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2005 Paper 1 Q5
D: 1484.0 B: 1528.7
integration power rule logarithm limiting cases integration by parts approximation definite integral

  1. Evaluate the integral \[ \int_0^1 \l x + 1 \r ^{k-1} \; \mathrm{d}x \] in the cases \(k\ne0\) and \(k = 0\,\). Deduce that \(\displaystyle \frac{2^k - 1}{k} \approx \ln 2\) when \(k \approx 0\,\).
  2. Evaluate the integral \[ \int_0^1 x \l x + 1 \r ^m \; \mathrm{d}x \; \] in the different cases that arise according to the value of \(m\).

Solution:

  1. Case \(k \neq 0\): \begin{align*} && \int_0^1 (x+1)^{k-1} \d x &= \left [\frac{1}{k}(x+1)^k \right]_0^1 \\ &&&= \frac{2^k-1}{k} \\ \end{align*} Case \(k = 0\): \begin{align*} && \int_0^1 (x+1)^{k-1} \d x &= \int_0^1 (x+1)^{-1} \d x \\ &&&= \left [\ln(x+1) \right]_0^1 \\ &&&= \ln 2 \end{align*} Therefore for \(k \approx 0\), we must have both integrals being close to each other, since the function is nice on this interval, ie \(\frac{2^k-1}{k} \approx \ln 2\)
  2. Case \(m = 0\). \(I = \frac12\) Case \(m \neq 0, -1, -2\) \begin{align*} u = x+1, \d u = \d x && \int_0^1 x(x+1)^m \d x &= \int_{u=1}^{u=2} (u-1)u^m \d u \\ &&&=\left[ \frac{u^{m+2}}{m+2} - \frac{u^{m+1}}{m+1} \right]_1^2 \\ &&&= 2^{m+1}\left ( \frac{2}{m+2} - \frac1{m+1} \right) - \frac{1}{m+2} + \frac{1}{m+1} \\ &&&= 2^{m+1} \frac{m}{(m+1)(m+2)} + \frac{1}{(m+1)(m+2)} \\ &&&= \frac{m2^{m+1}+1}{(m+1)(m+2)} \\ \end{align*} Case \(m = -1\). \begin{align*} && \int_0^1 \frac{x}{x+1} \d x &= \int_0^1 1 - \frac{1}{x+1} \d x \\ &&&= 1 - \ln2 \\ \end{align*} Case \(m = -2\): \begin{align*} && \int_0^1 \frac{x}{(x+1)^2} \d x &= \int_0^1\frac{x+1-1}{(x+1)^2} \d x \\ &&&= \left [ \ln (x+1) +(1+x)^{-1} \right]_0^1 \\ &&&= \ln 2 + \frac12 - 1 \\ &&&= \ln 2 - \frac12 \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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