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

inch

step-by-step explanation:

yard and mile are too big.

Answer from: Quest

x= 1

step-by-step explanation: convert 5\frac{11}{12}5

to improper fraction. use this rule: a \frac{b}{c}=\frac{ac+b}{c}a

ac+b

3.25+0.5x+1\frac{1}{3}+\frac{5}{6}x=\frac{5\times 12+11}{12}3.25+0.5x+1

5×12+11

2 simplify 5\times 125×12 to 6060

3.25+0.5x+1\frac{1}{3}+\frac{5}{6}x=\frac{60+11}{12}3.25+0.5x+1

60+11

3 simplify 60+1160+11 to 7171

3.25+0.5x+1\frac{1}{3}+\frac{5}{6}x=\frac{71}{12}3.25+0.5x+1

4 simplify \frac{5}{6}x

x to \frac{5x}{6}

3.25+0.5x+1\frac{1}{3}+\frac{5x}{6}=\frac{71}{12}3.25+0.5x+1

5 convert 1\frac{1}{3}1

to improper fraction. use this rule: a \frac{b}{c}=\frac{ac+b}{c}a

ac+b

3.25+0.5x+\frac{1\times 3+1}{3}+\frac{5x}{6}=\frac{71}{12}3.25+0.5x+

1×3+1

6 simplify 1\times 31×3 to 33

3.25+0.5x+\frac{3+1}{3}+\frac{5x}{6}=\frac{71}{12}3.25+0.5x+

3+1

7 simplify 3+13+1 to 44

3.25+0.5x+\frac{4}{3}+\frac{5x}{6}=\frac{71}{12}3.25+0.5x+

8 simplify 3.25+0.5x+\frac{4}{3}+\frac{5x}{6}3.25+0.5x+

to \frac{13.75}{3}+\frac{4x}{3}

13.75

\frac{13.75}{3}+\frac{4x}{3}=\frac{71}{12}

13.75

9 simplify \frac{13.75}{3}

13.75

to 4.5833334.583333

4.583333+\frac{4x}{3}=\frac{71}{12}4.583333+

10 subtract 4.5833334.583333 from both sides

\frac{4x}{3}=\frac{71}{12}-4.583333

−4.583333

11 simplify \frac{71}{12}-4.583333

−4.583333 to \frac{4}{3}

\frac{4x}{3}=\frac{4}{3}

12 multiply both sides by 33

4x=\frac{4}{3}\times 34x=

×3

13 simplify \frac{4}{3}\times 3

×3 to \frac{12}{3}

4x=\frac{12}{3}4x=

14 simplify \frac{12}{3}

to 44

4x=44x=4

15 divide both sides by 44

x=1

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