## Task Condition

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

$$
x=(y-2)^{3}, x=4 y-8
$$
## Solution

$$
\begin{aligned}
& (y-2)^{3}=4(y-2) \\
& (y-2)\left[4-(y-2)^{2}\right]=0 \\
& \begin{array}{l}
4-y^{2}+4 y-4=0 \\
y(y-4)=0
\end{array} \\
& \frac{S}{2}=\int_{2}^{4}(4 y-8) d y-\int_{2}^{4}(y-3)^{3} d y= \\
& =\left.2 y^{2}\right|_{4} ^{2}-\left.8 y\right|_{4} ^{2}-\int_{2}^{4}\left(y^{3}-6 y^{2}+12 y-8\right) d y= \\
& =32-8-32+16-\left(\left.\frac{y^{4}}{4}\right|_{4} ^{2}-\left.6 \frac{y^{3}}{3}\right|_{4} ^{2}+\left.6 y^{2}\right|_{4} ^{2}-\left.8 y\right|_{4} ^{2}\right)= \\
& =8-\left(\frac{256}{4}-4-128+16+96-24-32+16\right)=4 \\
& S=2 \cdot \frac{S}{2}=8
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_05_22_aa6951b1e92815ea34b6g-40.jpg?height=446&width=1371&top_left_y=1065&top_left_x=571)

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 $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+14-21$ »

Categories: Kuznetsov's Problem Book Integrals Problem 14 | Integrals

Ukrainian Banner Network

- Last edited on this page: 08:22, 11 June 2010.
- Content is available under CC-BY-SA 3.0.

Created by GeeTeatoo

## Problem Kuznetsov Integrals 14-22

## Material from Plusi