Problem 40:  Estimating the value of an integral ($✓$ $✓$) 2000 Paper I

(i)
Show that, for $0\le x\le 1$, the largest value of $\phantom{\rule{1em}{0ex}}\frac{{x}^{6}}{{\left({x}^{2}+1\right)}^{4}}\phantom{\rule{1em}{0ex}}$ is $\phantom{\rule{1em}{0ex}}\frac{1}{16}\phantom{\rule{0.3em}{0ex}}$.

What is the smallest value?

(ii)
Find constants $A\phantom{\rule{0.3em}{0ex}}$, $B\phantom{\rule{0.3em}{0ex}}$, $C\phantom{\rule{0.3em}{0ex}}$ and $D$ such that, for all $x\phantom{\rule{0.3em}{0ex}}$,
$\frac{1}{{\left({x}^{2}+1\right)}^{4}}\equiv \frac{\mathrm{d}\phantom{\rule{1em}{0ex}}}{\mathrm{d}x}\left(\frac{A{x}^{5}+B{x}^{3}+Cx}{{\left({x}^{2}+1\right)}^{3}}\right)+\frac{D{x}^{6}}{{\left({x}^{2}+1\right)}^{4}}\phantom{\rule{2.77695pt}{0ex}}.$

(iii)
Hence, or otherwise, prove that
$\frac{11}{24}\le {\int }_{0}^{1}\frac{1}{{\left({x}^{2}+1\right)}^{4}}\phantom{\rule{0.3em}{0ex}}\mathrm{d}x\le \frac{11}{24}+\frac{1}{16}\phantom{\rule{2.77695pt}{0ex}}.$

You should think about part (i) graphically, though it is not necessary to draw the graph: just set about it as if you were going to (starting point, ﬁnishing point, turning points, etc).

The equivalence sign in part (ii) indicates an equality that holds for all $x$ — you are not being asked to solve the equation for $x$.

For part (iii), you need to know that inequalities can be integrated: this is ‘obvious’ if you think about integration in terms of area, though a formal proof requires a formal deﬁnition of integration and this is the sort of thing you would do in a ﬁrst university course in mathematical analysis.

Solution to problem 40

(i) Let $\phantom{\rule{1em}{0ex}}\mathrm{f}\phantom{\rule{1.00006pt}{0ex}}\left(x\right)=\frac{{x}^{6}}{{\left({x}^{2}+1\right)}^{4}}\phantom{\rule{1em}{0ex}}.$ Then

${\mathrm{f}\phantom{\rule{1.00006pt}{0ex}}}^{\prime }\left(x\right)=\frac{6{x}^{5}}{{\left({x}^{2}+1\right)}^{4}}-\frac{8{x}^{7}}{{\left({x}^{2}+1\right)}^{5}}=\frac{2\left(3-{x}^{2}\right){x}^{5}}{{\left({x}^{2}+1\right)}^{5}}$

which is positive for $0<{x}^{2}<3$. Therefore, $\mathrm{f}\phantom{\rule{1.00006pt}{0ex}}\left(x\right)$ increases in value from $0$ at $x=0$ to $\frac{1}{16}$ at $x=1$.

(ii) Doing the differentiation gives

$\frac{1}{{\left({x}^{2}+1\right)}^{4}}\equiv \frac{5A{x}^{4}+3B{x}^{2}+C}{{\left({x}^{2}+1\right)}^{3}}-\frac{6x\left(A{x}^{5}+B{x}^{3}+Cx\right)}{{\left({x}^{2}+1\right)}^{4}}+\frac{D{x}^{6}}{{\left({x}^{2}+1\right)}^{4}}\phantom{\rule{2.77695pt}{0ex}}$

and multiplying by ${\left({x}^{2}+1\right)}^{4}$ gives the following identity:

 $1\equiv \left(5A{x}^{4}+3B{x}^{2}+C\right)\left({x}^{2}+1\right)-6x\left(A{x}^{5}+B{x}^{3}+Cx\right)+D{x}^{6}$ ($\ast$)

i.e.

