What is the arclength of $f(x)=x-sqrt(x+3)$ on $x in [1,3]$ ?

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Answer 1 · 2018-06-21T20:57:39

$L=(4sqrt6-1)/(8sqrt2)sqrt((4sqrt6-1)^2+1)-35/8+1/(8sqrt2)ln((4sqrt6-1+sqrt((4sqrt6-1)^2+1))/(7+5sqrt2))$ units.

$f(x)=x-sqrt(x+3)$

$f'(x)=1-1/(2sqrt(x+3))$

Arclength is given by:

$L=int_1^3sqrt(1+(1-1/(2sqrt(x+3)))^2)dx$

Apply the substitution $2sqrt(x+3)=u$ :

$L=int_4^(2sqrt6)sqrt(1+(1-1/u)^2)(1/2udu)$

Simplify:

$L=1/2int_4^(2sqrt6)sqrt(2u^2-2u+1)du$

Complete the square in the square root:

$L=1/(2sqrt2)int_4^(2sqrt6)sqrt((2u-1)^2+1)du$

Apply the substitution $2u-1=tantheta$ :

$L=1/(4sqrt2)intsec^3thetad theta$

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:

$L=1/(8sqrt2)[secthetatantheta+ln|sectheta+tantheta|]$

Reverse the substitution:

$L=1/(8sqrt2)[(2u-1)sqrt((2u-1)^2+1)+ln|(2u-1)+sqrt((2u-1)^2+1)|]_4^(2sqrt6)$

Insert the limits of integration:

$L=(4sqrt6-1)/(8sqrt2)sqrt((4sqrt6-1)^2+1)-35/8+1/(8sqrt2)ln((4sqrt6-1+sqrt((4sqrt6-1)^2+1))/(7+5sqrt2))$

What is the arclength of f(x)=x-sqrt(x+3)f(x)=x-sqrt(x+3) on x in [1,3]x in [1,3]?