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

May 22, 2018

$L = 4 {e}^{5} - 3 {e}^{3} + \frac{1}{2} {\sum}_{n = 1}^{\infty} \left(\begin{matrix}\frac{1}{2} \\ n\end{matrix}\right) {\int}_{3}^{5} {\left({e}^{4 - u} / u\right)}^{2 n - 1} \mathrm{dx}$

#### Explanation:

$f \left(x\right) = x {e}^{2 x - 3}$

$f ' \left(x\right) = \left(2 x + 1\right) {e}^{2 x - 3}$

Arc length is given by:

$L = {\int}_{3}^{4} \sqrt{1 + {\left(f ' \left(x\right)\right)}^{2}} \mathrm{dx}$

Rearrange:

$L = {\int}_{3}^{4} f ' \left(x\right) \sqrt{1 + {\left(f ' \left(x\right)\right)}^{-} 2} \mathrm{dx}$

For $x \in \left[3 , 4\right]$, ${\left(f ' \left(x\right)\right)}^{-} 2 < 1$. Take the series expansion of the square root:

$L = {\int}_{3}^{4} f ' \left(x\right) \left\{{\sum}_{n = 0}^{\infty} \left(\begin{matrix}\frac{1}{2} \\ n\end{matrix}\right) {\left(f ' \left(x\right)\right)}^{- 2 n}\right\} \mathrm{dx}$

Isolate the $n = 0$ term and simplify:

$L = {\int}_{3}^{4} f ' \left(x\right) \mathrm{dx} + {\sum}_{n = 1}^{\infty} \left(\begin{matrix}\frac{1}{2} \\ n\end{matrix}\right) {\int}_{3}^{4} {\left(f ' \left(x\right)\right)}^{1 - 2 n} \mathrm{dx}$

Hence

$L = f \left(4\right) - f \left(3\right) + {\sum}_{n = 1}^{\infty} \left(\begin{matrix}\frac{1}{2} \\ n\end{matrix}\right) {\int}_{3}^{4} {\left(\left(2 x + 1\right) {e}^{2 x - 3}\right)}^{1 - 2 n} \mathrm{dx}$

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

$L = 4 {e}^{5} - 3 {e}^{3} + \frac{1}{2} {\sum}_{n = 1}^{\infty} \left(\begin{matrix}\frac{1}{2} \\ n\end{matrix}\right) {\int}_{3}^{5} {\left({e}^{4 - u} / u\right)}^{2 n - 1} \mathrm{dx}$