## Task Condition

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

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

Let's find the limits of integration:

$$
\begin{aligned}
& \left(y^{2}-2 y\right)-\left(4-y^{2}\right)=0 \\
& 2 y^{2}-2 y-4=0 \\
& y^{2}-y-2=0
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_05_22_aa6951b1e92815ea34b6g-45.jpg?height=1060&width=1105&top_left_y=952&top_left_x=815)

Then the area of the figure will be:

$$
\begin{aligned}
& S=\int_{-1}^{2}\left(4-y^{2}\right)-\left(y^{2}-2 y\right) d y= \\
& =\int_{-1}^{2}\left(-2 y^{2}+2 y+4\right) d y=
\end{aligned}
$$

$$
\begin{aligned}
& =\left.\left(-2 \frac{y^{3}}{3}+2 \frac{y^{2}}{2}+4 y\right)\right|_{-1} ^{2}= \\
& =\left(-2 \frac{8}{3}+4+8\right)-\left(2 \frac{1}{3}+1-4\right)=-\frac{16}{3}+4+8-\frac{2}{3}-1+4=9
\end{aligned}
$$

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 $\% \mathrm{D} 1 \% 84 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+14-24$ » Categories: Kuznetsov's Problem Book Integrals Problem 14 | Integrals

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## Problem Kuznetsov Integrals $14-25$

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