LFX=AZQ
XLF=QAZ
FXL=ZQA
XFL=QZA
See the graph attached.
The midpoint rule states that you can calculate the area under a curve by using the formula:
![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})]](https://tex.z-dn.net/?f=M_%7Bn%7D%20%3D%20%5Cfrac%7Bb%20-%20a%7D%7B2%7D%20%5B%20f%28%5Cfrac%7Bx_%7B0%7D%20%2B%20x_%7B1%7D%20%7D%7B2%7D%29%20%2B%20%20f%28%5Cfrac%7Bx_%7B1%7D%20%2B%20x_%7B2%7D%20%7D%7B2%7D%29%20%2B%20...%20%2B%20%20f%28%5Cfrac%7Bx_%7Bn-1%7D%20%2B%20x_%7Bn%7D%20%7D%7B2%7D%29%5D)
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} = \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})]](https://tex.z-dn.net/?f=M_%7B4%7D%20%3D%20%5Cfrac%7B1%20-%200%7D%7B4%7D%20%5B%20f%28%5Cfrac%7B0%20%2B%20%20%5Cfrac%7B1%7D%7B4%7D%20%7D%7B2%7D%29%20%2B%20%20f%28%5Cfrac%7B%20%5Cfrac%7B1%7D%7B4%7D%20%2B%20%5Cfrac%7B1%7D%7B2%7D%20%7D%7B2%7D%29%20%2B%20%20f%28%5Cfrac%7B%5Cfrac%7B1%7D%7B2%7D%20%2B%20%5Cfrac%7B3%7D%7B4%7D%20%7D%7B2%7D%29%20%2B%20f%28%5Cfrac%7B%5Cfrac%7B3%7D%7B4%7D%20%2B%201%7D%20%7B2%7D%29%5D)
![M_{4} = \frac{1}{4} [ f(\frac{1}{8}) + f(\frac{3}{8}) + f(\frac{5}{8}) + f(\frac{7}{8})]](https://tex.z-dn.net/?f=M_%7B4%7D%20%3D%20%5Cfrac%7B1%7D%7B4%7D%20%5B%20f%28%5Cfrac%7B1%7D%7B8%7D%29%20%2B%20%20f%28%5Cfrac%7B3%7D%7B8%7D%29%20%2B%20%20f%28%5Cfrac%7B5%7D%7B8%7D%29%20%2B%20f%28%5Cfrac%7B7%7D%7B8%7D%29%5D)
Now, to evaluate your f(x), you need to look at the graph and notice that:
f(x) = x - x³
Therefore:
![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})]](https://tex.z-dn.net/?f=M_%7B4%7D%20%3D%20%5Cfrac%7B1%7D%7B4%7D%20%5B%28%5Cfrac%7B1%7D%7B8%7D%20-%20%28%5Cfrac%7B1%7D%7B8%7D%29%5E%7B3%7D%29%20%2B%20%28%5Cfrac%7B3%7D%7B8%7D%20-%20%28%5Cfrac%7B3%7D%7B8%7D%29%5E%7B3%7D%29%20%2B%20%28%5Cfrac%7B5%7D%7B8%7D%20-%20%28%5Cfrac%7B5%7D%7B8%7D%29%5E%7B3%7D%29%20%2B%20%28%5Cfrac%7B7%7D%7B8%7D%20-%20%28%5Cfrac%7B7%7D%7B8%7D%29%5E%7B3%7D%29%5D)
![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})]](https://tex.z-dn.net/?f=M_%7B4%7D%20%3D%20%5Cfrac%7B1%7D%7B4%7D%20%5B%28%5Cfrac%7B1%7D%7B8%7D%20-%20%5Cfrac%7B1%7D%7B512%7D%29%20%2B%20%28%5Cfrac%7B3%7D%7B8%7D%20-%20%5Cfrac%7B27%7D%7B512%7D%29%20%2B%20%28%5Cfrac%7B5%7D%7B8%7D%20-%20%5Cfrac%7B125%7D%7B512%7D%29%20%2B%20%28%5Cfrac%7B7%7D%7B8%7D%20-%20%5Cfrac%7B343%7D%7B512%7D%29%5D)
M₄ = 1/4 · (2 - 478/512)
= 0.2666
Hence, the <span>area of the region bounded by y = x³ and y = x</span> is approximately
0.267 square units.
Twenty four multiplied by seven is one hundred sixty eight
By using the arc length formula, we will see that the length of the curve is L = 1.48
<h3>
How to use the arc length formula?</h3>
Here we have the curve:
y = 2 - x^2 with 0 ≤ x ≤ 1
And we want to find the length of the curve.
The arc length formula for a curve y in the interval [x₁, x₂] is given by:

For our curve, we have:
dy/dx = -2x
And the interval is [0, 1]
Replacing that we get:

This integral is not trivial, using a table you can see that this is equal to:

evaluated from x = 1 to x = 0, when we do that, we will get:

That is the length of the curve.
If you want to learn more about curve length, you can read:
brainly.com/question/2005046