Trapezna formula

Izračunajte integral $\displaystyle I=\int\limits_{1}^{2}\ln x dx$ pomoću trapezne formule podijelom na intervala i usporedite s točnim rješenjem.

Rješenje.

Prema [M2, poglavlje 2.7.2] je

$\displaystyle n=5\Rightarrow \Delta x_{i}=\frac{b-a}{n}=\frac{2-1}{5}=0.2=h$

pa je

$\displaystyle x_{i}=a+ih,$ $\displaystyle h=0.2,$ $\displaystyle i=0,1,\ldots, n.$

Slijedi

$\displaystyle x_{0}=1,$ $\displaystyle x_{1}=1.2,$ $\displaystyle x_{2}=1.4,$ $\displaystyle x_{3}=1.6,$ $\displaystyle x_{4}=1.8,$ $\displaystyle x_{5}=2,$

pa je

$\displaystyle I$	$\displaystyle =\int\limits_{1}^{2}\ln x dx\approx 0.2\left[ \frac{f\left( x_{0... ...ight) +f\left( x_{2}\right) +f\left( x_{3}\right) +f\left( x_{4}\right) \right]$
	$\displaystyle =0.2\left[ \frac{0+0.69314}{2}+0.18232+0.33647+0.47+0.58778\right] =0.38463.$

Točno rješenje je $2\ln 2-1\approx 0.38629$ .