Question

How do you find the length of the curve `y=x^5/6+1/(10x^3)` between `1<=x<=2` ?

Answer 1

We can find the arc length to be `1261/240` by the integral
`L=int_1^2sqrt{1+({dy}/{dx})^2}dx`

Let us look at some details.

By taking the derivative,
`{dy}/{dx}={5x^4)/6-3/{10x^4}`

So, the integrand looks like:
`sqrt{1+({dy}/{dx})^2}=sqrt{({5x^4)/6)^2+1/2+(3/{10x^4})^2`
by completing the square
`=sqrt{({5x^4)/6+3/{10x^4})^2}={5x^4)/6+3/{10x^4}`

Now, we can evaluate the integral.
`L=int_1^2({5x^4)/6+3/{10x^4})dx=[x^5/6-1/{10x^3}]_1^2=1261/240`