Past exam of the year 1 of the computer science course of the University of Oxford / 2013 / Discrete mathematics, probability and continuous mathematics / 5

Differentiate the product $4(x-1{)}^{-2}(x+1{)}^{-1}$ using the product rule and chain rule:Combining terms over a common denominator gives

For a partial fraction decomposition, writeMultiplying by $(x-1{)}^2(x+1)$ gives $4=A(x^2-1)+B(x+1)+C(x-1{)}^2$. Setting $x=1,-1$ gives $B=2,C=1$, and comparing the coefficient of $x^2$ gives $A=-1$. ThusIntegrating each term gives the antiderivativeThe constant of integration can be chosen independently on each interval separated by $-1$ and $1$.

The power rule for a definite integral gives

With step size $h=1/4$, the composite trapezoidal rule gives half weight to the two endpoints:ThereforeIt exceeds the exact definite integral by $1/96$, as expected because $x^2$ is a convex function and each trapezoid lies above the curve.

For $N$ equal intervals, the composite trapezoidal rule isAdding the missing $N^2/N^3$ term therefore gives the exact relationSince $x^2$ is continuous, its Riemann sums, and hence its endpoint-averaged trapezoidal sums, converge to its definite integral. Thus $T_N\to 1/3$ and $1/(2N)\to 0$, givingNo pre-existing formula for the sum of squares is needed.