Question

Найдите площадь фигуры, ограниченной графиком функции у=х^2+8х+16 и осями координат

Answer 1

Что-то накалякал я, но ответ получил ====> см. прикрепленный файл.

Answer 2

Finding the Area of the Figure Bounded by the Graph of the Function y = x^2 + 8x + 16 and the Coordinate Axes

To find the area of the figure bounded by the graph of the function y = x^2 + 8x + 16 and the coordinate axes, we need to determine the points where the graph intersects the x-axis and the y-axis. These points will define the boundaries of the figure.

The graph of the function y = x^2 + 8x + 16 is a parabola that opens upwards. To find the x-intercepts, we set y = 0 and solve for x:

x^2 + 8x + 16 = 0

Using the quadratic formula, we can find the solutions for x:

x = (-b ± √(b^2 - 4ac)) / (2a)

In this case, a = 1, b = 8, and c = 16. Plugging in these values, we get:

x = (-8 ± √(8^2 - 4(1)(16))) / (2(1))

Simplifying further:

x = (-8 ± √(64 - 64)) / 2

x = (-8 ± √0) / 2

x = -8 / 2

x = -4

Therefore, the graph intersects the x-axis at x = -4. This means that the left boundary of the figure is x = -4.

To find the y-intercept, we set x = 0 and solve for y:

y = (0)^2 + 8(0) + 16

y = 16

Therefore, the graph intersects the y-axis at y = 16. This means that the top boundary of the figure is y = 16.

Now, we have the boundaries of the figure: x = -4 (left), x = 0 (right), y = 0 (bottom), and y = 16 (top).

To find the area of the figure, we need to calculate the area under the curve y = x^2 + 8x + 16 between x = -4 and x = 0. This can be done by integrating the function over this interval:

A = ∫[from -4 to 0] (x^2 + 8x + 16) dx

Integrating the function, we get:

A = [(1/3)x^3 + 4x^2 + 16x] [from -4 to 0]

Substituting the values, we get:

A = [(1/3)(0)^3 + 4(0)^2 + 16(0)] - [(1/3)(-4)^3 + 4(-4)^2 + 16(-4)]

Simplifying further:

A = [0 + 0 + 0] - [(-64/3) + 64 - 64]

A = 0 - (-64/3)

A = 64/3

Therefore, the area of the figure bounded by the graph of the function y = x^2 + 8x + 16 and the coordinate axes is 64/3 square units.