What is the arc length of #f(x)= e^(4x-1) # on #x in [2,4] #?

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Answer 1 · 2018-03-15T08:58:27

#L=sum_(n=0)^oo((1/2),(n))1/16^n(e^(15(1-2n))-e^(7(1-2n)))/(1-2n)# units.

#f(x)=e^(4x-1)#

#f'(x)=4e^(4x-1)#

Arc length is given by:

#L=int_2^4sqrt(1+(4e^(4x-1))^2)dx#

Rearrange:

#L=int_2^4(4e^(4x-1))sqrt(1+(4e^(4x-1))^-2)dx#

For #x in [2,4]#, #(4e^(4x-1))^-2<1#. Take the series expansion of the square root:

#L=4int_2^4e^(4x-1){sum_(n=0)^oo((1/2),(n))(4e^(4x-1))^(-2n)}dx#

Simplify:

#L=4sum_(n=0)^oo((1/2),(n))1/16^nint_2^4e^((1-2n)(4x-1))dx#

Integrate directly:

#L=4sum_(n=0)^oo((1/2),(n))1/16^n[e^((1-2n)(4x-1))]_2^4/(4(1-2n))#

Insert the limits of integration:

#L=sum_(n=0)^oo((1/2),(n))1/16^n(e^(15(1-2n))-e^(7(1-2n)))/(1-2n)#