Computational Methods

(b) 2.621

Let $f(x)=x^3-18$, so $x_{n+1}=x_n-\dfrac{x_n^3-18}{3x_n^2}$. Starting from $x_0=2.5$:

$x_1=2.5+\frac{2.375}{18.75}\approx2.6267,\qquad x_2\approx2.6267-\frac{0.1227}{20.698}\approx2.6207.$

So the root is $\approx2.621$ to three decimal places.

Trapezoidal rule: A numerical method that approximates a definite integral by dividing $[a,b]$ into $n$ equal strips of width $h=\dfrac{b-a}{n}$ and treating each strip as a trapezium:

$\int_a^b f(x)\,dx\approx\frac h2\left[(y_0+y_n)+2(y_1+y_2+\cdots+y_{n-1})\right].$

Here $a=1$, $b=3$, $n=4$, so $h=0.5$ and $f(x)=\dfrac1{1+x}$.

$\int_1^3\frac{dx}{1+x}\approx\frac{0.5}{2}\left[(0.5+0.25)+2(0.4+0.3333+0.2857)\right]=0.25(0.75+2.038)=0.25\times2.788\approx0.697.$

(The exact value is $\ln2\approx0.6931$.)