## Problem Statement

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

$$
y=-\arccos \sqrt{x}+\sqrt{x-x^{2}}, 0 \leq x \leq \frac{1}{4}
$$
## Solution

The length of the arc of a curve defined 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:

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

Then, using the above formula, we get:

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

Source — «http://pluspi.org/wiki/index.php/\�\�\�\�\�\�\�\�\�\�\�\� $\% \mathrm{D} 0 \% 9 \mathrm{~A} \% \mathrm{D} 1 \% 83 \% \mathrm{D} 0 \% \mathrm{~B} 7 \% \mathrm{D} 0 \% \mathrm{BD} \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 1 \% 86 \% \mathrm{D} 0 \% \mathrm{BE} \% \mathrm{D} 0 \% \mathrm{~B} 2 \mathrm{O} 0 \% 98 \% \mathrm{D} 0 \% \mathrm{BD} \% \mathrm{D1} \% 82$ $\% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+17-14$ " Categories: Kuznetsov's Problem Book Integrals Problem 17| Integrals | Problems for checking

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## Problem Kuznetsov Integrals 17-15

## Material from Plusi

## Contents

- 1 Problem Statement
- 2 Solution
- 3 Then
- 4 Using the substitution
- 5 Therefore
- 6 Using the substitution we get