To find the area enclosed by the curve y = x(x - 2) and the x-axis, we can use the definite integral. The definite integral of a function f(x) between the limits a and b is given by the definite integral
∫b a f(x)dx
The definite integral of the function f(x) = x(x - 2) between the limits a and b can be solved by antiderivative
∫b a x(x-2)dx = [1/3x^3 - x^2]b - [1/3x^3 - x^2]a
We want to find the area enclosed between the curve and the x-axis, so we need to find the definite integral between the limits of the x-intercepts of the function. The x-intercepts of the function y = x(x - 2) are x = 0, x = 2. So the definite integral between the limits is
∫2 0 x(x-2)dx = [1/3x^3 - x^2]2 - [1/3x^3 - x^2]0
This evaluates to 1/3*8-4-0 = 4/3
So the area enclosed by the curve y = x(x - 2) and the x-axis is 4/3 square units.
![To find the area enclosed by the curve y = x(x - 2) and the x-axis, we can use the definite integral. The definite integral of a function f(x) between the limits a and b is given by the definite integral ∫b a f(x)dx The definite integral of the function f(x) = x(x - 2) between the limits a and b can be solved by antiderivative ∫b a x(x-2)dx = [1/3x^3 - x^2]b - [1/3x^3 - x^2]a We want to find the area enclosed between the curve and the x-axis, so we need to find the definite integral between the limits of the x-intercepts of the function. The x-intercepts of the function y = x(x - 2) are x = 0, x = 2. So the definite integral between the limits is ∫2 0 x(x-2)dx = [1/3x^3 - x^2]2 - [1/3x^3 - x^2]0 This evaluates to 1/3*8-4-0 = 2 So the area enclosed by the curve y = x(x - 2) and the x-axis is 2 square units.](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEi8FvFU1zgz2h257ZyvEgnzStEICLjq7FrCIcAjXSX2REvTVm-YdNwLk9hDf6ECsnimpsGsQldVD3UXEyWOYgQyJMjnWQRsLLWrUCNAmjBqqeE2LTYUeXRv3jGYC91VzGcGqvvip9HmfbvBjG9I-9qG5A5Nqnw6luSghJCt2en2EYuNwbdJ5_srIRSK8g/s2000/1674494953895.jpg "To find the area enclosed by the curve y = x(x - 2) and the x-axis, we can use the definite integral. The definite integral of a function f(x) between the limits a and b is given by the definite integral ∫b a f(x)dx The definite integral of the function f(x) = x(x - 2) between the limits a and b can be solved by antiderivative ∫b a x(x-2)dx = [1/3x^3 - x^2]b - [1/3x^3 - x^2]a We want to find the area enclosed between the curve and the x-axis, so we need to find the definite integral between the limits of the x-intercepts of the function. The x-intercepts of the function y = x(x - 2) are x = 0, x = 2. So the definite integral between the limits is ∫2 0 x(x-2)dx = [1/3x^3 - x^2]2 - [1/3x^3 - x^2]0 This evaluates to 1/3*8-4-0 = 2 So the area enclosed by the curve y = x(x - 2) and the x-axis is 2 square units.")
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