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 $2x+1=u$ :

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

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