## Task Condition

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

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

From the graph, it can be seen that the area between the curves consists of two identical parts:

$$
S_{(0,4)}=2 S_{(2,4)}
$$

We will find the area of the part where \( x \in (2,4) \) as the difference of two integrals:

$$
\begin{aligned}
& S=\int_{2}^{4}(4 x-8) d x-\int_{2}^{4}(x-2)^{3} d x= \\
& =\left.\left(4 \frac{x^{2}}{2}-8 x\right)\right|_{2} ^{4}-\int_{2}^{4}\left(x^{3}-6 x^{2}+12 x-8\right) d x= \\
& =\left.\left(2 x^{2}-8 x\right)\right|_{2} ^{4}-\left.\left(\frac{x^{4}}{4}-6 \frac{x^{3}}{3}+12 \frac{x^{2}}{2}-8 x\right)\right|_{2} ^{4}=
\end{aligned}
$$

Integrals \(14-1\)

![](https://cdn.mathpix.com/cropped/2024_05_22_aa6951b1e92815ea34b6g-01.jpg?height=1368&width=600&top_left_y=978&top_left_x=1345)

$$
\begin{aligned}
& =(32-32)-(8-16)-\left(\frac{4^{4}}{4}-2 \cdot 4^{3}+6 \cdot 4^{2}-8 \cdot 4\right)+\left(\frac{2^{4}}{4}-2 \cdot 2^{3}+6 \cdot 2^{2}-8 \cdot 2\right)= \\
& =0+8-(64-128+96-32)+(4-16+24-16)=8-0+(-4)=4 \\
& S_{(0,4)}=2 \cdot S_{(2,4)}=2 \cdot 4=8
\end{aligned}
$$

%egraly_14-1»

Categories: Kuznetsov Integral Problems Problem 14 | Integrals

Ukrainian Banner Network

- Last edited on this page: 04:54, June 9, 2010.
- Content is available under CC-BY-SA 3.0.

Created by Geeteatoo

## Problem Kuznetsov Integrals 14-2

## Material from Plusi