# Arc length. Set up the integral required to find the length of from [0,1] analytically. Use midpoint rule with at least 10 sub-intervals to APPROXIMATE the length of the curve. You may...

**Arc length**.

- Set up the integral required to find the length of from [0,1] analytically.
- Use midpoint rule with at least 10 sub-intervals to APPROXIMATE the length of the curve. You may use a spreadsheet.
- Determine the anti-derivative using calculus techniques, and use it to find the exact value, than approximate to three decimal places.

### 1 Answer | Add Yours

**Set up the integral required to find the length of from [0,1] analytically. **

********Since you did not provide a function, I can only write out the formulas for you.

-The arc length formula is the integral of `ds` .

`L=\int_a^b ds`

Such that in terms of `dx` and `dy` :

`ds =sqrt(1+((dy)/(dx))^2) dx`

`ds =sqrt(1+((dx)/(dy))^2) dy`

Find the length of **...insert function here... ** from [0,1]. Since your function isn't given, I cannot continue on with this problem.

**Use midpoint rule with at least 10 sub-intervals to APPROXIMATE the length of the curve. You may use a spreadsheet.**

- (*I will choose to use the minimum 10 sub-intervals)

- The midpoint formula to approximate integrals is:

`int_a^b f(x) dx ~~Delta x[f(x_1)+f(x_2)+f(x_3)+...+f(x_n)]`

Where [a,b]=[0,1], and delta x is:

`Delta x = (b-a)/n = (1-0)/10 = 1/10`

We are using 10 `n` sub-intervals, from zero to one defined by [0,1]:

0,0.1,0.2,0.3,0.4,0.5,0.6,0.7,0.8,0.9, and 1.0

Then, our midpoints will be in between our sub-intervals. Substituting into the formula, we will have:

`int_0^1 f(x) dx ~~(1/10)[f(0.5)+f(1.5)+f(2.5)+...+f(9.5)]`

You will need to evaluate the function at: `f(0.5), f(1.5), f(2.5)...f(9.5)`

After you have found the values at their midpoints, you can substitute those values into the equation, add the values, and multiply the quantity by one-tenth.

**D****etermine the anti-derivative using calculus techniques, and use it to**

**find the exact value, than approximate to three decimal places.**

I do not know what type of function your question has given, so I cannot continue.