Answer :
Answer:
The integral for the arc of length is:
[tex]\displaystyle\int_1^3\sqrt{1+\frac{1}{x^2}}dx [/tex]
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:
[tex]\displaystyle\int_a^b \sqrt{1+[f'(x)]^2}dx[/tex]
We need the derivative of the function:
[tex]f'(x)=\frac{1}{x}[/tex]
And we need it squared:
[tex][f'(x)]^2=\frac{1}{x^2}[/tex]
Then the integral is:
[tex]\displaystyle\int_1^3\sqrt{1+\frac{1}{x^2}}dx[/tex]
Now, the Simposn’s rule with n=4 is:
[tex]\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)[/tex]
In this problem:
[tex]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}} [/tex]
So, the Simposn’s rule formula becomes:
[tex]\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)[/tex]
Then simplifying a bit:
[tex]\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)[/tex]
Then we just do those computations and we finally get the approximation via Simposn's rule:
[tex]\displaystyle\int_1^3\sqrt{1+\frac{1}{x^2}}dx\approx 1.5355453[/tex]
While when we do the integral by using technology we get: 2.3020.
The approximation with Simpon’s rule is close but about 33% smaller:
[tex]\displaystyle\frac{2.3020-1.5355453}{2.3020}\cdot100\%\approx 33\% [/tex]