}} \ln \l \frac{\sqrt{2}\cosh a -1}{\sqrt{2}\cosh a +1}\r + \frac{1}{2 \sqrt{2}} \ln \l \frac{\sqrt{2}+1}{\sqrt{2}-1}\r \] and find \[ \int_0^a \frac{\cosh x}{1+2\sinh^2 x} \, \mathrm{d} x \, . \] Hence show that \[ \int_0^\infty \frac{\cosh x - \sinh x}{1+2\sinh^2 x} \, \mathrm{d} x = \frac{\pi}{2\sqrt{2}} - \frac{1}{2 \sqrt{2}} \ln \l \frac{\sqrt{2}+1}{\sqrt{2}-1}\r \, . \] By substituting \(u = \e^x\) in this result, or otherwise, find \[ \int_1^\infty \frac{1}{1+u^4} \, \mathrm{d} u \, . \]

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2003 Paper 2 Q7
D: 1600.0 B: 1500.0
integration integration by parts sequences series inequality summation logarithmic improper integral geometric series convergence

Show that, if \(n>0\,\), then $$ \int_{e^{1/n}}^\infty\,{{\ln x} \over {x^{n+1}}}\,\d x = {2 \over {n^2\e}}\;. $$ You may assume that \(\ds \frac{\ln x} x \to 0\;\) as \(x\to\infty\,\). Explain why, if \(1 < a < b\,\), then $$ \int_b^\infty\,{{\ln x} \over {x^{n+1}}}\,\d x < \int_a^\infty\,{{\ln x} \over {x^{n+1}}}\,\d x\;. $$ Deduce that $$ \sum_{n=1}^{N}{1 \over n^2} < {\e \over 2}\int_{\e^{1/N}}^{\infty} \left({1-x^{-N}} \over {x^2-x}\right) \ln x\,\d x\;, $$ where \(N\,\) is any integer greater than \(1\).

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2002 Paper 3 Q1
D: 1700.0 B: 1500.0
integration integration by parts volume of revolution logarithmic function improper integral area under curve limits at infinity convergence

Find the area of the region between the curve \(\displaystyle y = {\ln x \over x}\,\) and the \(x\)-axis, for \(1 \le x \le a\). What happens to this area as \(a\) tends to infinity? Find the volume of the solid obtained when the region between the curve \(\displaystyle y = {\ln x \over x}\,\) and the \(x\)-axis, for \(1 \le x\le a\), is rotated through \(2 \pi\) radians about the \(x\)-axis. What happens to this volume as \(a\) tends to infinity?

Solution: \begin{align*} && \int_1^a \frac{\ln x}{x} \d x &= \left [ \ln x \cdot \ln x\right ]_1^a - \int_1^a \frac{\ln x}{x} \d x \\ \Rightarrow && \int_1^a \frac{\ln x}{x} \d x &= \frac12 \left ( \ln a \right) ^2 \\ && \int_1^\infty \frac{\ln x}{x} \d x &= \lim_{a \to \infty} \frac12 (\ln a)^2 \\ &&&= \infty \end{align*} \begin{align*} && \pi \int_1^a \left ( \frac{\ln x}{x} \right)^2 \d x &= \pi \int_{u=0}^{u=\ln a} \left ( \frac{u}{e^u} \right)^2 e^u \d u \\ &&&= \pi \int_0^{\ln a} u^2 e^{-u} \d u \\ &&&= \pi \left [-u^2e^{-u} \right]_0^{\ln a} +\pi \int_0^{\ln a} 2u e^{-u} \d u \\ &&&= -\frac{\pi}{a} (\ln a)^2 + \pi \left [-2u e^{-u} \right]_0^{\ln a} + \pi \int_0^{\ln a} e^{-u} \d u \\ &&&= -\frac{\pi}{a} (\ln a)^2- \frac{2 \pi}{a}\ln a+\pi \left (1 - \frac{1}{a} \right) \\ \\ && \pi \int_1^{\infty} \left ( \frac{\ln x}{x} \right)^2 \d x &= \lim_{a \to \infty} \left ( -\frac{\pi}{a} (\ln a)^2- \frac{2 \pi}{a}\ln a+\pi \left (1 - \frac{1}{a} \right) \right) \\ &&&= \pi \end{align*}

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1991 Paper 1 Q6
D: 1500.0 B: 1484.8
curve sketching stationary points rational function critical analysis integration discontinuity improper integral chain rule power rule

Criticise each step of the following arguments. You should correct the arguments where necessary and possible, and say (with justification) whether you think the conclusion are true even though the argument is incorrect.

