Question

Найти площадь фигуры ограниченной графиком функции y=-x2+4 и прямой у=0

Answer 1

Найдем общие точки у графиков:
$-x^2+4=0$
$x^2=4$
$x=(-2),2$

Теперь вычислим определенный интеграл от -2 до 2:
$\int\limits^{-2}_2 {-x^2+4} \, dx = \frac{-x^3}{3}+4x \Big|_{2}^{-2}=(- \frac{1}{3}*(-8)-8)-( -\frac{1}{3}*8+8)=( \frac{8-24}{3})-( \frac{24-8}{3})= \frac{-16}{3}-\frac{16}{3}= -\frac{32}{3}=-10\frac{2}{3}$
Так как нам нужна именно площадь а не интеграл, то мы представляем данное значение как модуль (площадь всегда положительна):
$|-10 \frac{2}{3}|=10 \frac{2}{3}$

Answer 2

Finding the Area of the Figure Bounded by the Graphs of y = -x^2 + 4 and y = 0

To find the area of the figure bounded by the graphs of the functions y = -x^2 + 4 and y = 0, we need to determine the points of intersection between the two functions and then calculate the area between these points.

The first step is to find the points of intersection by setting the two equations equal to each other:

-y = -x^2 + 4

Simplifying the equation, we get:

x^2 - y = 4

Since y = 0 for the second equation, we can substitute it into the first equation:

x^2 - 0 = 4

This simplifies to:

x^2 = 4

Taking the square root of both sides, we find:

x = ±2

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

Now we can calculate the area between these points. Since the graph of y = -x^2 + 4 is below the x-axis, we need to take the absolute value of the function to calculate the area.

The area can be calculated using the definite integral:

A = ∫[a, b] |f(x)| dx

Where a and b are the x-coordinates of the points of intersection.

Substituting the values, we have:

A = ∫[-2, 2] |-x^2 + 4| dx

Integrating the absolute value function, we get:

A = ∫[-2, 2] (x^2 - 4) dx

Evaluating the integral, we have:

A = [x^3/3 - 4x] [-2, 2]

Plugging in the values, we get:

A = [(2^3/3 - 4(2)) - ((-2)^3/3 - 4(-2))]

Simplifying further, we have:

A = [(8/3 - 8) - (-8/3 + 8)]

A = [8/3 - 8 + 8/3 - 8]

A = [16/3 - 16]

A = -16/3

Therefore, the area of the figure bounded by the graphs of y = -x^2 + 4 and y = 0 is -16/3 square units.