Answer:
The approximate area of the region using the midpoint rule with n = 2 is about 2.8418 square units.
Step-by-step explanation:
We want to use the midpoint rule with n = 2 to approximate the area of the region bounded by the graphs of:
[tex]\displaystyle y=\sqrt[3]{16-x^3}, y=x\text{ and } x=0[/tex]
This is shown in the graph below. Our area of interest is between the green, red, and blue lines.
First, find the intersection point of the two graphs:
[tex]\displaystyle x=\sqrt[3]{16-x^3}\Rightarrow x=2\Rightarrow (2, 2)[/tex]
Therefore, we will want to approximate from x = 0 to x = 2.
Note that the red graph (cubic root) is greater than the linear function for all x in the interval [0, 2].
Since the cubic function is greater than the linear function for all x in [0, 2], the area of the region will be the linear function subtracted from the cubic function. Hence:
[tex]\displaystyle \text{Area}=\int_{0}^2\sqrt[3]{16-x^3}-x\, dx[/tex]
We will let:
[tex]f(x)=\sqrt[3]{16-x^3}-x[/tex]
The midpoint rule is given by:
[tex]\displaystyle \text{Area}\approx \sum_{i=1}^nf\left(\frac{x_{i-1}+x_i}{2}\right)\Delta x[/tex]
Since the area is on the interval [0, 2] and n = 2, find the width:
[tex]\displaystyle \Delta x=\frac{2-0}{2}=1[/tex]
Hence, our x-coordinates are x₀ = 0, x₁ = 1, and x₂ = 2.
Therefore, by the midpoint rule:
[tex]\displaystyle \text{Area}\approx \sum_{i=1}^2f\left(\frac{x_{i-1}+x_i}{2}\right)(1)[/tex]
Substitute:
[tex]\displaystyle \text{Area}\approx (1)\left(f\left(\frac{0+1}{2}\right)+f\left(\frac{1+2}{2}\right)\right)[/tex]
Evaluate:
[tex]\displaystyle \text{Area}\approx\displaystyle f\left(0.5\right)+f(1.5)[/tex]
Evaluate:
[tex]\displaystyle \text{Area}\approx(\sqrt[3]{16-(.5)^3}-(.5))+(\sqrt[3]{16-(1.5)^3}-(1.5))}\approx 2.8418[/tex]
Hence, the approximate area of the region is about 2.8418 square units.
Notes:
The exact area is 2.80436... square units.
