Question

Use the mid-point rule with n = 4 to approximate the area of the region bounded by y = x3 and y = x. (10 points)

Answer 1

See the graph attached.

The midpoint rule states that you can calculate the area under a curve by using the formula:

`M_(n) = (b - a)/(2) [ f((x_(0) + x_(1) )/(2)) + f((x_(1) + x_(2) )/(2)) + ... + f((x_(n-1) + x_(n) )/(2))]`

In your case:
a = 0
b = 1
n = 4
x₀ = 0
x₁ = 1/4
x₂ = 1/2
x₃ = 3/4
x₄ = 1

Therefore, you'll have:

`M_(4) = (1 - 0)/(4) [ f((0 + (1)/(4) )/(2)) + f(( (1)/(4) + (1)/(2) )/(2)) + f(((1)/(2) + (3)/(4) )/(2)) + f(\frac{(3)/(4) + 1} {2})]`

`M_(4) = (1)/(4) [ f((1)/(8)) + f((3)/(8)) + f((5)/(8)) + f((7)/(8))]`

Now, to evaluate your f(x), you need to look at the graph and notice that:
f(x) = x - x³

Therefore:

`M_(4) = (1)/(4) [((1)/(8) - ((1)/(8))^(3)) + ((3)/(8) - ((3)/(8))^(3)) + ((5)/(8) - ((5)/(8))^(3)) + ((7)/(8) - ((7)/(8))^(3))]`


`M_(4) = (1)/(4) [((1)/(8) - (1)/(512)) + ((3)/(8) - (27)/(512)) + ((5)/(8) - (125)/(512)) + ((7)/(8) - (343)/(512))]`

M₄ = 1/4 · (2 - 478/512)
= 0.2666

Hence, the area of the region bounded by y = x³ and y = x is approximately 0.267 square units.