Example. Solve the equation

$$
\varphi(x)=\sin x+\int_{0}^{1}(1-x \cos x t) \varphi(t) d t
$$

by replacing its kernel with a degenerate one.
Solution. Expanding the kernel $K(x, t)=1-x \cos x t$ into a series, we obtain

$$
K(x, t)=1-x+\frac{x^{3} t^{2}}{2}-\frac{x^{3} t^{4}}{24}+\ldots
$$

Taking the first three terms of the expansion (3) as the degenerate kernel $L(x, t)$:

$$
L(x, t)=1-x+\frac{x^{3} t^{2}}{2}
$$

we will solve the new equation

$$
\widetilde{\varphi}(x)=\sin x+\int_{0}^{1}\left(1-x+\frac{x^{3} t^{2}}{2}\right) \tilde{\varphi}(t) d t
$$

From (4), we get

$$
\tilde{\varphi}(x)=\sin x+C_{1}(1-x)+C_{2} x^{3}
$$

where

$$
C_{1}=\int_{0}^{1} \tilde{\varphi}(t) d t, \quad C_{2}=\frac{1}{2} \int_{0}^{1} t^{2} \widetilde{\varphi}(t) d t
$$

Substituting (5) into (6), we obtain a system for determining $C_{1}$ and $C_{2}$. We have

$$
\begin{aligned}
& C_{1}=\int_{0}^{1}\left[\sin t+C_{1}(1-t)+C_{2} t^{3}\right] d t=\frac{1}{2} C_{1}+\frac{1}{4} C_{2}+1-\cos 1 \\
& C_{2}=\frac{1}{2} \int_{0}^{1}\left[t^{2} \sin t+C_{1}\left(t^{2}-t^{3}\right)+C_{2} t^{5}\right] d t=\frac{1}{24} C_{1}+\frac{1}{12} C_{2}+\sin 1-1+\frac{1}{2} \cos 1
\end{aligned}
$$

or

$$
\left\{\begin{aligned}
\frac{1}{2} C_{1}-\frac{1}{4} C_{2} & =1-\cos 1 \\
-\frac{1}{24} C_{1}+\frac{11}{12} C_{2} & =\sin 1+\frac{1}{2} \cos 1-1
\end{aligned}\right.
$$

Solving this system, we find

$$
C_{1}=1.0031, \quad C_{2}=0.1674
$$

then

$$
\tilde{\varphi}(x)=1.0031(1-x)+0.1674 x^{3}+\sin x
$$

The exact solution of the equation is $\varphi(x) \equiv 1$.

Now let's estimate $\|\varphi-\widetilde{\varphi}\|$ using formula (2). In the metric of the space $L_{2}$, we get

$$
\|\Lambda\| \leqslant \frac{1}{24}\left\{\int_{0}^{1} \int_{0}^{1} x^{10} t^{8} d x d t\right\}^{1 / 2}=\frac{1}{72 \sqrt{11}}<\frac{1}{238}
$$

$$
\begin{gathered}
\|K\| \leqslant\left\{\int_{0}^{1} \int_{0}^{1}(1-x \cos x t)^{2} d x d t\right\}^{1 / 2}= \\
=\left\{2 \cos 1-\frac{1}{8} \cos 2+\frac{1}{16} \sin 2-\frac{5}{6}\right\}^{1 / 2}<\frac{3}{5} \\
\|L\| \leqslant\left\{\int_{0}^{1} \int_{0}^{1}\left(1-x+\frac{x^{3} t^{2}}{2}\right)^{2} d x d t\right\}^{1 / 2}=\sqrt{\frac{5}{14}}<\frac{3}{5} \\
\|f\|=\left\{\int_{0}^{1} \sin ^{2} x d x\right\}^{1 / 2}=\frac{\sqrt{2-\sin 2}}{2}<\frac{3}{5}
\end{gathered}
$$

