## Problem Statement

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

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

$$
\begin{aligned}
f^{\prime}(x)=\left(\arccos \sqrt{x}-\sqrt{x-x^{2}}+4\right)^{\prime} & =-\frac{1}{\sqrt{1-x}} \cdot \frac{1}{2 \sqrt{x}}-\frac{1}{2 \sqrt{x} \sqrt{1-x}} \cdot(1-2 x)= \\
& =-\frac{1}{2 \sqrt{x} \sqrt{1-x}}-\frac{1}{2 \sqrt{x} \sqrt{1-x}}+\frac{2 x}{2 \sqrt{x} \sqrt{1-x}}= \\
& =-\frac{1}{\sqrt{x} \sqrt{1-x}}+\frac{x}{\sqrt{x} \sqrt{1-x}}= \\
& =-\frac{1-x}{\sqrt{x} \sqrt{1-x}}=-\sqrt{\frac{1-x}{x}}
\end{aligned}
$$

Then, using the formula above, we get:

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

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 $\% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+17-28>$

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

- Last modified: 08:00, 26 May 2010.
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## Problem Kuznetsov Integrals 17-29

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