Find the area enclosed between the curve y=x^3 and the straight line y=x
To find the area enclosed between the curve y=x^3 and the straight line y=x, we need to determine the points of intersection between the two curves and then integrate the difference between the two functions over that interval.
First, let's find the points of intersection.
Setting y = x for both curves, we have:
x = x^3
This simplifies to:
0 = x^3 - x
Factoring out an x, we have:
0 = x(x^2 - 1)
So, we have three possible solutions:
x = 0, x = 1, and x = -1
Now, we need to determine the interval over which we'll integrate. Since the curve y = x^3 is below the line y = x for x < -1 and above the line y = x for x > 1, we only need to consider the interval from x = -1 to x = 1.
So, the area enclosed between the curve y = x^3 and the line y = x is given by the integral:
∫[from -1 to 1] (x - x^3) dx
Integrating this expression gives:
∫[from -1 to 1] (x - x^3) dx = [x^2/2 - x^4/4] [from -1 to 1]
= (1/2 - 1/4) - (1/2 - 1/4)
= 1/4 - 1/4
= 0
Therefore, the area enclosed between the curve y = x^3 and the line y = x is 0 units squared.
To find the area enclosed between the curve y = x^3 and the straight line y = x, we need to determine the points of intersection between the two curves.
First, let's set the equations equal to each other to find the points of intersection:
x^3 = x
Rearranging the equation, we get:
x^3 - x = 0
Factoring out an x, we have:
x(x^2 - 1) = 0
This equation will be satisfied when either x = 0 or x^2 - 1 = 0.
For x = 0, we have one point of intersection (0, 0).
For x^2 - 1 = 0, we can solve for x:
x^2 = 1
Taking the square root, we get:
x = ±1
Therefore, there are two more points of intersection (1, 1) and (-1, -1).
Now, we need to determine which curve is on top between these points. We can do this by comparing the y-values of the two curves.
For the curve y = x^3, we substitute the x-values of the points of intersection:
At (0, 0), y = (0)^3 = 0
At (1, 1), y = (1)^3 = 1
At (-1, -1), y = (-1)^3 = -1
For the line y = x, we substitute the x-values of the points of intersection:
At (0, 0), y = 0
At (1, 1), y = 1
At (-1, -1), y = -1
We can see that for all points, the curve y = x^3 is on top.
To calculate the area enclosed, we integrate the difference between the two curves with respect to x between the points of intersection. The integral expression is as follows:
A = ∫[from -1 to 1] (x^3 - x) dx
Evaluating this integral, we have:
A = [1/4 * x^4 - 1/2 * x^2] [from -1 to 1]
Substituting the upper and lower limits of integration, we get:
A = [1/4 * (1)^4 - 1/2 * (1)^2] - [1/4 * (-1)^4 - 1/2 * (-1)^2]
Simplifying further, we have:
A = [1/4 * 1 - 1/2 * 1] - [1/4 * 1 - 1/2 * 1]
A = [1/4 - 1/2] - [1/4 - 1/2]
A = -1/4 - (-1/4)
A = -1/4 + 1/4
A = 0
Therefore, the area enclosed between the curve y = x^3 and the straight line y = x is zero.