What is the area of the figure shown?
2 answers:
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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.
The midpoint rule states that you can calculate the area under a curve by using the formula:
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:
Now, to evaluate your f(x), you need to look at the graph and notice that:
f(x) = x - x³
Therefore:
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.
Plotting the data (attached photo) roughly shows that the data is skewed to the left. In other words, data is skewed negatively and that the long tail will be on the negative side of the peak.
In such a scenario, interquartile range is normally the best measure to compare variations of data.
Therefore, the last option is the best for the data provided.
In such a scenario, interquartile range is normally the best measure to compare variations of data.
Therefore, the last option is the best for the data provided.