## Problem Statement

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

$$
y=2-e^{x}, \ln \sqrt{3} \leq x \leq \ln \sqrt{8}
$$
## Solution

$$
\begin{aligned}
& f^{\prime}(x)=-\mathrm{e}^{\mathrm{x}} \\
& \left(f^{\prime}(x)\right)^{2}=\mathrm{e}^{2 \mathrm{x}}
\end{aligned}
$$

## Then

$$
1=\int_{\ln \sqrt{3}}^{\ln \sqrt{8}} \sqrt{1+\mathrm{e}^{2 \mathrm{x}}} d x
$$

## Using the substitution

$$
\begin{aligned}
& \mathrm{t}=\sqrt{1+\mathrm{e}^{2 \mathrm{x}}} \\
& x=\ln \sqrt{8} \quad \mathrm{t}=\sqrt{1+8}=3 \\
& x=\ln \sqrt{3} \quad \mathrm{t}=\sqrt{1+3}=2 \\
& \mathrm{t}^{2}=1+\mathrm{e}^{2 \mathrm{x}} \quad \mathrm{e}^{2 \mathrm{x}}=\mathrm{t}^{2}-1 \quad 2 x=\ln \left(\mathrm{t}^{2}-1\right)
\end{aligned}
$$

## Therefore

$$
d x=\frac{\mathrm{t} d t}{\mathrm{t}^{2}-1}
$$

## Using the substitution we get

$$
\begin{aligned}
& \mathrm{l}=\int_{2}^{3} \frac{\mathrm{t}^{2} d t}{\mathrm{t}^{2}-1}=\int_{2}^{3} \frac{\left(\mathrm{t}^{2}-1+1\right) d t}{\mathrm{t}^{2}-1}=\int_{2}^{3} d t+\int_{2}^{3} \frac{d t}{\mathrm{t}^{2}-1}= \\
& =\left.\mathrm{t}\right|_{2} ^{3}+\left.\frac{1}{2} \ln \left|\frac{\mathrm{t}-1}{\mathrm{t}+1}\right|\right|_{2} ^{3}=3-2+\frac{1}{2}\left(\ln \frac{1}{2}-\ln \frac{1}{3}\right)=1+\frac{1}{2} \ln \frac{3}{2}
\end{aligned}
$$

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 $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+17-15$ » Categories: Kuznetsov's Problem Book Integrals Problem 17 | Integrals

- Last edited: 11:49, 18 March 2010.
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## Problem Kuznetsov Integrals $17-16$

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