Arc Length

Problem 235: Find the length of the curve defined by

(1) $\begin{equation*} y = \frac{1}{2}\left(e^x + e^{-x} \right) \end{equation*}$

from $x=0$ to $x= 1$ .

Solution: The length of a curve is given by

(2) $\begin{equation*} L = \int_{a}^{b} \sqrt{1+[f'(x) ]^2} \ dx.\end{equation*}$

That is,

(3) $\begin{align*} f(x) & = \frac{1}{2}\left(e^x + e^{-x} \right) \\ f'(x) & = \frac{1}{2}\left(e^x -e^{-x}\right). \end{align*}$

Then

(4) $\begin{align*} L & = \int_{a}^{b} \sqrt{1+[f'(x) ]^2} \ dx \\ & = \int_{0}^{1} \sqrt{1+\left( \frac{1}{2}e^x-\frac{1}{2}e^{-x}\right)^2} \ dx \\ & = \int_{0}^{1} \sqrt{\left(\frac{e^x}{2}-\frac{e^{-x}}{2} \right)^2 +1} \ dx \\ & = \int_{0}^{1} \sqrt{(\sinh x)^2 + 1} \ dx \\ & = \int_{0}^{1} \sqrt{\cosh^2 x} \ dx \\ & = \int_{0}^{1} \cosh x \ dx \\ & = \sinh x \bigg|_{0}^{1} \\ & = \frac{e^{-x}(e^{2x}-1)}{2}\bigg|_{0}^{1} \\ & = \frac{e^1(e^2 - 1)}{2} - \frac{e^0(e^0-1)}{2} \\ & = \frac{e^1-e^{-1}}{2}.\end{align*}$

That is, the length of the curve between 0 and 1 is

$\frac{1}{2}(e^1 - e^{-1}).$

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