Question 9.2: Evaluation of a single definite integral using fourth-order ......

Step 1: Since the limits of integration are [0,3], the integral has to be transformed to the form \int_{-1}^{1} f(t) d t. In the present problem a = 0 and b = 3. Substituting these values in Eq. (9.31)

x=\frac{1}{2}[t(b-a)+a+b] \quad \text { and } \quad d x=\frac{1}{2}(b-a) d t         (9.31)

gives: x=\frac{1}{2}[t(b-a)+a+b]=\frac{1}{2}[t(3-0)+0+3]=\frac{3}{2}(t+1) \quad and \quad d x=\frac{1}{2}(b-a) d t=\frac{1}{2}(3-0) d t=\frac{3}{2} d t Substituting these values in the integral gives:

I=\int_{0}^{3} e^{-x^{2}} d x=\int_{-1}^{1} f(t) d t=\int_{-1}^{1} \frac{3}{2} e^{-\left[\frac{3}{2}(t+1)\right]^{2}} d t

Step 2: Use four-point Gauss quadrature to evaluate the integral. From Eq. (9.30),

\int_{-1}^1 f(x) d x \approx C_1 f\left(x_1\right)+C_2 f\left(x_2\right)+C_3 f\left(x_3\right)+\ldots+C_n f\left(x_n\right)         (9.30)

and using Table 9-1:

Table 9-1: Weight coefficients and Gauss points coordinates

The exact value of the integral (when carried out analytically) is 0.8862073 . The error is only about 1%.