Question

Найдите площадь фигуры, ограниченной графиком заданных функции y=8+2x-x^2, y=x+6, Помогите пожалуйста

Answer 1

РЕШЕНИЕ
Пределы интегрирования из решения уравнения
 -x²+2x+8=x+6
b = - 1, a = 2
Площадь фигуры - интеграл разности функций.
$S(x)= \int\limits^a_2 {(-2 -x +x^2) } \, dx= \frac{-2x}{1} - \frac{x^2}{2}+ \frac{x^3}{3}$
S= S(2)-S(-1) = 3 1/3 - 1 1/6 = 4 1/2 = 4.5 - площадь - ОТВЕТ 

Answer 2

Finding the Area of the Enclosed Figure

To find the area of the figure enclosed by the given functions \(y=8+2x-x^2\) and \(y=x+6\), we need to determine the points of intersection of the two functions and then calculate the area between them.

First, let's find the points of intersection by solving the system of equations formed by the two functions:

1. Set the two functions equal to each other: \(8+2x-x^2 = x+6\). 2. Rearrange the equation to form a quadratic equation: \(x^2 - x - 2 = 0\). 3. Solve the quadratic equation to find the x-coordinates of the points of intersection.

Now, let's calculate the area between the two functions using the points of intersection and the definite integral formula:

\[A = \int_{a}^{b} |f(x) - g(x)| dx\]

Where: - \(f(x) = 8+2x-x^2\) - \(g(x) = x+6\) - \(a\) and \(b\) are the x-coordinates of the points of intersection.

Let's proceed with solving the quadratic equation and calculating the area between the functions.