Question

Compute integral 0 to 4 f(x)dx using the Romberg integral technique on the trapezoidal integrals evaluated by the trapezoidal rule taking h = 1 and h = 0.5. The tabulated
values are given below.
x 0 0.5 1 1.5 2.0 2.5 3.0 3.5 4.0
f(x) 1 4 3 2 2.5 2.9 3.6 4 1.8

Answer 1

The Romberg's method is $I = I_{2}+\dfrac{1}{3}(I_{2}-I_{1})$ , where $I_{1}$ and $I_{2}$ are obtained by using Trapezoidal rule taking h = 1 and h = 0.5.

Taking h=1, the tabulated values are

$x ~~~~~~~~~~~~~: 0~~~~1~~~~2~~~~~~3~~~~~~~4\\ y=f(x):1~~~3~~~2.5~~~3.6~~~~1.8$

Using Trapezoidal rule,

$I_{1} = \displaystyle\int_{0}^{4}f(x)dx = \dfrac{h}{2}\left((y_{0}+y_{n})+2(y_{1}+y_{2}+\cdots+y_{n-1})\right)\\ =\dfrac{1}{2}\left((1+1.8)+2(3+2.5+3.6)\right)=10.5$

Taking h=0.5, the tabulated values are

$x ~~~~~~~~~~~~~: 0~~~~0.5~~~~1~~~~1.5~~~~~2~~~~~~2.5~~~~~3~~~~~3.5~~~~~4\\ y=f(x):1~~~~~~4~~~~~3~~~~~~2~~~~~2.5~~~~2.9~~~~3.6~~~~~4~~~~~1.8$

Using Trapezoidal rule,

$I_{2} = \displaystyle\int_{0}^{4}f(x)dx = \dfrac{h}{2}\left((y_{0}+y_{n})+2(y_{1}+y_{2}+\cdots+y_{n-1})\right)\\ =\dfrac{0.5}{2}\left((1+1.8)+2(4+3+2+2.5+2.9+3.6+4)\right)=11.7$

Now, using Romberg's formula with $I_{1}$ and $I_{2}$

$I = I_{2}+\dfrac{1}{3}(I_{2}-I_{1}) = 11.7 + \dfrac{1}{3}(11.7-10.5) = 12.1$

Hence approximate value of $\displaystyle\int_{0}^{4}f(x)dx~ \text{is}~ 12.1$