  1. The function \(g\) defined by \[ \mathrm{g}(x)=\frac{2x^{3}+3}{x^{4}+4} \] satisfies \(\mathrm{g}'(x)=0\) only for \(x=0\) or \(x=\pm1.\) Hence the stationary values are given by \(x=0\), \(\mathrm{g}(x)=\frac{3}{4}\) and \(x=\pm1,\) \(\mathrm{g}(x)=1.\) Since \(\frac{3}{4}<1,\) there is a minimum at \(x=0\) and maxima at \(x=\pm1.\) Thus we must have \(\frac{3}{4}\leqslant\mathrm{g}(x)\leqslant1\) for all \(x\).
  2. \({\displaystyle \int(1-x)^{-3}\,\mathrm{d}x=-3(1-x)^{-4}}\quad\) and so \(\quad{\displaystyle \int_{-1}^{3}(1-x)^{-3}\,\mathrm{d}x=0.}\)

Solution:

  1. \begin{align*} && g(x) &= \frac{2x^3+3}{x^4+4} \\ \Rightarrow && g'(x) &= \frac{6x^2(x^4+4) - (2x^3+3)(4x^3)}{(x^4+4)^2} \\ &&&= \frac{-2x^6-12x^3+24x^2}{(x^4+4)} \\ &&&= \frac{-2x^2(x^4+6x-12)}{(x^4+4)} \end{align*} So \(g'(x)\) is not \(0\) for \(x = \pm 1\). We can also note that \(g(-1) = \frac1{5} \neq 1\) Even if the other turning point was \(1\), we would also need to check the behaviour as \(x \to \pm \infty\). We can also note that \(g(-1) = \frac{1}{5} < \frac34\) so the conclusion is also not true.
  2. There are several errors. \[ \int (1-x)^{-3} \d x = \underbrace{\frac{1}{4}}_{\text{correct constant}}(1-x)^{-4} + \underbrace{C}_{\text{constant of integration}} \] We cannot integrate through the asymptote at \(1\). There is a sense in which we could argue \(\displaystyle \int_{-1}^3 (1-x)^{-3} \d x = 0\), specifically using Cauchy principal value \begin{align*} \mathrm {p.v.} \int_{-1}^3 (1-x)^{-3} &=\lim_{\epsilon \to 0} \left [ \int_{-1}^{1-\epsilon} (1-x)^{-3} \d x+ \int_{1+\epsilon}^{3} (1-x)^{-3} \d x\right] \\ &=\lim_{\epsilon \to 0} \left [ \left[ \frac14 (1-x)^{-4}\right]_{-1}^{1-\epsilon}+ \left[ \frac14 (1-x)^{-4}\right]_{1+\epsilon}^3\right] \\ &=\lim_{\epsilon \to 0} \left [ \frac14 \epsilon^{-4}-\frac14 \frac1{2^4} + \frac14 \frac1{2^4} - \frac14 \epsilon^{-4} \right] \\ &= \lim_{\epsilon \to 0} 0 \\ &= 0 \end{align*} However, in many normal ways of treating this integral it would be undefined.

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1988 Paper 3 Q1
D: 1700.0 B: 1500.0
curve sketching asymptotes stationary points integration inequality improper integral bounding integrals exponential function

Sketch the graph of \[ y=\frac{x^{2}\mathrm{e}^{-x}}{1+x}, \] for \(-\infty< x< \infty.\) Show that the value of \[ \int_{0}^{\infty}\frac{x^{2}\mathrm{e}^{-x}}{1+x}\,\mathrm{d}x \] lies between \(0\) and \(1\).

Solution:

TikZ diagram
First notice the integrand is always positive over the range we are integrating, so the integral is greater than \(0\). Since \(\frac{x}{1+x} \leq 1\) for \(x \geq 0\) we can note that: \begin{align*} \int_0^{\infty} \frac{x^2e^{-x}}{1+x} \d x &=\int_0^{\infty} \frac{x}{1+x}xe^{-x} \d x \\ &< \int_0^\infty xe^{-x} \d x \\ &= \left [ -xe^{-x} \right]_0^{\infty} + \int_0^{\infty} e^{-x} \d x \\ &= 0 + 1 \\ &= 1 \end{align*} and so we are done.

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