The norms of the resolvents $\boldsymbol{R}_{\boldsymbol{K}}$ and $\boldsymbol{R}_{\boldsymbol{L}}$ are estimated by the formulas

$$
\left\|R_{K}\right\| \leqslant \frac{\|K\|}{1-|\lambda| \cdot\|K\|}, \quad\left\|R_{L}\right\| \leqslant \frac{\|L\|}{1-|\lambda| \cdot\|L\|}
$$

where $|\lambda|=1$. Thus, $\left\|R_{K}\right\| \leqslant \frac{3}{2},\left\|R_{L}\right\| \leqslant \frac{3}{2}$, and then

$$
\|\varphi-\widetilde{\varphi}\|<\frac{1}{238}\left(1+\frac{3}{2}\right)\left(1+\frac{3}{2}\right) \frac{3}{5}<0.016
$$

## Problems for Independent Solution

Find the solutions of integral equations using the replacement of the kernel with a degenerate one and give an estimate of the error of the solution.

312. $\varphi(x)=e^{x}-x-\int_{0}^{1} x\left(e^{x t}-1\right) \varphi(t) d t$.
313. $\varphi(x)=x+\cos x+\int_{0}^{1} x(\sin x t-1) \varphi(t) d t$.
314. $\varphi(x)=\frac{1}{2}\left(e^{-x}+3 x-1\right)+\int_{0}^{1}\left(e^{-x t^{2}}-1\right) x \varphi(t) d t$.
315. $\varphi(x)=\frac{x}{2}+\frac{1}{2} \sin x+\int_{0}^{1}\left(1-\cos x t^{2}\right) x \varphi(t) d t$.

## § 24. Replacing the Integral with a Finite Sum

Suppose we have a Fredholm integral equation of the second kind

$$
\varphi(x)-\lambda \int_{a}^{b} K(x, t) \varphi(t) d t=f(x)
$$

where $K(x, t)$ and $f(x)$ have continuous derivatives of the required order, and $\lambda$ is a given number.

Take any quadrature formula

$$
\int_{a}^{b} \Phi(x) d x \approx \sum_{k=1}^{n} A_{k} \Phi\left(x_{k}\right)
$$

where $x_{1}, x_{2}, \ldots, x_{n}$ are the abscissas of points on the interval $[a, b]$, and the coefficients $A_{1}, A_{2}, \ldots, A_{n}$ do not depend on the form of the function $\Phi(x)$.

Setting $x=x_{k}(k=1,2, \ldots, n)$ in equation (1), we get

$$
\varphi\left(x_{k}\right)-\lambda \int_{a}^{b} K\left(x_{k}, t\right) \varphi(t) d t=f\left(x_{k}\right), \quad k=1,2, \ldots, n
$$

Replace the integral on the left side of (3) with a sum using the quadrature formula (2):

$$
\varphi\left(x_{k}\right)-\lambda \sum_{m=1}^{n} A_{m} K\left(x_{k}, x_{m}\right) \varphi\left(x_{m}\right)=f\left(x_{k}\right), \cdots k=1 ; 2, \ldots, n
$$

System (4) is a linear system of $n$ algebraic equations with $n$ unknowns $\varphi\left(x_{1}\right), \varphi\left(x_{2}\right), \ldots, \varphi\left(x_{n}\right)$, which are the approximate values of the solution $\varphi(x)$ at the nodes $x_{1}, x_{2}, \ldots, x_{n}$. The approximate solution of equation (1) on the interval $[a, b]$ can be taken as the function

$$
\widetilde{\varphi}(x)=f(x)+\lambda \sum_{m=1}^{n} A_{m} K\left(x, x_{m}\right) \varphi\left(x_{m}\right)
$$

which takes the values $\varphi\left(x_{1}\right), \varphi\left(x_{2}\right), \ldots, \varphi\left(x_{n}\right)$ at the points $x_{1}, x_{2}, \ldots, x_{n}$.

The values of the coefficients $A_{k}$ and the abscissas $x_{k}$ of the quadrature formula (2) will be:

1) for the rectangle formula:

$$
\begin{gathered}
x_{1}=a, \quad x_{2}=a+h, \quad \ldots, \quad x_{n}=a+(n-1) h \\
A_{1}=A_{2}=\ldots=A_{n}=h, \quad \text { where } \quad h=\frac{b-a}{n}
\end{gathered}
$$

2) for the trapezoidal formula:

$$
\begin{gathered}
x_{1}=a, \quad x_{2}=