Mathematics, 06.08.2019 04:30 bloodry

Use simpson's rule with n = 10 to estimate the arc length of the curve. compare your answer with the value of the integral produced by your calculator. (round your answers to six decimal places.) y = ln(6 + x3), 0 ≤ x ≤ 5

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Using Simpson's Rule with n = 10 to estimate the arc length of the curve. The value of the integral produced is 8.986792. The value of produced by my graphing calculator is 5.990861.

Let's consider the given function;

y = ㏑ (6 + x³), 0 ≤ x ≤ 5

The arc length (L) can be computed by using the formula:

$\mathbf{L= \int^b_a \sqrt{1+ \Big(\dfrac{dy}{dx}\Big)^2\ dx}}$

Since the interval varies from 0 to 5,

Then, the integral for the arc length will be;

$\mathbf{L= \int^5_0 \sqrt{1+ \Big(\dfrac{3x^2}{6+x^3}\Big)^2\ dx}}$

Let us assume that arc length (L) can be written as:

$\mathbf{L= \int^5_0 f(x) \ dx}$

where,

the function f(x) is equal to:

$\mathbf{f(x) = \sqrt{1+ \Big(\dfrac{3x^2}{6+x^3} \Big)^2}}$

The resolution of the arc length can be given as:

$\mathbf{\Delta x = \dfrac{b-a}{N}}$

$\mathbf{\Delta x = \dfrac{5-0}{10}}$

$\mathbf{\Delta x = \dfrac{5}{10}}$

$\mathbf{\Delta x =0.5}$

However, intervals of $\mathbf{X_i}$ can be expressed as:

$\mathbf{X_i = \{0, \ 0.5, \ 1, \ 1.5, \ 2, \ 2.5, \ 3, \ 3.5,\ 4, \ 4.5, \ 5 \}}$

Now, by applying Simpson's rule;

$\mathbf{L=\dfrac{\Delta x}{2}(35.9471687354)}$

$\mathbf{L=\dfrac{0.5}{2}(35.9471687354)}$

L = 8.986792 to six decimal places

By using graphing calculator;

$\mathbf{L= \int^5_0 \sqrt{1+ \Big(\dfrac{3x^2}{6+x^3}\Big)^2}\ dx}$

L = 5.99086027171

L = 5.990861

Therefore, we can conclude that the value of the integral produced is 8.986792, the value of produced by my graphing calculator is 5.990861.

Learn more about Simpson's rule here:

link

Answer from: jenadkrson62

$y=\ln(6+x^3)\implies y'=\dfrac{3x^2}{6+x^3}$

The arc length of the curve is

$\displaystyle\int_0^5\sqrt{1+\frac{9x^4}{(6+x^3)^2}}\,\mathrm dx$

which has a value of about 5.99086.

Let $f(x)=\sqrt{1+\frac{9x^4}{(6+x^3)^2}}$ . Split up the interval of integration into 10 subintervals,

[0, 1/2], [1/2, 1], [1, 3/2], ..., [9/2, 5]

The left and right endpoints are given respectively by the sequences,

$\ell_i=\dfrac{i-1}2$

$r_i=\dfrac i2$

with $1\le i\le10$ .

These subintervals have midpoints given by

$m_i=\dfrac{\ell_i+r_i}2=\dfrac{2i-1}4$

Over each subinterval, we approximate f(x) with the quadratic polynomial

$p_i(x)=f(\ell_i)\dfrac{(x-m_i)(x-r_i)}{(\ell_i-m_i)(\ell_i-r_i)}+f(m_i)\dfrac{(x-\ell_i)(x-r_i)}{(m_i-\ell_i)(m_i-r_i)}+f(r_i)\dfrac{(x-\ell_i)(x-m_i)}{(r_i-\ell_i)(r_i-m_i)}$

so that the integral we want to find can be estimated as

$\displaystyle\sum_{i=1}^{10}\int_{\ell_i}^{r_i}p_i(x)\,\mathrm dx$

It turns out that

$\displaystyle\int_{\ell_i}^{r_i}p_i(x)\,\mathrm dx=\frac{f(\ell_i)+4f(m_i)+f(r_i)}6$

so that the arc length is approximately

$\displaystyle\sum_{i=1}^{10}\frac{f(\ell_i)+4f(m_i)+f(r_i)}6\approx5.99086$

Answer from: Quest

14 is the best answer can i get brainliest?

Answer from: Quest

stew reads at the fastest rate of 265 words per minute.

step-by-step explanation:

it's a given that felicia reads at 260 word per minute and on connie's graph we can see that she reads 200 words per minute. since stew's table tells us that he read 795 in 3 minutes, we can divide 795 by 3 to find out how many words he reads per minute.

795/3 = 265

stew reads 265 words per minute - he reads at the fastest rate since felecia reads at 260 words per minute and connie reads at 200 words per minute.

stew (265) > felecia (260) > connie (200)

Need 45 ! compare the rates at which each of the three students read: which student read at a fa

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