Question

Вычислить площадь фигуры,ограниченной линиями y=x^2-6x+7, y=-x^2+4x-1.Сделать чертеж

Answer 1

РЕШЕНИЕ
Сразу рисунок к задаче в приложении. 
Площадь фигуры - интеграл разности функций ограничивающих её.
Находим пределы интегрирования решив уравнение.
x²-6x+7= -x²+4x-1
Упрощаем
2*x² -2x+8 = 0
Решаем квадратное уравнение и находим корни.
Пределы интегрирования - a = 4, b = 1.
Находим разность функций.
Y = -x²+4x-1 - (x²-6x+7) = -8 +10*x - 2x²
Находим интеграл разности функций
$S(x)= \int\limits^4_1 ({-8 +10x-2x^2}) \, dx= \frac{-8x}{1}+ \frac{10x^2}{2}- \frac{2x^3}{3}=9$
ОТВЕТ: Площадь = 9.

Answer 2

Calculating the Area of the Figure

To calculate the area of the figure bounded by the lines \(y=x^2-6x+7\) and \(y=-x^2+4x-1\), we can start by finding the points of intersection of the two curves. These points will define the boundaries of the region, and then we can integrate to find the area.

First, let's find the points of intersection by setting the two equations equal to each other and solving for x.

The equations are: 1. \(y=x^2-6x+7\) 2. \(y=-x^2+4x-1\)

Setting them equal to each other: \(x^2-6x+7 = -x^2+4x-1\)

Solving for x, we get: \(2x^2-10x+8 = 0\)

This is a quadratic equation, and we can solve for x using the quadratic formula: \[x = \frac{{-b \pm \sqrt{{b^2-4ac}}}}{{2a}}\]

After finding the x-values, we can then find the corresponding y-values by substituting these x-values into either of the original equations.

Let's calculate the points of intersection and then proceed to find the area of the bounded figure.

Solution

The points of intersection are: \[x_1 = 2 \quad \text{and} \quad x_2 = 2\]

Substituting these x-values into the first equation: \[y_1 = (2)^2-6(2)+7 = 3\]

Substituting these x-values into the second equation: \[y_2 = -(2)^2+4(2)-1 = 3\]

The points of intersection are (2, 3) and (2, 3).

Now, we can calculate the area of the bounded figure by integrating the absolute difference between the two curves from \(x=2\) to \(x=2\):

\[A = \int_{2}^{2} |(x^2-6x+7) - (-x^2+4x-1)| \, dx\]

\[A = \int_{2}^{2} |2x^2-10x+8| \, dx\]

\[A = 0\]

Therefore, the area of the figure bounded by the given lines is 0.

Sketch

Unfortunately, I'm unable to provide a sketch at the moment. However, you can visualize the figure by plotting the two given equations on a graphing tool or software.