Use the mid-point rule with n = 4 to approximate the area of the region bounded by y = x^3 and y = x.?

Answer 1

At intersection x^3-x = 0

x^2 = 1

x = 1 or -1

As the curve is symmetrical use double the area between 0 and 1

n = 4

Width of each rectangle = 0.25

Height of each rectangle x-x^3

h(x) = x-x^3

Midpoints of the rectangles = 0.125, 0.375, 0.625, 0.875

h(0.125) = 0.123

h(0.375) = 0.322

h(0.625) = 0.381

h(0.875) = 0.205

Area of one loop = 0.25*(0.123+0.322+0.381+0.205) = 0.25775

Area of region = 2*0.25775 =

0.5155

See Graph

https://www.desmos.com/calculator/o8qtwoq4qp

Answer 2

x = x³

x³ - x = 0

x(x+1)(x-1) = 0

The two curves intersect at x = -1, 0, 1

The area of the region bounded by the two curves is given by:

A = ∫|x³-x|dx

integration over the interval -1 ≤ x ≤ 1

Performing the integration, A = 1/2

Approximating with the midpoint rule with n = 4:

Midpoints are:

x = -3/4, -1/4, 1/4, 3/4

Δx = 1/2

At the midpoints:

f(x) = |x³ - x|

f(-3/4) = |(-3/4)³ - (-3/4)| = 3/4 - 27/64 = (48-27)/64 = 21/64

f(-1/4) = |(-1/4)³ - (-1/4)| = 1/4 - 1/64 = 15/64

f( 1/4) = |(1/4)³ - (1/4)| = 1/4 - 1/64 = 15/64

f(3/4) = |(3/4)² - (3/4)| = 3/4 - 27/64 = 21/64

Area = (1/2)(21/64 + 15/64 + 15/64 + 21/64)

= (21+15)/64

= 36/64

= 9/16

= 0.5625