Integral of y^2e^(2y)

\int y^2 e^{2y}dy

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Answer

$\frac{1}{2}y^2e^{2y}+\frac{1}{4}e^{2y}-\frac{1}{2}ye^{2y}+C_0$

Step by step solution

Problem

$\int y^2 e^{2y}dy$
1

Use the integration by parts theorem to calculate the integral $\int y^2e^{2y}dy$, using the following formula

$\displaystyle\int u\cdot dv=u\cdot v-\int v \cdot du$
2

First, identify $u$ and calculate $du$

$\begin{matrix}\displaystyle{u=y^2}\\ \displaystyle{du=2ydy}\end{matrix}$
3

Now, identify $dv$ and calculate $v$

$\begin{matrix}\displaystyle{dv=e^{2y}dy}\\ \displaystyle{\int dv=\int e^{2y}dy}\end{matrix}$
4

Solve the integral

$v=\int e^{2y}dy$
5

Solve the integral $\int e^{2y}dy$ applying u-substitution. Let $u$ and $du$ be

$\begin{matrix}u=2y \\ du=2dy\end{matrix}$
6

Isolate $dy$ in the previous equation

$\frac{du}{2}=dy$
7

Substituting $u$ and $dy$ in the integral

$\int\frac{e^u}{2}du$
8

Taking the constant out of the integral

$\frac{1}{2}\int e^udu$
9

The integral of the exponential function is given by the following formula $\displaystyle \int a^xdx=\frac{a^x}{\ln(a)}$, where $a > 0$ and $a \neq 1$

$\frac{1}{2}e^u$
10

Substitute $u$ back for it's value, $2y$

$\frac{1}{2}e^{2y}$
11

Now replace the values of $u$, $du$ and $v$ in the last formula

$\frac{1}{2}y^2e^{2y}-\int ye^{2y}dy$
12

Use the integration by parts theorem to calculate the integral $\int ye^{2y}dy$, using the following formula

$\displaystyle\int u\cdot dv=u\cdot v-\int v \cdot du$
13

First, identify $u$ and calculate $du$

$\begin{matrix}\displaystyle{u=y}\\ \displaystyle{du=dy}\end{matrix}$
14

Now, identify $dv$ and calculate $v$

$\begin{matrix}\displaystyle{dv=e^{2y}dy}\\ \displaystyle{\int dv=\int e^{2y}dy}\end{matrix}$
15

Solve the integral

$v=\int e^{2y}dy$
16

Solve the integral $\int e^{2y}dy$ applying u-substitution. Let $u$ and $du$ be

$\begin{matrix}u=2y \\ du=2dy\end{matrix}$
17

Isolate $dy$ in the previous equation

$\frac{du}{2}=dy$
18

Substituting $u$ and $dy$ in the integral

$\frac{1}{2}y^2e^{2y}-\int y\frac{e^u}{2}dy$
19

Taking the constant out of the integral

$\frac{1}{2}y^2e^{2y}-\int y\frac{e^u}{2}dy$
20

The integral of the exponential function is given by the following formula $\displaystyle \int a^xdx=\frac{a^x}{\ln(a)}$, where $a > 0$ and $a \neq 1$

$\frac{1}{2}y^2e^{2y}-\int y\frac{e^u}{2}dy$
21

Substitute $u$ back for it's value, $2y$

$\frac{1}{2}y^2e^{2y}-\int y\frac{e^u}{2}dy$
22

Now replace the values of $u$, $du$ and $v$ in the last formula

$\frac{1}{2}y^2e^{2y}-\left(\frac{1}{2}ye^{2y}-\frac{1}{2}\int e^{2y}dy\right)$
23

Solve the integral $\int e^{2y}dy$ applying u-substitution. Let $u$ and $du$ be

$\begin{matrix}u=2y \\ du=2dy\end{matrix}$
24

Isolate $dy$ in the previous equation

$\frac{du}{2}=dy$
25

Substituting $u$ and $dy$ in the integral

$\frac{1}{2}y^2e^{2y}-\left(\frac{1}{2}ye^{2y}-\frac{1}{2}\int\frac{e^u}{2}du\right)$
26

Taking the constant out of the integral

$\frac{1}{2}y^2e^{2y}-\left(\frac{1}{2}ye^{2y}-\frac{1}{2}\cdot \frac{1}{2}\int e^udu\right)$
27

Multiply $\frac{1}{2}$ times $-\frac{1}{2}$

$\frac{1}{2}y^2e^{2y}-\left(\frac{1}{2}ye^{2y}-\frac{1}{4}\int e^udu\right)$
28

The integral of the exponential function is given by the following formula $\displaystyle \int a^xdx=\frac{a^x}{\ln(a)}$, where $a > 0$ and $a \neq 1$

$\frac{1}{2}y^2e^{2y}-\left(\frac{1}{2}ye^{2y}-\frac{1}{4}e^u\right)$
29

Substitute $u$ back for it's value, $2y$

$\frac{1}{2}y^2e^{2y}-\left(\frac{1}{2}ye^{2y}-\frac{1}{4}e^{2y}\right)$
30

Multiply $\left(\frac{1}{2}ye^{2y}+-\frac{1}{4}e^{2y}\right)$ by $-1$

$\frac{1}{2}y^2e^{2y}+\frac{1}{4}e^{2y}-\frac{1}{2}ye^{2y}$
31

Add the constant of integration

$\frac{1}{2}y^2e^{2y}+\frac{1}{4}e^{2y}-\frac{1}{2}ye^{2y}+C_0$

Answer

$\frac{1}{2}y^2e^{2y}+\frac{1}{4}e^{2y}-\frac{1}{2}ye^{2y}+C_0$

Problem Analysis

Main topic:

Integration by parts

Time to solve it:

0.38 seconds

Views:

225