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

(i)

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

What is the smallest value?

(ii)

Find constants $A$, $B$, $C$ and $D$ such that, for all $x$,

$$\frac{1}{{\left(x^2+1\right)}^4}\equiv \frac{\mathrm{d}}{\mathrm{d}x}\left(\frac{A{x}^5+B{x}^3+Cx}{{\left(x^2+1\right)}^3}\right)+\frac{D{x}^6}{{\left(x^2+1\right)}^4}.$$

(iii)

Hence, or otherwise, prove that

$$\frac{11}{24}\le {\int \; <!--nolimits-->}_0^1\frac{1}{{\left(x^2+1\right)}^4}\mathrm{d}x\le \frac{11}{24}+\frac{1}{16}.$$

Comments

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 $\mathrm{f}\left(x\right)=\frac{x^6}{{\left(x^2+1\right)}^4}.$ Then

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

(ii) Doing the differentiation gives

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.

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 $0\le x\le 1$

Inequalities can be integrated, so

i.e.

from which the required result follows immediately.

Post-mortem

Instead of equating coefficients in ($\ast$), you could obtain equations for $A$, $B$, $C$ 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 $x^2+1$. 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 $\frac{11}{48}+\frac{5}{64}\pi$ so the inequalities can be used to give (rather bad) estimates for $\pi$ namely $2.933\le \pi \le 3.733$.