## Problem Statement

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

$$
y=\sqrt{x-x^{2}}-\arccos \sqrt{x}+5, \frac{1}{9} \leq x \leq 1
$$
## 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:

$$
\begin{aligned}
f^{\prime}(x)=\left(\sqrt{x-x^{2}}-\arccos \sqrt{x}+5\right)^{\prime} & =\frac{1}{2} \cdot \frac{1}{\sqrt{x-x^{2}}} \cdot(1-2 x)+\frac{1}{2} \cdot \frac{1}{\sqrt{1-x}} \cdot \frac{1}{\sqrt{x}}= \\
& =\frac{1}{2 \sqrt{x} \sqrt{1-x}}-\frac{2 x}{2 \sqrt{x} \sqrt{1-x}}+\frac{1}{2 \sqrt{x} \sqrt{1-x}}= \\
& =\frac{1-2 x+1}{2 \sqrt{x} \sqrt{1-x}}=\sqrt{\frac{1-x}{x}}
\end{aligned}
$$

Then, using the formula above, we get:

$$
\begin{aligned}
L & =\int_{1 / 9}^{1} \sqrt{1+\left(\sqrt{\frac{1-x}{x}}\right)^{2}} d x= \\
& =\int_{1 / 9}^{1} \sqrt{1+\frac{1-x}{x}} d x= \\
& =\int_{1 / 9}^{1} \sqrt{\frac{x+1-x}{x}} d x= \\
& =\int_{1 / 9}^{1} \frac{1}{\sqrt{x}} d x=\left.2 \sqrt{x}\right|_{1 / 9} ^{1}=2-\frac{2}{\sqrt{9}}=\frac{4}{3}
\end{aligned}
$$

Source — «http://pluspi.org/wiki/index.php/\�\�\�\�\�\�\�\�\�\�\�\�_\�\� \%D1\%83\%D0\%B7\%D0\%BD\%D0\%B5\%D1\%86\%D0\%BE\%D0\%B2_\%D0\%98\%D0\%BD\%D1\%82\%D0\%B5 $\% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+17-19$ »

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

- Last modified: 08:32, 26 May 2010.
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## Problem Kuznetsov Integrals $17-20$

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