## Problem Statement

Calculate the area of the figure bounded by the graphs of the functions:

$$
y=x \sqrt{4-x^{2}}, y=0, (0 \leq x \leq 2)
$$
## Solution

Find the abscissas of the points of intersection of the graphs of the functions

$$
\begin{gathered}
y=x \sqrt{4-x^{2}}, y=0 \\
x \sqrt{4-x^{2}}=0 \\
x=0, x= \pm 2
\end{gathered}
$$

Calculate the area:

![](https://cdn.mathpix.com/cropped/2024_05_22_aa6951b1e92815ea34b6g-34.jpg?height=1240&width=1063&top_left_y=939&top_left_x=879)

Substitution:

$$
\begin{aligned}
& t=4-x^{2} \Rightarrow d t=-2 x d x \\
& x=0 \Rightarrow t=4 \\
& x=2 \Rightarrow t=0
\end{aligned}
$$

We get:

$$
=-\frac{1}{2} \cdot \int_{4}^{0} \sqrt{t} d t=-\left.\frac{1}{2} \cdot \frac{t^{\frac{3}{2}}}{\left(\frac{3}{2}\right)}\right|_{4} ^{0}=-\frac{1}{3} \cdot\left(0^{\frac{3}{2}}-4^{\frac{3}{2}}\right)=
$$

$$
=-\frac{1}{3} \cdot(0-8)=\frac{8}{3}
$$

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_14-18" Categories: Kuznetsov Problem Book Integrals Problem 14 | Integrals

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## Problem Kuznetsov Integrals 14-19

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