## Problem Statement

Calculate the lengths of the arcs of the curves given by equations in a rectangular coordinate system.

$$
y=\frac{1-e^{x}-e^{-x}}{2}, 0 \leq x \leq 3
$$
## Solution

The length of the arc of a curve defined by the equation $y=f(x) ; a \leq x \leq b$ is given by the formula

$$
L=\int_{a}^{b} \sqrt{1+\left(f^{\prime}(x)\right)^{2}} d x
$$

Let's find the derivative of the given function:

$$
f^{\prime}(x)=\left(\frac{1-e^{x}-e^{-x}}{2}\right)^{\prime}=\frac{1}{2} \cdot\left(-e^{x}+e^{-x}\right)
$$

Then, using the formula above, we get:

$$
\begin{aligned}
L & =\int_{0}^{3} \sqrt{1+\frac{1}{4} \cdot\left(-e^{x}+e^{-x}\right)^{2}} d x= \\
& =\frac{1}{2} \int_{0}^{3} \sqrt{4+\left(-e^{x}+e^{-x}\right)^{2}} d x= \\
& =\frac{1}{2} \int_{0}^{3} \sqrt{4+e^{2 x}-2+e^{-2 x}} d x= \\
& =\frac{1}{2} \int_{0}^{3} \sqrt{e^{2 x}+2+e^{-2 x}} d x= \\
& =\frac{1}{2} \int_{0}^{3} \sqrt{\left(e^{x}+e^{-x}\right)^{2}} d x= \\
& =\frac{1}{2} \int_{0}^{3}\left(e^{x}+e^{-x}\right) d x= \\
& =\left.\frac{1}{2}\left(e^{x}-e^{-x}\right)\right|_{0} ^{3}=\frac{1}{2}\left(\left(e^{3}-e^{0}\right)-\left(e^{-3}-e^{0}\right)\right)=\frac{1}{2}\left(e^{3}-e^{-3}\right)
\end{aligned}
$$

Source — «http://pluspi.org/wiki/index.php/\�\�\�\�\�\�\�\�\�\�\�\� \%D0\%9A\%D1\%83\%D0\%B7\%D0\%BD\%D0\%B5\%D1\%86\%D0\%BE\%D0\%B2_\%D0\%98\%D0\%BD $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+17-31$ » Categories: Kuznetsov's Problem Book Integrals Problem 17 | Integrals

- Last edited: 12:24, 26 May 2010.
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