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

given that triangle abc is transformed two times to get triangle

a"b"c"

first abc is transformed in to a'b'c' as follows:

the coordinates of a are (1,-1) and transformed into a'(-1,1)

similarly b (4,-2) became b'(-4,2) and

c(7,-2) became c'(-7,2)

i.e. (x,y) becomes (-x,-y)

this is nothing but reflection about a point here origin.

thus first transformation is reflection on the origin.

next is exactly shifting 3 units down

source:

Answer from: Quest

there are infinitely many solutions.

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