Set up the integral that represents the arc length of the curve f(x) = ln(x) + 5 on [1, 3], and then use Simpson's Rule with n =

Question

3 years ago

7

Set up the integral that represents the arc length of the curve f(x) = ln(x) + 5 on [1, 3], and then use Simpson's Rule with n =

4 to approximate the arc length. Then compare the approximation to the actual arc length found by integrating with technology.

Mathematics

1 answer:

marta [7] · Answer 1 · 2022-05-12T08:01:13+03:00

Answer:

The integral for the arc of length is:

$\displaystyle\int_1^3\sqrt{1+\frac{1}{x^2}}dx$

By using Simpon’s rule we get: 1.5355453

And using technology we get: 2.3020

The approximation is about 33% smaller than the exact result.

Explanation:

The formula for the length of arc of the function f(x) in the interval [a,b] is:

$\displaystyle\int_a^b \sqrt{1+[f'(x)]^2}dx$

We need the derivative of the function:

$f'(x)=\frac{1}{x}$

And we need it squared:

$[f'(x)]^2=\frac{1}{x^2}$

Then the integral is:

$\displaystyle\int_1^3\sqrt{1+\frac{1}{x^2}}dx$

Now, the Simposn’s rule with n=4 is:

$\displaystyle\int_a^b g(x)}dx\approx\frac{\Delta x}{3}\left( g(a)+4g(a+\Delta x)+2g(a+2\Delta x) +4g(a+3\Delta x)+g(b) \right)$

In this problem:

$a=1,b=3,n=4, \displaystyle\Delta x=\frac{b-a}{n}=\frac{2}{4}=\frac{1}{2},g(x)= \sqrt{1+\frac{1}{x^2}}$

So, the Simposn’s rule formula becomes:

$\displaystyle\int_1^3\sqrt{1+\frac{1}{x^2}}dx\\\approx \frac{\frac{1}{3}}{3}\left( \sqrt{1+\frac{1}{1^2}} +4\sqrt{1+\frac{1}{\left(1+\frac{1}{2}\right)^2}} +2\sqrt{1+\frac{1}{\left(1+\frac{2}{2}\right)^2}} +4\sqrt{1+\frac{1}{\left(1+\frac{3}{2}\right)^2}} +\sqrt{1+\frac{1}{3^2}} \right)$

Then simplifying a bit:

$\displaystyle\int_1^3\sqrt{1+\frac{1}{x^2}}dx \approx \frac{1}{9}\left( \sqrt{1+\frac{1}{1^2}} +4\sqrt{1+\frac{1}{\left(\frac{3}{2}\right)^2}} +2\sqrt{1+\frac{1}{\left(2\right)^2}} +4\sqrt{1+\frac{1}{\left(\frac{5}{2}\right)^2}} +\sqrt{1+\frac{1}{3^2}} \right)$