Chapter 7

With the usual notation for the Newton-Cotes quadrature formula and using the equally spaced quadrature points \(x_{k}=a+k h\) for \(k=0,1, \ldots, n\) and \(n \geq 1\), show that \(w_{k}=w_{n-k}\) for \(k=0,1, \ldots, n\).

A quadrature formula on the interval \([-1,1]\) uses the quadrature points \(x_{0}=-\alpha\) and \(x_{1}=\alpha\), where \(0<\alpha \leq 1\) : $$ \int_{-1}^{1} f(x) \mathrm{d} x \approx w_{0} f(-\alpha)+w_{1} f(\alpha) . $$ The formula is required to be exact whenever \(f\) is a polynomial of degree 1 . Show that \(w_{0}=w_{1}=1\), independent of the value of \(\alpha\). Show also that there is one particular value of \(\alpha\) for which the formula is exact also for all polynomials of degree 2. Find this \(\alpha\), and show that, for this value, the formula is also exact for all polynomials of degree 3 .

Write down the errors in the approximation of $$ \int_{0}^{1} x^{4} \mathrm{~d} x \text { and } \int_{0}^{1} x^{5} \mathrm{~d} x $$ by the trapezium rule and Simpson's rule. Hence find the value of the constant \(C\) for which the trapezium rule gives the correct result for the calculation of $$ \int_{0}^{1}\left(x^{5}-C x^{4}\right) \mathrm{d} x $$ and show that the trapezium rule gives a more accurate result than Simpson's rule when \(\frac{15}{14}