See the graph attached.
The midpoint rule states that you can calculate the area under a curve by using the formula:
[tex]M_{n} = \frac{b - a}{2} [ f(\frac{x_{0} + x_{1} }{2}) + f(\frac{x_{1} + x_{2} }{2}) + ... + f(\frac{x_{n-1} + x_{n} }{2})][/tex]
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:
[tex]M_{4} = \frac{1 - 0}{4} [ f(\frac{0 + \frac{1}{4} }{2}) + f(\frac{ \frac{1}{4} + \frac{1}{2} }{2}) + f(\frac{\frac{1}{2} + \frac{3}{4} }{2}) + f(\frac{\frac{3}{4} + 1} {2})][/tex]
[tex]M_{4} = \frac{1}{4} [ f(\frac{1}{8}) + f(\frac{3}{8}) + f(\frac{5}{8}) + f(\frac{7}{8})][/tex]
Now, to evaluate your f(x), you need to look at the graph and notice that:
f(x) = x - x³
Therefore:
[tex]M_{4} = \frac{1}{4} [(\frac{1}{8} - (\frac{1}{8})^{3}) + (\frac{3}{8} - (\frac{3}{8})^{3}) + (\frac{5}{8} - (\frac{5}{8})^{3}) + (\frac{7}{8} - (\frac{7}{8})^{3})][/tex]
[tex]M_{4} = \frac{1}{4} [(\frac{1}{8} - \frac{1}{512}) + (\frac{3}{8} - \frac{27}{512}) + (\frac{5}{8} - \frac{125}{512}) + (\frac{7}{8} - \frac{343}{512})][/tex]
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.
