## Task Condition

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

$$
x=4-(y-1)^{2}, x=y^{2}-4 y+3
$$
## Solution

Find the ordinates of the points of intersection of the graphs of the functions $x=4-(y-1)^{2}$

, $x=y^{2}-4 y+3:$

$$
4-(y-1)^{2}=y^{2}-4 y+3
$$

![](https://cdn.mathpix.com/cropped/2024_05_22_aa6951b1e92815ea34b6g-59.jpg?height=1008&width=1013&top_left_y=941&top_left_x=927)

$$
\begin{aligned}
& 4-y^{2}+2 y-1=y^{2}-4 y+3 \\
& 2 y^{2}-6 y=0 \\
& 2 y(y-3)=0 \\
& y=0, y=3
\end{aligned}
$$

Calculate the area:

$$
\begin{aligned}
& S=\int_{0}^{3}\left(\left(4-(y-1)^{2}\right)-\left(y^{2}-4 y+3\right)\right) d y= \\
& =\int_{0}^{3}\left(4-y^{2}+2 y-1-y^{2}+4 y-3\right) d y=
\end{aligned}
$$

$$
\begin{aligned}
& =\int_{0}^{3}\left(-2 y^{2}+6 y\right) d y=\frac{-2 y^{3}}{3}+\left.3 y^{2}\right|_{0} ^{3}= \\
& =\left(\frac{-2 \cdot 3^{3}}{3}+3 \cdot 3^{2}\right)-\left(\frac{-2 \cdot 0^{3}}{3}+3 \cdot 0^{2}\right)= \\
& =(-18+27)-(0+0)=9
\end{aligned}
$$

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 $\% \mathrm{D} 1 \% 84 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+14-31$ » Categories: Kuznetsov's Problem Book Integrals Problem 14 | Integrals

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