## 7 Examples
After solving by means of (5) and (6) for \(\lambda_{A}=2\), and \(n=2\), we obtain the solution  
\[\bar{u}_{2}(x)=-\frac{3436e-11712}{2391}-\frac{(1268e-5892)x}{2391}-\frac{(109 6e-4248)x^{2}}{2391}.\]  
Similarly, for \(n=4\) and \(n=8\), we acquire the higher order solutions  
\[\bar{u}_{4}(x) = 1.026427+1.024317x+0.515273x^{2}+0.172608x^{3}+0.043388x^{4},\] \[\bar  
the components of \(G\) and the function \(f(x)\). For \(\lambda_{A}=2\), and \(n=2\), \(n=4\) and \(n=8\), we obtain the corresponding solutions_  
\[\tilde{u}_{2}(x) = 1.131517+0.972464x+0.969511x^{2},\] \[\tilde{u}_{4}(x) = 1.029984+1.025406x+0.526503x^{2}+0.145148x^{3}+0.072597x^{4},\] \[\tilde{u}_{8}(x) = 1.007331+1.006524x+0.503972x^{2}+0.168191x^{3}+0.042115x^{4}\] \[+0.008410x^{5}+0.001444x^{6}+0.000167x^{7}+0.000041x^{8}.\]  
_In Table 7, we list the maximum errors between these approximate solutions and the exact solution. Additionally, in Table 8, we present the results for the value of the parameter \(\lambda_{A}=50\). The results are very close to those obtained above with the composite Trapezoidal rule and Taylor series._  
_PM: Collocation with Polynomials:_  
_Finally, we solve the integral equation in (48) by employing a projection method such as the simple collocation method given in Section 6, where the ordinary Lagrange basis functions are used. For \(\lambda_{A}=1\), by means of (41) and polynomials of degree \(n=3\), \(n=4\) and \(n=5\) interpolating \(u(x)\) on the equally spaced nodes \(x_{j}=\left(j-1\right)/n,\ j=1,2,\ldots,n+1\), we obtain the following solutions_  
\[\tilde{u}_{3}(x) = 0.999256+1.013577x+0.425543x^{2}+0.278580x^{3},\] \[\tilde{u}_{4}(x) = 0.999985+0.998801x+0.509787x^{2}+0.140276x^{3}+0.069415x^{4},\] \[\int_{a}^{b}y(x)dx\approx\frac{h}{2}\Bigg{\{}y(x_{1})+2\sum_{j=2}^{n}y(x_{j})+y(x _{n+1})\Bigg{\}}, \tag{49}\]0 \[\int_{a}^{b}y(x)dx\approx\frac{h}{2}\Bigg{\{}y(x_{1})+2\sum_{j=2}^{n}y(x_{j})+y(x _{n+1})\Bigg{\}}, \tag{49}\]1  
_respectively. In Table 9, we give the maximum errors between these approximate solutions and the exact solution. By way of illustration, we also compare these results with the maximum errors obtained in [24; 25], where an advanced quasi-projection method based on B-spline tight framelets is utilized._ \begin{table}
\begin{tabular}{c c c c c} \hline \(n\) & \(\mathbf{e}_{n}=\|\mathbf{u}-\tilde{\mathbf{u}}_{n}\|_{\infty}\) & \(\eta=\frac{\mathbf{e}_{n}}{|\mathbf{u}|}\) & \(\mathbf{100\%\times\eta}\) & Ratio \\ \hline
2 & \(3.55\times 10^{-1}\) & \(1.31\times 10^{-1}\) & \(13.07\%\) & \\
4 & \(8.14\times 10^{-2}\) & \(2.99\times 10^{-2}\) & \(2.99\%\) & \(4.4\) \\
8 & \(1.99\times 10^{-2}\) & \(7.33\times 10^{-3}\) & \(0.73\%\) & \(4.1\) \\
16 & \(4.95\times 10^{-3}\) & \(1.82\times 10^{-3}\) & \(0.18\%\) & \(4.1\) \\ \hline \end{tabular}
\end{table} 
Table 7: IE (48), \(\lambda_{A}=2\). QM: Trapezoidal rule and Polynomials. **Example 3**. : _Consider the integral equation_  
\[u(x)-\frac{1}{\lambda_{A}}\int_{0}^{1}\frac{cu(t)}{c^{2}+(x-t)^{2}}dt=f(x),\quad 0 \leq x\leq 1, \tag{50}\]  
_where the parameters \(\lambda_{A}\
eq 0\) and \(c>0\). This equation appears in electrostatics and it is known as [PERSON]'s equation [26]. The kernel function_  
\[K(x,t)=\frac{c}{c^{2}+(x-t)^{2}}, \tag{51}\]  
_is continuous on \([0,1]\times[0,1]\) with a peak at \(x=t\) when \(c\) is small. Figure 2 shows the shape of \(K(x,t)\) for various values of \(c\) and \(x=0.5\). It is understood that as \(c\) diminishes to zero, the more difficult it is to construct the solution to the problem. Let \(\lambda_{A}=-0.5\), \(c=0.1\) and_  
\[f(x) = \frac{1}{50}(10x-4)\ln\biggl{(}\frac{100x^{2}-200x+101}{100x^{2}+ 1}\biggr{)}+\] \[\frac{1}{50}(100x^{2}-80x+5)(\arctan(10x)-\arctan(10x-10))+\] \[\frac{1}{50}(50x^{2}-40x+13),\]  
_so that (50) has the exact solution \(u(x)=0.06-0.8x+x^{2}\)[26]._