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2011 Paper 3 Q4
D: 1700.0 B: 1516.0
integration inverse function inequality Young's inequality arcsin geometric proof curve sketching power function

The following result applies to any function \(\f\) which is continuous, has positive gradient and satisfies \(\f(0)=0\,\): \[ ab\le \int_0^a\f(x)\, \d x + \int_0^b \f^{-1}(y)\, \d y\,, \tag{\(*\)}\] where \(\f^{-1}\) denotes the inverse function of \(\f\), and \(a\ge 0\) and \(b\ge 0\).

  1. By considering the graph of \(y=\f(x)\), explain briefly why the inequality \((*)\) holds. In the case \(a>0\) and \(b>0\), state a condition on \(a\) and \(b\) under which equality holds.
  2. By taking \(\f(x) = x^{p-1}\) in \((*)\), where \(p>1\), show that if \(\displaystyle \frac 1p + \frac 1q =1\) then \[ ab \le \frac{a^p}p + \frac{b^q}q\,. \] Verify that equality holds under the condition you stated above.
  3. Show that, for \(0\le a \le \frac12 \pi\) and \(0\le b \le 1\), \[ ab \le b\arcsin b + \sqrt{1-b^2} \, - \cos a\,. \] Deduce that, for \(t\ge1\), \[ \arcsin (t^{-1}) \ge t - \sqrt{t^2-1}\,. \]

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1998 Paper 1 Q2
D: 1500.0 B: 1516.0
integration substitution change of variable definite integral algebraic manipulation power function

Show, by means of a suitable change of variable, or otherwise, that \[ \int_{0}^{\infty}\mathrm{f}((x^{2}+1)^{1/2}+x)\,{\mathrm d}x =\frac{1}{2} \int_{1}^{\infty}(1+t^{-2})\mathrm{f}(t)\,{\mathrm d}t. \] Hence, or otherwise, show that \[ \int_{0}^{\infty}((x^{2}+1)^{1/2}+x)^{-3}\,{\mathrm d}x =\frac{3}{8}. \]

Solution: \begin{align*} && t &= (x^2+1)^{1/2}+x \\ && 1&=t^2-2tx \\ && x &= \frac{t^2-1}{2t} = \frac12 \left (t - \frac1t\right) \\ && \frac{\d x}{\d t} &= \frac12 \left ( 1+ \frac{1}{t^2} \right) \\ \Rightarrow && \int_0^{\infty} f((x^2+1)^{1/2}+x) \d x &= \int_{t=1}^{t = \infty}f(t) \frac12(1 + t^{-2}) \d t\\ &&&= \frac12 \int_1^{\infty}(1+t^{-2})f(t) \d t \end{align*} \begin{align*} \int_{0}^{\infty}((x^{2}+1)^{1/2}+x)^{-3}\,{\mathrm d}x &= \frac12 \int_1^{\infty}(1+t^{-2})t^{-3} \d t \\ &= \frac12 \left [\frac{-1}{2}t^{-2}-\frac{1}{4}t^{-4} \right]_{1}^{\infty} \\ &= \frac12 \left ( \frac12 + \frac14\right) = \frac38 \end{align*}

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