Question

Найдите площадь фигуры, ограниченной параболой y=4-x^2 и прямой y=x+2Объясните ход решения, если можно!!

Answer 1

Найдем пределы интегрирования
4-х²=х+2
х²+х-2=0
х1+х2=-1 и х1*х2=-2⇒х1=-2 и х2=1
Первый график парабола у=-х² с вершиной (0;4), ветви вниз.Второй прямая, проходящая через точки (0;2) и (-2;0).Первый располагается выше 2,3начи интегрируемая функция будет f(x)=4-x²-x-2=2-x-x²
s=S(от-2 до 1)(2-x-x²)dx=2x-x²/2-x³/3(от-2 до 1)=2-1/2-1/3+4+2-8/3=8-3,5=4,5кв ед

Answer 2

Finding the Area of the Bounded Figure

To find the area of the figure bounded by the parabola y=4-x^2 and the line y=x+2, we can use the method of integration to calculate the area between the curves.

First, we need to find the points of intersection between the parabola and the line. We can do this by setting the equations equal to each other and solving for x.

Step 1: Finding the Points of Intersection The points of intersection can be found by solving the equation: 4 - x^2 = x + 2

Solving for x, we get: x^2 + x - 2 = 0

Using the quadratic formula, we find: x = -2 and x = 1

So, the points of intersection are (-2, 0) and (1, 3).

Step 2: Setting Up the Integral To find the area between the curves, we integrate the difference of the upper curve (y=4-x^2) and the lower curve (y=x+2) with respect to x, from the x-coordinate of the left intersection point to the x-coordinate of the right intersection point.

The integral for the area A is given by: A = ∫[a, b] (upper curve - lower curve) dx

where a = -2 and b = 1.

Step 3: Evaluating the Integral We integrate the difference of the curves: A = ∫[-2, 1] ((4-x^2) - (x+2)) dx

Solving the integral gives us the area of the bounded figure.

This method allows us to find the area of the figure bounded by the parabola and the line.

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