## Task Condition

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

$$
y=4-x^{2}, y=x^{2}-2 x
$$
## Solution

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

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

![](https://cdn.mathpix.com/cropped/2024_05_22_aa6951b1e92815ea34b6g-04.jpg?height=1010&width=1034&top_left_y=937&top_left_x=911)

$$
\begin{aligned}
& D=(-1)^{2}-4 \cdot 1 \cdot(-2)=9 \\
& x_{1}=\frac{1-\sqrt{9}}{2 \cdot 1}=-1 \\
& x_{2}=\frac{1+\sqrt{9}}{2 \cdot 1}=2
\end{aligned}
$$

Calculate the area:

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

$$
\begin{aligned}
& -\left(-\frac{2 \cdot(-1)^{3}}{3}+(-1)^{2}+4 \cdot(-1)\right)=-\frac{16}{3}+4+8-\frac{2}{3}-1+4= \\
& =-\frac{18}{3}+15=-6+15=9
\end{aligned}
$$

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 $\% \mathrm{D} 1 \% 84 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+14-4$ » Categories: Kuznetsov's Problem Book Integrals Problem 14 | Integrals

Ukrainian Banner Network

- Last edited on this page: 16:15, 29 May 2010.
- Content is available under CC-BY-SA 3.0.

Created by GeeTeatoo

## Problem Kuznetsov Integrals 14-4

## Material from PlusPi