## 7 Examples
\begin{table}
\begin{tabular}{c c c c c c} \hline  & & [24]\(\mathbf{e_{n}}=\|\mathbf{u-\tilde{u}_{n}}\|_{2}\) & \(\mathbf{e_{n}}=\|\mathbf{u-\tilde{u}_{n}}\|_{\infty}\) \\ \hline \(\mathbf{n}\) & \(\mathbf{N^{*}}\) & \(\mathbf{B_{2}}\)**-OEP** & \(\mathbf{B_{4}}\)**-OEP** & \(\mathbf{N^{*}}\) & \\ \hline
2 & 40 & \(1.03\times 10^{-3}\) & \(4.45\times 10^{-7}\) & 3 & \(1.96\times 10^{-2}\) \\
3 & 112 & \(1.08\times 10^{-4}\) & \(1.78\times 10^{-7}\) & 4 & \(1.47\times 10^{-3}\) \\
4 & 288 & \(1.25\times 10^{-4}\) & \(4.35\times 10^{-8}\) & 5 & \(6.02\times 10^{-5}\) \\
5 & 704 & \(9.78\times 10^{-5}\) & \(1.58\times 10^{-9}\) & 6 & \(2.87\times 10^{-6}\) \\ \hline \end{tabular}  
* \(N\) = order of system of linear algebraic equations to solve.  
\end{table} 
Table 9: IE (48), \(\lambda_{A}=1\): PM: Collocation with Polynomials. We noticed that Taylor series and interpolation polynomials are not generally efficient in solving problems with kernel functions of the type (51) when \(c\) is small (\(c<0.5\)). Therefore, we apply here the quadrature method in Section 5 by using the Trapezoidal and Simpson's rules. QM: Trapezoidal rule: Let us consider \(n+1\) equally spaced quadrature points \(0=x_{1}<x_{2}<\ldots<x_{n+1}=1\), \(x_{j}=(j-1)h\), \(j=1,2,\ldots,n+1\), where \(h=1/n\), in the interval \([0,1]\). We use the quadrature formula in (49) to approximate the integral in (50) and then, through the steps in Section 5, write Equation (50) in the form (34), i.e.,  
\[T_{q}\tilde{u}_{n}=\tilde{u}_{n}-G\Psi(\tilde{u}_{n})=f, \tag{52}\]  
where the vectors  
\[G=-\frac{1}{n}\Big{(} \begin{array}{cc}\frac{1}{10\left(\frac{1}{100}+x^{2} \right)}&\frac{1}{5\left(\frac{1}{100}+(x-x_{2})^{2}\right)}&\cdots&\frac{1}{1 0\left(\frac{1}{100}+(x-1)^{2}\right)}\\ \end{array} \Big{)},\;\Psi(\tilde{u}_{n})=\left( \begin{array}{c}\tilde{u}_{n }(0)\\ \tilde{u}_{n}(x_{2})\\ \vdots\\ \tilde{u}_{n}(1)\\ \end{array} \right).\]  
Solving (52) via (5) and (6), we obtain an approximate analytic solution to (50) in the interval \([0,1]\). For example, for \(n=2\), we have  
\[\tilde{u}_{2}(x)=f(x)-\frac{4.052649\times 10^{-4}}{0.01+x^{2}}+\frac{0.004199} {0.01+(x-0.5)^{2}}-\frac{0.005479}{0.01+(x-1.0)^{2}}.\]  
In Table 10, we record the maximum errors for various values of \(n\). The maximum error occurs at the point \(x=1\) except in the instance \(n=8\), which is a poor approximation anyway, where it is located at \(x=0.935\). QM: Simpson's rule: We utilize now the quadrature formula in (45), with \(n\) being an even number, and repeat the steps above. As an illustration, for \(n=16\), we obtain the solution  
\[\tilde{u}_{16}(x) = f(x)-\frac{2.554573\times 10^{-4}}{x^{2}+0.01}-\frac{2.269247\times 1 0^{-4}}{(x-0.0625)^{2}+0.01}+\frac{2.119289\times 10^{-4}}{(x-0.125)^{2}+0.01}\] \[+\frac{8.937448\times 10^{-4}}{(x-0.1875)^{2}+0.01}+\frac{6.691821 \times 10^{-4}}{(x-0.25)^{2}+0.01}+\frac{0.001507}{(x-0.3125)^{2}+0.01}+\frac{8. 573774\times 10^{-4}}{(x-0.375)^{2}+0.01}\] \[+\frac{0.00169}{(x-0.4375)^{2}+0.01}+\frac{7.766971\times 10^{-4}}{( x-0.5)^{2}+0.01}+\frac{0.001200}{(x-0.5625)^{2}+0.01}+\frac{4.271220\times 10^{ -4}}{(x-0.625)^{2}+0.01}\] \[+\frac{2.803521\times 10^{-4}}{(x-0.6875)^{2}+0.01}-\frac{1.912995 \times 10^{-4}}{(x-0.75)^{2}+0.01}-\frac{0.001151}{(x-0.8125)^{2}+0.01}-\frac{0.00 1080}{(x-0.875)^{2}+0.01}\] \[-\frac{0.003089}{(x-0.9375)^{2}+0.01}-\frac{0.001106}{(x-1.0)^{2}+ 0.01}.\]  
Table 11 shows the errors of the approximate solutions for varying values of \(n\). It is noted that the maximum error occurs at the point \(x=1\), with the exception of the instances \(n=8\) and \(n=16\), where it is located at the points \(x=0.95\) and \(x=0.995\), respectively. In the same Table,