Question

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

Answer 1

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

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

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