Question

Set up but do not evaluate an integral for the length of x^2/16.

Answer 1

`L=\int\limits^b_a {\sqrt{1+(x^2)/(64)}} \, dx`

Explanation:

Recall the formula for the arc length of a function given a closed interval
`[a,b]`:
`L=\int\limits^b_a {\sqrt{1+((dy)/(dx))^2}} \, dx`

Given the function
`y=(x^2)/(16)`, then
`(dy)/(dx)=(x)/(8)` by the product rule.

Thus, the set-up integral would be:

`L=\int\limits^b_a {\sqrt{1+((x)/(8))^2} } \, dx\\ \\L=\int\limits^b_a {\sqrt{1+(x^2)/(64)}} \, dx`