# Weighted Optimal Quadrature Formulas in Sobolev Space and Their Applications
## 7 Numerical Results  
The following examples illustrate the solution of integral equations using the quadrature method based on the optimal quadrature formula with derivatives for \(m=2,4,8\). The obtained results are compared with the exact solutions and with those reported by other researchers, using the absolute error as the measure of accuracy. \[E_{m}=\max_{x}|y(x)-y_{ex}(x)|.\]  
Here \(E_{m}\) is the maximum absolute error, \(y(x)\) is the approximate solution and \(y_{ex}(x)\) is the exact solution. It should be noted that a program was created with the Maple language for the above algorithm. All calculations are performed using 32 significant digits. **Example 1**. : _In (50) \(K(x,s)=-x\times(e^{x\times s}-1),f(x)=e^{x}-x,\lambda=1,a=0,b=1\). Then integral Equation (61) will take the following form:_  
\[y(x)+\int_{0}^{1}x\times(e^{x\times s}-1)\times y(s)ds=e^{x}-x,x\in[0,1] \tag{54}\]  
_The exact solution of integral Equation (54) is as follows:_  
\[y_{ex}(x)=1.\]  
The results for this example are summarized in Table 1.  
Here ES is the exact solution, OQF is the optimal quadrature formula and MAE is the maximum absolute error. The authors of [29] solved this example using a modified multistage average integral method and obtained a result with the maximum absolute error \(E_{m}=3.44\times 10^{-15}\) for \(N=13\) equidistant collocation nodes. On the basis of using the method of integral replacement with the twelfth-order quadrature formula, the authors of [45] obtained the result with the maximum absolute error \(E_{m}=4.44\times 10^{-16}\) at \(N=20\) integration intervals. Our method gave the result with the value of the maximum absolute error \(E_{m}=4.0\times 10^{-31}\) at \(m=1\). **Example 2**. : _In (50), \(K(x,s)=x\times s,f(x)=\cos(x),\lambda=1,a=0,b=1\). Then integral Equation (50) takes the following form:_  
\[y(x)-\int_{0}^{1}x\times s\times y(s)ds=\cos(x),x\in[0,1]. \tag{55}\]  
_The exact solution of integral Equation (55) is as follows:_  
\[y_{ex}(x)=cos(x)+1.5\times(\sin(1)+\cos(1)-1)\times x.\]  
The results for this example are shown in Table 2.  
To solve the Fredholm integral equation of the second kind, the authors of [30] developed a method using spline technology. Using this method, they obtained the result for this example. The maximum absolute error was \(E_{m}=2.0\times 10^{-5}\) at \(N=11\), \(n=10\). Our method gave the result with the value of maximum absolute error \(E_{m}=1.62\times 10^{-8}-2.43\times 10^{-19}\) at \(m=\overline{3,8}\). \begin{table}
\begin{tabular}{c c c c c c c c} \hline \multirow{2}{*}{\(x_{i}\)} & \multirow{2}{*}{ES} & \multicolumn{3}{c}{**OQF**} & \multicolumn{3}{c}{**Absolute Error**} \\ \cline{3-8}  & & \(m=2\) & \(m=4\) & \(m=8\) & \(m=2\) & \(m=4\) & \(m=8\) \\ \hline
0.0 & 1.000 & 1.000 & 1.000 & 1.000 & 0.00\(E+00\) & 0.00\(E+00\) & 0.00\(E+00\)  
**Example 3**. : _In (50), \(K(x,s)=x^{2}-x-s^{2}+s,f(x)=-2x^{3}+3x^{2}-x,\lambda=-1,a=0,b=1\). Then integral Equation (50) will take the following form:_  
\[y(x)+\int_{0}^{1}(x^{2}-x-s^{2}+s)y(s)ds=-2x^{3}+3x^{2}-x,x\in[0,1]. \tag{56}\]  
_The exact solution of integral Equation (56) is as follows:_  
\[y_{ex}(x)=-2x^{3}+3x^{2}-x.\]  
The results for this example are given in Table 3.  
The authors of [42] developed a graph-theoretic polynomial using Hosoi polynomials and solved integral Equation (56). They obtained the result with the maximum absolute error \(E_{m}=8.88\times 10^{-16}\) at \(n=3\). The authors of [43] solved integral Equation (56) and obtained the result with the maximum absolute error \(E_{m}=4.97\times 10^{-4}\) at \(N=8\). Our method gave the result with the value of the maximum absolute error \(E_{m}=1.0\times 10^{-32}\) at \(m=1\). **Example 4**. : _In (50) \(K(x,s)=x\times e^{s},f(x)=e^{-x},\lambda=-1,a=0,b=1\).