[tex]y=\ln(6+x^3)\implies y'=\dfrac{3x^2}{6+x^3}[/tex]
The arc length of the curve is
[tex]\displaystyle\int_0^5\sqrt{1+\frac{9x^4}{(6+x^3)^2}}\,\mathrm dx[/tex]
which has a value of about 5.99086.
Let [tex]f(x)=\sqrt{1+\frac{9x^4}{(6+x^3)^2}}[/tex]. 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,
[tex]\ell_i=\dfrac{i-1}2[/tex]
[tex]r_i=\dfrac i2[/tex]
with [tex]1\le i\le10[/tex].
These subintervals have midpoints given by
[tex]m_i=\dfrac{\ell_i+r_i}2=\dfrac{2i-1}4[/tex]
Over each subinterval, we approximate [tex]f(x)[/tex] with the quadratic polynomial
[tex]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)}[/tex]
so that the integral we want to find can be estimated as
[tex]\displaystyle\sum_{i=1}^{10}\int_{\ell_i}^{r_i}p_i(x)\,\mathrm dx[/tex]
It turns out that
[tex]\displaystyle\int_{\ell_i}^{r_i}p_i(x)\,\mathrm dx=\frac{f(\ell_i)+4f(m_i)+f(r_i)}6[/tex]
so that the arc length is approximately
[tex]\displaystyle\sum_{i=1}^{10}\frac{f(\ell_i)+4f(m_i)+f(r_i)}6\approx5.99086[/tex]
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;
The arc length (L) can be computed by using the formula:
[tex]\mathbf{L= \int^b_a \sqrt{1+ \Big(\dfrac{dy}{dx}\Big)^2\ dx}}[/tex]
Since the interval varies from 0 to 5,
Then, the integral for the arc length will be;
[tex]\mathbf{L= \int^5_0 \sqrt{1+ \Big(\dfrac{3x^2}{6+x^3}\Big)^2\ dx}}[/tex]
Let us assume that arc length (L) can be written as:
[tex]\mathbf{L= \int^5_0 f(x) \ dx}[/tex]
where,
[tex]\mathbf{f(x) = \sqrt{1+ \Big(\dfrac{3x^2}{6+x^3} \Big)^2}}[/tex]
The resolution of the arc length can be given as:
[tex]\mathbf{\Delta x = \dfrac{b-a}{N}}[/tex]
[tex]\mathbf{\Delta x = \dfrac{5-0}{10}}[/tex]
[tex]\mathbf{\Delta x = \dfrac{5}{10}}[/tex]
[tex]\mathbf{\Delta x =0.5}[/tex]
However, intervals of [tex]\mathbf{X_i}[/tex] can be expressed as:
[tex]\mathbf{X_i = \{0, \ 0.5, \ 1, \ 1.5, \ 2, \ 2.5, \ 3, \ 3.5,\ 4, \ 4.5, \ 5 \}}[/tex]
Now, by applying Simpson's rule;
[tex]\mathbf{L=\dfrac{\Delta x}{2}(35.9471687354)}[/tex]
[tex]\mathbf{L=\dfrac{0.5}{2}(35.9471687354)}[/tex]
L = 8.986792 to six decimal places
[tex]\mathbf{L= \int^5_0 \sqrt{1+ \Big(\dfrac{3x^2}{6+x^3}\Big)^2}\ dx}[/tex]
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:
https://brainly.com/question/14915665?referrer=searchResults
