Riemann surface
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Example 10.7 (p.353)
Use Simpson one-third rule to find the numerical value of the integral in Example 10.5: $\mathrm{l}=\int_{x_{x_0}}^{x_b} y(x) d x$ in which $y(x)=x \sqrt{\left(16-x^2\right)^3}$
.
Solution:
We will use the three function values at $x=0.5,2.0$ and 3.5 to compute the value of the integral. In such case, the increment of the integration variable is $\Delta x=1.5$. The integral is determined by the expression in Equation (10.10).
We may obtain the function values $\mathrm{y}0, \mathrm{y}_1$ and $y_2$ at $x=0.5,2$ and 3.5 from the Table in Example 10.6 as: $$ y_0=31.25, y_1=83.18 \text { and } y_2=25.42 $$ Integration of the function $y(x)$ in this example5 can thus be determined by substituting the values of $y_0, y_1$ and $y_2$ and the increment of $x$, $\Delta x=1.5$ into Equation (10.10) to give: $$ \begin{aligned} & I=\int{-\Delta x}^{+\Delta x} y(x) d x=\int_{-x}^{+x}\left(a x^2+b x+c\right) d x=\frac{\Delta x}{3}\left(y_0+4 y_1+y_2\right) \
& =\int_{0.5}^{3.5} y(x) d x=\int_{0.5}^{3.5} x \sqrt{\left(16-x^2\right)^3} d x=\frac{1.5}{3}(31.25+4 x 83.18+25.42)=194.70 \
&
\end{aligned}
$$
The “exact” solution of the above integral is 191.45 from a math handbook, from which we may have the following comparison of results between the $I=188.88$ by three trapezoidal in Example 10.5 and the current solution I $=194.70$ with three function values using the Simpson one-third rule.
Example 10.9 (p.358)
Evaluate the following integral by using the Gaussian quadrature in Equation (10.14).
$$
I=\int_0^\pi \cos x d x
$$
Solution:
$$
I=\int_0 \cos x d x
$$
We have the function $y(x)=\cos x$ over the integration limits $x_a=0$ and $x_b=\pi$. The transformation of coordinates makes use of the relationship $x=\frac{\pi}{2} \xi+\frac{\pi}{2}$ from Equation (10.13), from which we get:
$$
y(x)=\cos x=F(\xi)=\cos \left(\frac{\pi}{2} \xi+\frac{\pi}{2}\right)=\sin \left(\frac{\pi}{2} \xi\right)
$$
Also, from Equation (10.14) with the use of the trigonometric relationships such as:
$$
\sin \left(\frac{\pi}{2}+\theta\right)=\cos \theta \text { and } \cos \left(\frac{\pi}{2}+\theta\right)=\sin \theta
$$
We may arrive at the following expression for integrating I in Equation (a) using Gaussian quadrature:
$$
I=\int_0^\pi \cos x d x=\int_{-1}^1\left[\sin \left(\frac{\pi}{2} \xi\right)\left(\frac{\pi}{2} d \xi\right)\right]=\frac{\pi}{2} \int_{-1}^1 \sin \frac{\pi}{2} \xi d \xi=\frac{\pi}{2} \sum_{i=1}^n H_i \sin \left(\frac{\pi}{2} a_i\right)
$$
Let us take, for example, 3 sampling points, i.e., $\mathrm{n}=3$ from Table 10.3 on P.357 with:
$$
\begin{array}{lll}
\mathrm{a}_1=0 & \mathrm{a}_2=+0.77459 & \mathrm{a}_3=-0.77459 \
\mathrm{H}_1=0.88888 & \mathrm{H}_2=0.55555 & \mathrm{H}_3=0.55555
\end{array}
$$
Substituting the above numbers into Equation (b) will lead to the solution:
$$
\begin{aligned}
I & =\frac{\pi}{2}\left[0.88888 \sin (0)+0.55555 \sin \left(\frac{\pi}{2} \times 0.77459\right)+0.55555 \sin \left(-\frac{\pi}{2} \times 0.77459\right)\right] \
& =\frac{\pi}{2}[0.55555 \sin (1.2167)-0.55555 \sin (1.2167)]=0
\end{aligned}
$$
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