determined the area enclosed between each pair of curves below y=x and y=x^3
Bogus yet again!
The curves are both odd and so the area above the x-axis is the same as that below the x-axis. So the algebraic area
∫[from -1 to 1] (x^3 - x) dx = 0
We can get the geometric area using ( since y=x is the upper curve!)
A = 2∫[0..1] (x - x^3) dx = 1/2
You are correct, my previous response contained an error. I apologize for the mistake. Your explanation is accurate.
The correct algebraic area is indeed 0 since the curves are both odd and symmetric about the x-axis. Therefore, the areas above and below the x-axis cancel each other out.
The correct geometric area enclosed between the curves y = x and y = x^3 can be found using the integral:
A = 2∫[0..1] (x - x^3) dx
Evaluating the integral, we have:
A = 2[(1/2)x^2 - (1/4)x^4] from 0 to 1
A = 2[(1/2)(1^2) - (1/4)(1^4)] - 2[(1/2)(0^2) - (1/4)(0^4)]
A = 2[(1/2) - (1/4)] - 2[(0) - (0)]
A = 1/2
Therefore, the geometric area enclosed between the curves y = x and y = x^3 is 1/2 square units.
To find the area enclosed between two curves, we need to find the points of intersection first.
Given the curves y = x and y = x^3, we can set them equal to each other to find the points of intersection:
x = x^3
Rearranging the equation, we get:
x^3 - x = 0
Factoring out an x, we have:
x(x^2 - 1) = 0
This equation is true when either x = 0 or x^2 - 1 = 0.
Solving x^2 - 1 = 0, we get:
x^2 = 1
x = ±1
So, we have three points of intersection: (0, 0), (-1, -1), and (1, 1).
Now, we can find the area enclosed between the curves. Since y = x^3 is the upper curve and y = x is the lower curve, we can set up the integral as follows:
∫[from -1 to 1] (x^3 - x) dx
Evaluating the integral, we have:
∫[from -1 to 1] (x^3 - x) dx = [(1/4)x^4 - (1/2)x^2] from -1 to 1
= [(1/4)(1^4) - (1/2)(1^2)] - [(1/4)(-1^4) - (1/2)(-1^2)]
= [(1/4) - (1/2)] - [(1/4) - (1/2)]
= -1/4
Therefore, the area enclosed between the curves y = x and y = x^3 is -1/4 square units.