Mathematics, 17.05.2021 01:00 kamand10

HELPPP! Use the mid-point rule with n = 2 to approximate the area of the region bounded by y equals the cube root of the quantity 16 minus x cubed y = x, and x = 0.
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Answer from: andrewrangel63

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:

$\displaystyle y=\sqrt[3]{16-x^3}, y=x\text{ and } x=0$

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:

$\displaystyle x=\sqrt[3]{16-x^3}\Rightarrow x=2\Rightarrow (2, 2)$

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:

$\displaystyle \text{Area}=\int_{0}^2\sqrt[3]{16-x^3}-x\, dx$

We will let:

$f(x)=\sqrt[3]{16-x^3}-x$

The midpoint rule is given by:

$\displaystyle \text{Area}\approx \sum_{i=1}^nf\left(\frac{x_{i-1}+x_i}{2}\right)\Delta x$

Since the area is on the interval [0, 2] and n = 2, find the width:

$\displaystyle \Delta x=\frac{2-0}{2}=1$

Hence, our x-coordinates are x₀ = 0, x₁ = 1, and x₂ = 2.

Therefore, by the midpoint rule:

$\displaystyle \text{Area}\approx \sum_{i=1}^2f\left(\frac{x_{i-1}+x_i}{2}\right)(1)$

Substitute:

$\displaystyle \text{Area}\approx (1)\left(f\left(\frac{0+1}{2}\right)+f\left(\frac{1+2}{2}\right)\right)$

Evaluate:

$\displaystyle \text{Area}\approx\displaystyle f\left(0.5\right)+f(1.5)$

Evaluate:

$\displaystyle \text{Area}\approx(\sqrt[3]{16-(.5)^3}-(.5))+(\sqrt[3]{16-(1.5)^3}-(1.5))}\approx 2.8418$

Hence, the approximate area of the region is about 2.8418 square units.

Notes:

The exact area is 2.80436... square units.

HELPPP!

Use the mid-point rule with n = 2 to approximate the area of the region bounded by y equals

Answer from: Quest

(2 x^3 + 7 x^2 + 5 x + 2)/(x (x + 2))

step-by-step explanation:

simplify the following:

x/x^2 + 3 x + 2 - (x (x + 1))/(x + 2)

combine powers. x/x^2 = x^(1 - 2):

x^(1 - 2) + 3 x + 2 - (x (x + 1))/(x + 2)

1 - 2 = -1:

x^(-1) + 3 x + 2 - (x (x + 1))/(x + 2)

put each term in 1/x + 3 x + 2 - (x (x + 1))/(x + 2) over the common denominator x (x + 2): 1/x + 3 x + 2 - (x (x + 1))/(x + 2) = (x + 2)/(x (x + 2)) + (3 x^2 (x + 2))/(x (x + 2)) + (2 x (x + 2))/(x (x + 2)) - (x^2 (x + 1))/(x (x + 2)):

(x + 2)/(x (x + 2)) + (3 x^2 (x + 2))/(x (x + 2)) + (2 x (x + 2))/(x (x + 2)) - (x^2 (x + 1))/(x (x + 2))

(x + 2)/(x (x + 2)) + (3 x^2 (x + 2))/(x (x + 2)) + (2 x (x + 2))/(x (x + 2)) - (x^2 (x + 1))/(x (x + 2)) = ((x + 2) + 3 x^2 (x + 2) + 2 x (x + 2) - x^2 (x + 1))/(x (x + 2)):

(2 + x + 3 x^2 (x + 2) + 2 x (x + 2) - x^2 (x + 1))/(x (x + 2))

3 x^2 (x + 2) = 3 x^3 + 6 x^2:

(2 + x + 3 x^3 + 6 x^2 + 2 x (x + 2) - x^2 (x + 1))/(x (x + 2))

2 x (x + 2) = 2 x^2 + 4 x:

(2 + x + 6 x^2 + 3 x^3 + 2 x^2 + 4 x - x^2 (x + 1))/(x (x + 2))

-x^2 (x + 1) = -x^2 - x^3:

(-x^2 - x^3 + 3 x^3 + 6 x^2 + 2 x^2 + 4 x + x + 2)/(x (x + 2))

grouping like terms, 3 x^3 - x^3 + 6 x^2 + 2 x^2 - x^2 + 4 x + x + 2 = (3 x^3 - x^3) + (6 x^2 + 2 x^2 - x^2) + (x + 4 x) + 2:

((3 x^3 - x^3) + (6 x^2 + 2 x^2 - x^2) + (x + 4 x) + 2)/(x (x + 2))

3 x^3 - x^3 = 2 x^3:

(2 x^3 + (6 x^2 + 2 x^2 - x^2) + (x + 4 x) + 2)/(x (x + 2))

6 x^2 + 2 x^2 - x^2 = 7 x^2:

(2 x^3 + 7 x^2 + (x + 4 x) + 2)/(x (x + 2))

x + 4 x = 5 x:

answer: (2 x^3 + 7 x^2 + 5 x + 2)/(x (x + 2))

Answer from: Quest

The answer is

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HELPPP! Use the mid-point rule with n = 2 to approximate the area of the region bounded by y equals...