What is the arc length of $f(x)=x^2-2x+35$ on $x in [1,7]$ ?

Question

1481 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-03-07T17:08:08

The arc length is $3sqrt(145)+1/4ln(12+sqrt145)$ units.

$y=x^2-2x+35$

$y'=2x-2$

Arc length is given by:

$L=int_1^7sqrt(1+(2x-2)^2)dx$

Apply the substitution $2x-2=tantheta$ :

$L=1/2intsec^3thetad theta$

This is a known integral. If you do not have it memorized apply integration by parts or look it up in a table of integrals:

$L=1/4[secthetatantheta+ln|sectheta+tantheta|]$

Reverse the substitution:

$L=1/4[(2x-2)sqrt(1+(2x-2)^2)+ln|(2x-2)+sqrt(1+(2x-2)^2)|]_1^7$

Insert the limits of integration:

$L=3sqrt(145)+1/4ln(12+sqrt145)$

What is the arc length of f(x)=x^2-2x+35f(x)=x^2-2x+35 on x in [1,7]x in [1,7]?