Question

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.

Answer 1

The integral for the arc of length is:

`\displaystyle\int_1^3\sqrt{1+(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 √(1+[f'(x)]^2)dx`

We need the derivative of the function:

`f'(x)=(1)/(x)`

And we need it squared:

`[f'(x)]^2=(1)/(x^2)`

Then the integral is:

`\displaystyle\int_1^3\sqrt{1+(1)/(x^2)}dx`

Now, the Simposn’s rule with n=4 is:

`\displaystyle\int_a^b g(x)}dx\approx(\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=(b-a)/(n)=(2)/(4)=(1)/(2),g(x)= \sqrt{1+(1)/(x^2)}`

So, the Simposn’s rule formula becomes:

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

Then simplifying a bit:

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

Then we just do those computations and we finally get the approximation via Simposn's rule:

`\displaystyle\int_1^3\sqrt{1+(1)/(x^2)}dx\approx 1.5355453`

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:

`\displaystyle(2.3020-1.5355453)/(2.3020)\cdot100\%\approx 33\%`