another one i'm stumped by
What is the area between the curves x = y and y = x^3 from -1 to y = 1
i got 0.75 for this one, am i correct
Nope. The region is symmetric, so you can just double the area on the interval [0,1]
See
https://www.wolframalpha.com/input/?i=2(integral%5B0..1%5D++(x-x%5E3)dx)
So, what did you do?
PS The above integral gives the total geometric area. If you want the net algebraic area, where regions below the x-axis are subtracted, then the result is zero.
ohhh alright i think i understand, thank you
To find the area between two curves, you can use integration. Here's how you can approach this problem:
1. Sketch the given curves: x = y and y = x^3.
2. Find the points of intersection between the curves by setting them equal to each other:
x = y ----(1)
y = x^3 ----(2)
Substitute equation (1) into equation (2):
x = x^3
This gives you a cubic equation: x^3 - x = 0.
Factor out an x:
x(x^2 - 1) = 0
The solutions to this equation are x = 0, x = -1, and x = 1.
So, the points of intersection are (-1, -1), (0, 0), and (1, 1).
3. Determine the integration limits: You are given the range -1 ≤ x ≤ 1. Since the curves intersect at x = -1 and x = 1, those will be the limits of integration.
4. Write the integral to find the area between the curves:
To find the area between two curves, you subtract one curve from the other and integrate. In this case, you want to integrate the curve y = x^3 minus the curve y = x:
A = ∫[x=-1 to x=1] (x^3 - x) dx
5. Evaluate the integral:
Integrating x^3 - x with respect to x gives you:
A = [1/4 * x^4 - 1/2 * x^2] evaluated from x = -1 to x = 1
Evaluating this expression at the limits of integration gives:
A = [1/4 * (1)^4 - 1/2 * (1)^2] - [1/4 * (-1)^4 - 1/2 * (-1)^2]
A = [1/4 - 1/2] - [1/4 - 1/2]
A = -1/4 + 1/2
A = 1/4
So, the correct answer is 1/4, not 0.75.
Therefore, the area between the curves x = y and y = x^3 from -1 to y = 1 is 1/4.