8. Приближенное значение интеграла вычисляется по квадратурной формуле
$$
\int_{0}^{1} f(x) d x \approx \alpha f\left(\frac{1}{2}-4 h\right)+\beta f\left(\frac{1}{2}+3 h\right) .
$$

Найти постоянные $\alpha, \beta$ и $h$, при которых формула точна на произвольном многочлене второй степени.

Question

8. Приближенное значение интеграла вычисляется по квадратурной формуле
$$
\int_{0}^{1} f(x) d x \approx \alpha f\left(\frac{1}{2}-4 h\right)+\beta f\left(\frac{1}{2}+3 h\right) .
$$

Найти постоянные $\alpha, \beta$ и $h$, при которых формула точна на произвольном многочлене второй степени.

Accepted Answer

∻Solution by Steps∻ ‖ step 1 ⋮ We start by integrating the constant function 1 from 0 to 1. The integral is given by: $$ \int_{0}^{1} 1 \, dx = 1 $$ ‖ ‖ step 2 ⋮ Next, we integrate the function $ x $ from 0 to 1. The integral is: $$ \int_{0}^{1} x \, dx = \left[ \frac{x^2}{2} \right]_{0}^{1} = \frac{1}{2} = 0.5 $$ ‖ ‖ step 3 ⋮ Then, we integrate the function $ x^2 $ from 0 to 1. The integral is: $$ \int_{0}^{1} x^2 \, dx = \left[ \frac{x^3}{3} \right]_{0}^{1} = \frac{1}{3} \approx 0.33333 $$ ‖ ‖ step 4 ⋮ We now have the results of the integrals: $$ \int_{0}^{1} 1 \, dx = 1 $$ $$ \int_{0}^{1} x \, dx = 0.5 $$ $$ \int_{0}^{1} x^2 \, dx = 0.33333 $$ ‖ ‖ step 5 ⋮ We need to solve the system of equations to find $ \alpha $, $ \beta $, and $ h $. The system is: $$ \alpha \left( \frac{1}{2} - 4h \right) + \beta \left( \frac{1}{2} + 3h \right) = 1 $$ $$ \alpha \left( \frac{1}{2} - 4h \right)^2 + \beta \left( \frac{1}{2} + 3h \right)^2 = \frac{1}{3} $$ $$ 2\alpha \left( \frac{1}{2} - 4h \right) + 2\beta \left( \frac{1}{2} + 3h \right) = 0.5 $$ ‖ ‖ step 6 ⋮ Solving the system of equations, we find: $$ \alpha = \frac{4}{7}, \quad \beta = \frac{3}{7}, \quad h = \frac{1}{14} $$ ‖ ∻Answer∻ ⚹ $ \alpha = \frac{4}{7}, \beta = \frac{3}{7}, h = \frac{1}{14} $ ⚹ ∻Key Concept∻ ⚹Quadrature formula for approximating integrals⚹ ∻Explanation∻ ⚹The quadrature formula is used to approximate the value of an integral by using a weighted sum of function values at specific points. In this problem, we determined the weights $ \alpha $ and $ \beta $, and the points $ \frac{1}{2} - 4h $ and $ \frac{1}{2} + 3h $ to make the formula exact for any quadratic polynomial.⚹◊What are the conditions that need to be satisfied for the formula to be accurate on a polynomial of second degree?⍭ Generate me a similar question◊

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