What is the arc length of #f(x)=xe^(2x-3) # on #x in [3,4] #?

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Answer 1 · 2018-05-22T06:32:56

#L=4e^5-3e^3+1/2sum_(n=1)^oo((1/2),(n))int_3^5(e^(4-u)/u)^(2n-1)dx#

#f(x)=xe^(2x-3)#

#f'(x)=(2x+1)e^(2x-3)#

Arc length is given by:

#L=int_3^4sqrt(1+(f'(x))^2)dx#

Rearrange:

#L=int_3^4f'(x)sqrt(1+(f'(x))^-2)dx#

For #x in [3,4]#, #(f'(x))^-2<1#. Take the series expansion of the square root:

#L=int_3^4f'(x){sum_(n=0)^oo((1/2),(n))(f'(x))^(-2n)}dx#

Isolate the #n=0# term and simplify:

#L=int_3^4f'(x)dx+sum_(n=1)^oo((1/2),(n))int_3^4(f'(x))^(1-2n)dx#

Hence

#L=f(4)-f(3)+sum_(n=1)^oo((1/2),(n))int_3^4((2x+1)e^(2x-3))^(1-2n)dx#

Apply the substitution #2x+1=u#:

#L=4e^5-3e^3+1/2sum_(n=1)^oo((1/2),(n))int_3^5(e^(4-u)/u)^(2n-1)dx#