Problems

1. Show that, for \(0\le x\le 1\), the largest value of \(\displaystyle \frac{x^6}{(x^2+1)^4}\) is \(\frac1{16}\).
2. Find constants \(A\), \(B\), \(C\) and \(D\) such that, for all \(x\), \[ \frac{1}{(x^2+1)^4}= \frac{\d \ }{\d x} \left( \frac{Ax^5+Bx^3+Cx}{(x^2+1)^3}\right) +\frac{Dx^6}{(x^2+1)^4}. \]
3. Hence, or otherwise, prove that \[ \frac{11}{24} \le \int_{0}^{1}\frac{1}{(x^{2}+1)^{4}}\, \d x \le \frac{11}{24} + \frac 1{16} \; . \]

The function \(\f\) satisfies \(0\leqslant\f(t)\leqslant K\) when \(0\leqslant t\leqslant x\). Explain by means of a sketch, or otherwise, why \[0\leqslant\int_{0}^{x} \f (t)\,{\mathrm d}t \leqslant Kx.\] By considering \(\displaystyle \int_{0}^{1}\frac{t}{n(n-t)}\,{\mathrm d}t\), or otherwise, show that, if \(n>1\), \[ 0\le \ln \left( \frac n{n-1}\right) -\frac 1n \le \frac 1 {n-1} - \frac 1n \] and deduce that \[ 0\le \ln N -\sum_{n=2}^N \frac1n \le 1. \] Deduce that as \(N\to \infty\) \[ \sum_{n=1}^N \frac1n \to\infty. \] Noting that \(2^{10}=1024\), show also that if \(N<10^{30}\) then \[ \sum_{n=1}^N \frac1n <101. \]

Show by means of a sketch, or otherwise, that if \(0\leqslant\mathrm{f}(y)\leqslant\mathrm{g}(y)\) for \(0\leqslant y\leqslant x\) then \[ 0\leqslant\int_{0}^{x}\mathrm{f}(y)\,\mathrm{d}y\leqslant\int_{0}^{x}\mathrm{g}(y)\,\mathrm{d}y. \] Starting from the inequality \(0\leqslant\cos y\leqslant1,\) or otherwise, prove that if \(0\leqslant x\leqslant\frac{1}{2}\pi\) then \(0\leqslant\sin x\leqslant x\) and \(\cos x\geqslant1-\frac{1}{2}x^{2}.\) Deduce that \[ \frac{1}{1800}\leqslant\int_{0}^{\frac{1}{10}}\frac{x}{(2+\cos x)^{2}}\,\mathrm{d}x\leqslant\frac{1}{1797}. \] Show further that if \(0\leqslant x\leqslant\frac{1}{2}\pi\) then \(\sin x\geqslant x-\frac{1}{6}x^{3}.\) Hence prove that \[ \frac{1}{3000}\leqslant\int_{0}^{\frac{1}{10}}\frac{x^{2}}{(1-x+\sin x)^{2}}\,\mathrm{d}x\leqslant\frac{2}{5999}. \]

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 \(1\).

Show Solution

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Solution:

+ - ↺

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.