Use the mid-point rule with n = 4 to approximate the area of the region bounded by y = x^3 and y = x.?
I know how to use the midpoint rule to get the area under a curve but I'm confused on how to get the area between the two curves. Do I subtract them somehow? How do I set it up?
To approximate the area between the curves y = x^3 and y = x using the midpoint rule, you need to first find the interval over which you are calculating the area.
Given that n = 4, the interval will be divided into four equal subintervals. To do this, you need to find the limits of integration for x.
To find the limits of integration, you need to determine the x-values at which the two curves intersect. This occurs when x^3 = x. So, set x^3 = x:
x^3 - x = 0
Factor out an x:
x(x^2 - 1) = 0
In solving this equation, you find two values for x: x = 0 and x = ±1. These are the x-values at which the two curves intersect.
Next, you need to determine the subinterval width (Δx) for the midpoint rule. Since n = 4 and you have three x-values to consider (0, -1, and 1), the subinterval width will be Δx = (1 - (-1))/4 = 1/2.
Now, we can set up the summation for the midpoint rule:
Area ≈ Σ[f(c_i)Δx]
where
f(c_i) is the function evaluated at the midpoint of each subinterval, and
Δx is the subinterval width.
The midpoint of each subinterval can be found by taking the average of the left and right endpoint of the subinterval.
For the given problem, we have four subintervals with the midpoints as follows:
c_1 = (-1 - 0)/2 = -1/2
c_2 = (0 + 1/2)/2 = 1/4
c_3 = (1/2 + 1)/2 = 3/4
c_4 = (1 + 2)/2 = 3/2
Now, evaluate the function at each midpoint:
f(c_1) = (-1/2)^3 = -1/8
f(c_2) = (1/4)^3 = 1/64
f(c_3) = (3/4)^3 = 27/64
f(c_4) = (3/2)^3 = 27/8
Finally, compute the area using the midpoint rule:
Area ≈ Σ[f(c_i)Δx]
≈ (-1/8) * (1/2) + (1/64) * (1/2) + (27/64) * (1/2) + (27/8) * (1/2)
≈ -1/16 + 1/128 + 27/128 + 27/16
≈ 1 + 27/16
≈ 43/16
Therefore, the approximate area, using the midpoint rule with n = 4, is 43/16.
yes. the height of each rectangle is the distance between the curves: x-x^3
or, you can consider finding the area below y=x and then subtracting from that the area below y=x^3