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What is the arc length of #f(x)=sqrt(x-1) # on #x in [2,6] #?

Answer 1

Emilia Aguilar

#L=sqrt5/2(sqrt21-1)+1/4ln((2sqrt5+sqrt21)/(2+sqrt5))#

Given

#f(x)=sqrt(x-1)#, #x in [2,6]#

Let #y=f(x)#, we thus have:

#x=y^2+1#, #y in [1,sqrt5]#

Take the derivative with respect to #y#:

#x'=2y#

Arc length is given by:

#L=int_1^sqrt5sqrt(1+4y^2)dy#

Apply the substitution #2y=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[2ysqrt(1+4y^2)+ln|2y+sqrt(1+4y^2)|]_1^sqrt5#

Insert the limits of integration:

#L=sqrt5/2(sqrt21-1)+1/4ln((2sqrt5+sqrt21)/(2+sqrt5))#

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Answer 2

Charles Anctil

The arc length of ( f(x) = \sqrt{x-1} ) on the interval ([2,6]) is ( \frac{3}{2} \left( \sqrt{10} + \sqrt{2} \right) ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.