## Problem Statement

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

$$
y=e^{x}+13, \ln \sqrt{15} \leq x \leq \ln \sqrt{24}
$$
## Solution

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

$$
f^{\prime}(x)=\left(e^{x}+13\right)^{\prime}=e^{x}
$$

Then, using the formula above, we get:

$$
\begin{aligned}
L & =\int_{\ln \sqrt{15}}^{\ln \sqrt{24}} \sqrt{1+e^{2 x}} d x=\left|\begin{array}{c}
\sqrt{1+e^{2 x}}=t \\
2 x=\ln \left(t^{2}-1\right) \\
d x=\frac{t d t}{t^{2}-1} \\
t_{1}=\sqrt{15+1}=4 ; \\
t_{2}=\sqrt{24+1}=5
\end{array}\right|=\int_{4}^{5} t \cdot \frac{t d t}{t^{2}-1}= \\
& =\int_{4}^{5} \frac{t^{2}}{t^{2}-1} d t= \\
& =\int_{4}^{5} \frac{t^{2}-1+1}{t^{2}-1} d t= \\
& =\int_{4}^{5} d t+\int_{4}^{5} \frac{d t}{t^{2}-1}= \\
& =\left.t\right|_{4} ^{5}+\left.\frac{1}{2} \ln \left|\frac{t-1}{t+1}\right|\right|_{4} ^{5}=5-4+\frac{1}{2} \ln \frac{4}{6}-\frac{1}{2} \ln \frac{3}{5}=1+\frac{1}{2} \ln \left(\frac{2}{3}: \frac{3}{5}\right)=1+\frac{1}{2} \ln \frac{10}{9}
\end{aligned}
$$

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\%D1\%82 $\% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B} \_17-13$ »

Categories: Kuznetsov's Problem Book Integrals Problem 17 | Integrals

- Last modified: 05:42, 26 May 2010.
- Content is available under CC-BY-SA 3.0.

Created by GeeTeatoo

## Problem Kuznetsov Integrals 17-14

## Material from PlusPi