# Weighted Optimal Quadrature Formulas in Sobolev Space and Their Applications
## 7 Numerical Results
Then the integral Equation (50) will take the following form:_  
\[y(x)+\int_{0}^{1}e^{x+s}\times y(s)ds=e^{-x},x\in[0,1] \tag{58}\]  
_The exact solution of integral Equation (58) is as follows:_  
\[y_{ex}(x)=e^{-x}+\frac{2e^{x}}{3-e^{2}}.\]  
The results for this example are shown in Table 5.  
The authors of [33] solved integral Equation (58) using the integral method of mean values and obtained the result with the maximum absolute error \(E_{m}=2.0\times 10^{-5}\) at \(N=11\). Our method gave the result with the value of maximum absolute error \(E_{m}=3.44\times 10^{-7}-5.17\times 10^{-18}\) at \(m=\overline{2,8}\). **Example 6**. : _In (50) \(K(x,s)=e^{x-s-12},f(x)=\cos(x)-\frac{e^{-x-12}}{2}[e\times\sin(1)+e\times\cos (1)-1],\lambda=-1,a=0,b=1\). Then integral Equation (50) will take the following form:_  
\[y(x)+\int_{0}^{1}e^{x-s-12}\times y(s)ds=\cos(x)-\frac{e^{-x-12}}{2}\times(e \times\sin(1)+e\times\cos(1)-1),x\in[0,1]. \tag{59}\]  
_The exact solution of integral Equation (59) is as follows:_  
\[y_{ex}(x)=\cos(x).\]  
The results for this example are given in Table 6. \begin{table}
\begin{tabular}{c c c c c c c} \hline \hline \multirow{2}{*}{\(x_{i}\)} & \multirow{2}{*}{ES} & \multicolumn{3}{c}{**OQF**} & \multicolumn{3}{c}{**Absolute Error**} \\ \cline{3-7}  & & \(m=2\) & \(m=4\) & \(m=8\) & \(m=2\) & \(m=4\) & \(m=8\) \\ \hline
0.0 & 0.544 & 0.544 & 0.544 & 0.544 & \(1.27E-07\) & \(3.01E-11\) & \(1.90E-18\) \\ \hline
0.1 & 0.401 & 0.401 & 0.401 & 0.401 & \(1.40E-07\) & \(3.33E-11\) & \(2.10E-18\) \\ \hline
0.2 & 0.262 & 0.262 & 0.262 & 0.262 & \(1.55E-07\) & \(3.68E-11\) & \(2.32E-18\) \\ \hline
0.3 & 0.126 & 0.126 & 0.126 & 0.126 & \(1.71E-07\) & \(4.07E-11\) & \(2.57E-18\) \\ \hline
0.4 & \(-0.009\) & \(-0.009\) & \(-0.009\) & \(-0.009\) & \(1.89E-07\) & \(4.50E-11\) & \(2.84E-18\) \\ \hline
0.5 & \(-0.145\) & \(-0.145\) & \(-0.145\) & \(-0.145\) & \(2.09E-07\) & \(4.97E-11\) & \(3.14E-18\) \\ \hline
0.6 & \(-0.281\) & \(-0.281\) & \(-0.281\) & \(-0.281\) & \(2.31E-07\) & \(5.49E-11\) & \(3.47E-18\) \\ \hline
0.7 & \(-0.421\) & \(-0.421\) & \(-0.421\) & \(-0.421\) & \(2.55E-07\) & \(6.07E-11\) & \(3.83E-18\) \\ \hline
0.8 & \(-0.565\) & \(-0.565\) & \(-0.565\) & \(-0.565\) & \(2.82E-07\) & \(6.71E-11\) & \(4.24E-18\) \\ \hline
0.9 & \(-0.714\) & \(-0.714\) & \(-0.714\) & \(-0.714\) & \(3.11E-07\) & \(7.42E-11\) & \(4.68E-18\) \\ \hline
1.0 & \(-0.871\) & \(-0.871\) & \(-0.871\) & \(-0.871\) & \(3.44E-07\) & \(8.19E-11\) & \(5.17E-18\) \\ \hline  & & & & **MAE** & \(3.44E-07\) & \(8.19E-11\) & \(5.17E-18\) \\ \hline \hline \end{tabular}
\end{table} 
Table 5: Results of solving the integral equation in Example 5. The authors of [34] solved integral Equation (59) by applying a new type of spline function of fractional order and obtained the result with the maximum absolute error \(E_{m}=3.1\times 10^{-6}\) at \(N=11\) and \(n=2\). Our method gave the result with the value of maximum absolute error \(E_{m}=7.06\times 10^{-9}-6.04\times 10^{-24}\) at \(m=\overline{1,8}\). **Example 7**. : _In (50) \(K(x,s)=e^{2x-\frac{5}{3}},f(x)=e^{2x+\frac{1}{3}},\lambda=-\frac{1}{3},a=0,b=1\). Then integral Equation (50) will take the following form:_  
\[y(x)+\frac{1}{3}\int_{0}^{1}e^{2x-\frac{5}{3}}\times y(s)ds=e^{2x+\frac{1}{3}},x\in[0,1]. \tag{60}\]  
_The exact solution of integral Equation (60) is as follows:_  
\[y_{ex}(x)=e^{2x}.\]  
The results for this example are given in Table 7.  
The authors of [35] solved integral Equation (60) based on a special representation of vector forms of triangular functions and obtained the result with the maximum absolute error \(E_{m}=6.4\times 10^{-7}\) at \(m=1024\). The authors of [39] using a combination of Taylor series and block-plus functions solved the same integral Equation (60). They obtained the result with the maximum absolute error \(E_{m}=4.6\times 10^{-4}\) at \(N=80\). The authors of [40] solved integral Equation (60) using Chebyshev polynomial approximation and obtained the result with the maximum absolute error \(E_{m}=0.49\times 10^{-4}\) at \(N=10\).