## Problem Statement

Calculate the volumes of solids formed by rotating the figures bounded by the graphs of the functions. The axis of rotation is $O x$.

$$
y=x e^{x}, y=0, x=1
$$
## Solution

Since the $O x$ axis is the axis of rotation, the volume is found using the formula:

$$
V=\pi \cdot \int_{a}^{b} y^{2} d x
$$

Let's find the limits of integration: From the problem statement, we already have: $x_{2}=1$. Now let's find the lower limit:

$$
y=x e^{x}=0 \Rightarrow x_{1}=0
$$

Let's find the volume of the solid of revolution:

$$
V=\pi \cdot \int_{0}^{1}\left(x e^{x}\right)^{2} d x=
$$

We will use the integration by parts formula:

$$
\int_{a}^{b} u d v=\left.u v\right|_{a} ^{b}-\int_{a}^{b} v d u
$$

Let's denote:

$$
\begin{gathered}
u=x^{2} ; d u=2 x d x \\
d v=e^{2 x} d x ; v=\frac{e^{2 x}}{2}
\end{gathered}
$$

![](https://cdn.mathpix.com/cropped/2024_05_22_cfec74fe4e2360126fc5g-13.jpg?height=1379&width=765&top_left_y=984&top_left_x=1162)

In this case, we get:

$$
=\pi \cdot\left(\left.\frac{x^{2} e^{2 x}}{2}\right|_{0} ^{1}-\int_{0}^{1} x e^{2 x} d x\right)=\pi \cdot\left(\frac{e^{2}}{2}-\int_{0}^{1} x e^{2 x} d x\right)=
$$

We will use the integration by parts formula again:

$$
\begin{aligned}
& u=x ; d u=d x \\
& d v=e^{2 x} d x ; v=\frac{e^{2 x}}{2}
\end{aligned}
$$

We get:

$$
\begin{aligned}
& =\pi \cdot\left(\frac{e^{2}}{2}-\left.\frac{x e^{2 x}}{2}\right|_{0} ^{1}+\frac{1}{2} \int_{0}^{1} e^{2 x} d x\right)= \\
& =\pi \cdot\left(\frac{e^{2}}{2}-\frac{e^{2}}{2}+\left.\frac{e^{2 x}}{4}\right|_{0} ^{1}\right)=\pi \cdot \frac{e^{2}-1}{4}
\end{aligned}
$$

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 $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B} \_21-7$ »

Categories: Kuznetsov's Problem Book Integrals Problem 21 | Integrals

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