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Задача Задача №37191
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08 ноября 2023 07:00 · публичный вопрос
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Find the area enclosed between the curve y=4-x²

Нейросеть08 ноября 2023 07:00

To find the area enclosed between the curve y=4-x², we need to integrate the function over a specific interval.

First, let's find the points where the curve intersects with the x-axis by setting y=0:

0 = 4 - x²

Rearranging the equation, we get:

x² = 4

Taking the square root of both sides, we find:

x = ±2

So, the curve intersects the x-axis at x=-2 and x=2.

To find the area enclosed between the curve and the x-axis, we can integrate the function 4-x² between these two points:

Area = ∫(4 - x²) dx (from x=-2 to x=2)

Integrating, we get:

Area = [4x - (x³/3)] (from x=-2 to x=2)

Evaluating the expression at the limits, we have:

Area = [4(2) - (2³/3)] - [4(-2) - (-2³/3)]

Simplifying, we get:

Area = [8 - (8/3)] - [-8 - (-8/3)]

Area = 8/3 - 8/3

Area = 0

Therefore, the area enclosed between the curve y=4-x² and the x-axis is 0 units².

find the area enclosed between the curve y=4-x²
To find the area enclosed between the curve y=4-x², we need to integrate the function over a specific interval. First, let's find the points where the curve intersects with the x-axis by setting y=0: 0 = 4 - x² Rearranging the equation, we get: x² = 4 Taking the square root of both sides, we find: x = ±2 So, the curve intersects the x-axis at x=-2 and x=2. To find the area enclosed between the curve and the x-axis, we can integrate the function 4-x² between these two points: Area = ∫(4 - x²) dx (from x=-2 to x=2) Integrating, we get: Area = [4x - (x³/3)] (from x=-2 to x=2) Evaluating the expression at the limits, we have: Area = [4(2) - (2³/3)] - [4(-2) - (-2³/3)] Simplifying, we get: Area = [8 - (8/3)] - [-8 - (-8/3)] Area = 8/3 - 8/3 Area = 0 Therefore, the area enclosed between the curve y=4-x² and the x-axis is 0 units².