## Problem Statement

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

$$
y=\frac{e^{2 x}+e^{-2 x}+3}{4}, 0 \leq x \leq 2
$$
## 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:

$$
f^{\prime}(x)=\left(\frac{e^{2 x}+e^{-2 x}+3}{4}\right)^{\prime}=\frac{1}{4} \cdot\left(2 e^{2 x}-2 e^{-2 x}\right)=\frac{1}{2} \cdot\left(e^{2 x}-e^{-2 x}\right)
$$

Then, using the above formula, we get:

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

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%D1%82%D0%B5%D0%B3%D1%80%D0%B0%D0%BB%D1%8B_17-29»

Categories: Kuznetsov's Problem Book Integrals Problem 17 | Integrals | Problems for Checking

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## Problem Kuznetsov Integrals 17-30

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