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

Mar 15, 2018

$L = {\sum}_{n = 0}^{\infty} \left(\begin{matrix}\frac{1}{2} \\ n\end{matrix}\right) \frac{1}{16} ^ n \frac{{e}^{15 \left(1 - 2 n\right)} - {e}^{7 \left(1 - 2 n\right)}}{1 - 2 n}$ units.

#### Explanation:

$f \left(x\right) = {e}^{4 x - 1}$

$f ' \left(x\right) = 4 {e}^{4 x - 1}$

Arc length is given by:

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

Rearrange:

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

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

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

Simplify:

$L = 4 {\sum}_{n = 0}^{\infty} \left(\begin{matrix}\frac{1}{2} \\ n\end{matrix}\right) \frac{1}{16} ^ n {\int}_{2}^{4} {e}^{\left(1 - 2 n\right) \left(4 x - 1\right)} \mathrm{dx}$

Integrate directly:

$L = 4 {\sum}_{n = 0}^{\infty} \left(\begin{matrix}\frac{1}{2} \\ n\end{matrix}\right) \frac{1}{16} ^ n {\left[{e}^{\left(1 - 2 n\right) \left(4 x - 1\right)}\right]}_{2}^{4} / \left(4 \left(1 - 2 n\right)\right)$

Insert the limits of integration:

$L = {\sum}_{n = 0}^{\infty} \left(\begin{matrix}\frac{1}{2} \\ n\end{matrix}\right) \frac{1}{16} ^ n \frac{{e}^{15 \left(1 - 2 n\right)} - {e}^{7 \left(1 - 2 n\right)}}{1 - 2 n}$