Числено интегриране

В числения анализ, числено интегриране определя група от алгоритми за намиране стойността на определен интеграл. Понятието се използва и при численото решаване на диференциални уравнения.

Идеята на численото интегриране е функцията f(x) да се приближи с подходяща функция φ(x), която по-лесно може да се интегрира. Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(x) = \phi(x) + r(x)} , където:

• Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \phi(x)} може да се интегрира точно
• $r(x)$ e остатъка (грешката - residual)

Най-често φ(x) е интерполационен полином построен по някакви възли в интервала $[a,b]$ за $f(x)$.

Числените методи за интегриране се налага да се използват:

• Когато не съществува примитивна функция за f(x) (интегралът не се изразява с елементарни функции)
• когато примитивната функция за f(x) е много сложен израз

Ако f(x) е плавно изменяща се функция, която може да се интегрира в малък брой измерения и има определени гранични стойности, съществуват редица методи с различна степен на точност за апроксимиране на интеграла $\int _{a}^{b}f(x)\,dx$.

Представяме интеграла по следния начин: $\int _{a}^{b}f(x)\,dx=\sum _{i=1}^{n}A_{i}f(x_{i})+R(f)$.

Формули на Нютон-Коутс за числено интегриране

Пример. Да се пресметне по формулата на десните правоъгълници $\int _{2}^{3}{\frac {ln(x)}{x}}\,dx,n=10$

Решение. По условие $[a,b]=[2,3];n=10$

Метод на правоъгълниците

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle h = \frac{b - a}{n} = \frac{3-2}{10} = 0.1}

Метод на правоъгълниците

Съгласно $I\approx \int _{a}^{b}=\int _{x_{0}}^{x_{1}}+\int _{x_{1}}^{x_{2}}+...+\int _{x_{n-1}}^{x_{n}}=y_{0}h+y_{1}h+...+y_{n-1}h=h\sum _{i=0}^{n-1}y_{i}$

x = {2, 2.1, 2.2, 2.3, 2.4, 2.5, 2.6, 2.7, 2.8, 2.9, 3}

y = {0.346574, 0.353303, 0.35839, 0.362134, 0.364779, 0.366516,0.367504, 0.367871, 0.367721, 0.367142, 0.366204}.

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle I \approx h \sum_{i=0}^{n-1} y_i = 0.362193}

Аналитично решение

$\int _{2}^{3}{\frac {log(x)}{x}}dx=1/2(log^{2}(3)-log^{2}(2))\approx 0.363248$

Решение с Матлаб

h = 0.1 % step
m = 0; % sum
for i = 2:h:3-h
m = log(i)/i + m
end
I = m*h
I =  0.36219

Оценка на грешката

Грешка от интегриране: Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vert r_0 \vert \leq \left | \int_{x_0}^{x_1} R_0\,dx\ \right | \leq M_1 \left | \int_{x_0}^{x_1} (x-x_0)\,dx\ \right | = M_1 \frac{ (x-x_0)^2 }{2} \Bigg|_{x_0}^{x_1} = M_1 \frac{h^2}{2} = O(h^2) }

Сумарна грешка:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle R \leq n*M_1 \frac{h^2}{2} = M_1 \frac{h(b-a)}{2} }

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle M_1=\max\limits_{[2,3]}\vert f'(\xi)\vert}

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f=\frac{ln(x)}{x}} за Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f'=\frac{1-ln(x)}{x^2}}

Максималната стойност в [2,3] на Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f'=\frac{1-ln(x)}{x^2}} е при x = 2

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle M_1 = \frac{1 - ln(2)}{2^2} = 0.077 }

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle R = M_1 \frac{h(b-a)}{2} = 0.077*0.1/2 = 0.004}

Анализ

Разликата от аналитичното решение и численото решение е Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 0.3632 - 0.3621 = 0.0011} , което е в рамките на максималната грешка.

Формула на трапеца

Геометрично извеждане

Идеята на геометричното извеждане е да замести площта под кривата y = f(x) за x = a до х = b с площта на трапец ограничена от точките (a, 0), (b, 0), [a, f (a)], и [b, f (b)].

Метод на трапеца

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int_a^b f(x) \,dx \approx \frac{b-a}{2} \left [ f(a) + f(b) \right ]}

Правилото на трапеца няма как да е точно за големи интервали, но ако разглежданият интервал се раздели на по-малки интервали и се сумират техните стойности ще се получи сравнително точно заместване. Ако функцията f има втора производна то грешката от интегриране намалява с Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle h^2 } , където h e големината на интеграла.

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int_a^b f(x)dx \approx h \left ( \frac{f(x_0)}{2} +f(x_1) + \dots + f(x_{n-1}) + \frac{f(x_n)}{2} \right ) }

Аналитично извеждане

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle L_1(x) = y_0 \frac{x-x_1}{x_0-x_1}+y_0 \frac{x-x_1}{x_1-x_0} <=> L_1(x) = y_0 \frac{x-x_1}{h}+y_0 \frac{x-x_1}{h} }

грешка на приближението Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle R_1(x) = \frac{f''(\xi)}{2}(x-x_0)(x-x_1) }

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle | R_1(x) | \leq \frac{M_2}{2} \left | (x-x_0)(x-x_1) \right | } , където

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle M_2 = \max\limits_{[a,b]} \left | f''(\xi) \right | }

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(x) = L_1(x) + R_1(x) }

Интегрираме в интервала Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle [x_0,x_1],\, I = \int_{x_0}^{x_1} f(x)\,dx = \int_{x_0}^{x_1} L_1 (x)\,dx + \int_{x_0}^{x_1} R_1 (x) \,dx }

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} I \approx \int_{x_0}^{x_1} L_1 (x)\,dx = \\ & = \int_{x_0}^{x_1} \left ( y_0 \frac{x-x_1}{-h} +y_1 \frac{x-x_0}{h} \right ) \,dx \\ & = \int_{x_0}^{x_1} y_0 \frac{x-x_1}{-h} \,dx + \int_{x_0}^{x_1} y_1 \frac{x-x_0}{h} \,dx \\ & = \frac{y_0}{-h}\int_{x_0}^{x_1} (x-x_1) \,dx + \frac{y_1}{h}\int_{x_0}^{x_1}(x-x_0) dx \\ & = \frac{y_0}{-h} \frac{(x-x_1)^2}{2} \Bigg |_{x_0}^{x_1} + \frac{y_1}{h} \frac{(x-x_0)^2}{2} \Bigg |_{x_0}^{x_1} \\ & = h \frac{(y_0+y_1)}{2} \end{align} }

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |r_1| \leq M_2\frac{h^3}{12} }

Формули

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle I \approx h \left ( \frac{y_0+y_n}{2} + \sum_{i=1}^{n-1}{y_i} \right ) }

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle | R(x) | \leq n \frac{M_2}{2} \frac{h^3}{12} = M_2\frac{(b-a)}{12}h^2}

Решение

Пример. Да се пресметне по формулата на десните правоъгълници Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int_2^3 \frac{ln(x)}{x}\,dx , n = 10 }

h = 0.1;
sum = 0;
for i = 2+h:h:3-h
sum = log(i)/i + sum
end
sum = sum + (log(2)/2+log(3)/3)/2
I = sum*h
I =  0.36317

Формула на Симпсън

Анализ

Simpson’s Rule and Newton-Cotes Formulas

промери

теория

лекция