## Problem Statement

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

$$
y=\frac{x^{2}}{4}-\frac{\ln x}{2}, 1 \leq x \leq 2
$$
## 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:

$$
f^{\prime}(x)=\left(\frac{x^{2}}{4}-\frac{\ln x}{2}\right)^{\prime}=\frac{x}{2}-\frac{1}{2 x}=\frac{x^{2}-1}{2 x}
$$

Then, using the formula above, we get:

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

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 $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+17-2$ »

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

- Last modified: 12:05, 25 May 2010.
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## Problem Kuznetsov Integrals 17-3

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