$1\equiv \left(D-A\right){x}^{6}+\left(5A-3B\right){x}^{4}+\left(3B-5C\right){x}^{2}+C\phantom{\rule{2.77695pt}{0ex}}.$

Equating coefficients of the different powers of $x$ on each side of the equivalence sign gives $1=C$,   $0=3B-5C$,   $0=5A-3B$,   $0=D-A$, so $A=1$,   $B=\frac{5}{3}$,   $C=1$ and   $D=1$.

(iii) Using the results of parts (i) and (ii), we see that for $0\le x\le 1$

$\frac{\mathrm{d}\phantom{\rule{1em}{0ex}}}{\mathrm{d}x}\left(\frac{{x}^{5}+\frac{5}{3}{x}^{3}+x}{{\left({x}^{2}+1\right)}^{3}}\right)\le \frac{1}{{\left({x}^{2}+1\right)}^{4}}\le \frac{\mathrm{d}\phantom{\rule{1em}{0ex}}}{\mathrm{d}x}\left(\frac{{x}^{5}+\frac{5}{3}{x}^{3}+x}{{\left({x}^{2}+1\right)}^{3}}\right)+\frac{1}{16}\phantom{\rule{2.77695pt}{0ex}}.$

Inequalities can be integrated, so

${\int }_{0}^{1}\frac{\mathrm{d}\phantom{\rule{1em}{0ex}}}{\mathrm{d}x}\left(\frac{{x}^{5}+\frac{5}{3}{x}^{3}+x}{{\left({x}^{2}+1\right)}^{3}}\right)\mathrm{d}x\le {\int }_{0}^{1}\frac{1}{{\left({x}^{2}+1\right)}^{4}}\phantom{\rule{0.3em}{0ex}}\mathrm{d}x\le {\int }_{0}^{1}\frac{\mathrm{d}\phantom{\rule{1em}{0ex}}}{\mathrm{d}x}\left(\frac{{x}^{5}+\frac{5}{3}{x}^{3}+x}{{\left({x}^{2}+1\right)}^{3}}\right)\mathrm{d}x+{\int }_{0}^{1}\frac{1}{16}\phantom{\rule{0.3em}{0ex}}\mathrm{d}x\phantom{\rule{2.77695pt}{0ex}}$

i.e.

${\left[\frac{{x}^{5}+\left(5∕3\right){x}^{3}+x}{{\left({x}^{2}+1\right)}^{3}}\right]}_{0}^{1}\le {\int }_{0}^{1}\frac{1}{{\left({x}^{2}+1\right)}^{4}}\phantom{\rule{0.3em}{0ex}}\mathrm{d}x\le {\left[\frac{{x}^{5}+\left(5∕3\right){x}^{3}+x}{{\left({x}^{2}+1\right)}^{3}}\right]}_{0}^{1}+{\left[\frac{x}{16}\right]}_{0}^{1}\phantom{\rule{0.3em}{0ex}}$

from which the required result follows immediately.

Post-mortem

Instead of equating coefficients in ($\ast$), you could obtain equations for $A\phantom{\rule{0.3em}{0ex}}$, $B\phantom{\rule{0.3em}{0ex}}$, $C\phantom{\rule{0.3em}{0ex}}$ and $D$ by putting four carefully chosen values of $x$ into the equation. An obvious choice is $x=0$, but (thinking ﬂexibly!) you could try $x=i$ to eliminate terms with factors of ${x}^{2}+1\phantom{\rule{0.3em}{0ex}}$. Then it becomes more difficult to ﬁnd good choices.

You might wonder why, in part (ii), the term inside the derivative has only odd powers of $x$. Would it not make it more general to include even powers as well? You could in fact include even powers but you would ﬁnd that their coefficients would be zero: all the other terms in the equation are even in $x$, so the derivative has to be an even function which means that the function being differentiated must be odd.

Although the idea of this question is good, the ﬁnal result is a bit feeble. It only gives the value of the integral to an accuracy of about $\frac{1}{16}∕\frac{11}{24}$ which is about $15%$. The actual value of the integral can be found fairly easily using the substitution $x=tant$ and is $\frac{11}{48}+\frac{5}{64}\pi \phantom{\rule{0.3em}{0ex}}$ so the inequalities can be used to give (rather bad) estimates for $\pi \phantom{\rule{0.3em}{0ex}}$ namely $2.933\le \pi \le 3.733\phantom{\rule{0.3em}{0ex}}$.