Integrate x^2e^(-2x)

\int x^2 e^{-2x}dx

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Answer

$-\frac{1}{2}xe^{-2x}-\frac{1}{4}e^{-2x}-\frac{1}{2}x^2e^{-2x}+C_0$

Step by step solution

Problem

$\int x^2 e^{-2x}dx$
1

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

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

First, identify $u$ and calculate $du$

$\begin{matrix}\displaystyle{u=x^2}\\ \displaystyle{du=2xdx}\end{matrix}$
3

Now, identify $dv$ and calculate $v$

$\begin{matrix}\displaystyle{dv=e^{-2x}dx}\\ \displaystyle{\int dv=\int e^{-2x}dx}\end{matrix}$
4

Solve the integral

$v=\int e^{-2x}dx$
5

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

$\begin{matrix}u=-2x \\ du=-2dx\end{matrix}$
6

Isolate $dx$ in the previous equation

$\frac{du}{-2}=dx$
7

Substituting $u$ and $dx$ 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, $-2x$

$-\frac{1}{2}e^{-2x}$
11

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

$\int xe^{-2x}dx-\frac{1}{2}x^2e^{-2x}$
12

Use the integration by parts theorem to calculate the integral $\int xe^{-2x}dx$, using the following formula

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

First, identify $u$ and calculate $du$

$\begin{matrix}\displaystyle{u=x}\\ \displaystyle{du=dx}\end{matrix}$
14

Now, identify $dv$ and calculate $v$

$\begin{matrix}\displaystyle{dv=e^{-2x}dx}\\ \displaystyle{\int dv=\int e^{-2x}dx}\end{matrix}$
15

Solve the integral

$v=\int e^{-2x}dx$
16

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

$\begin{matrix}u=-2x \\ du=-2dx\end{matrix}$
17

Isolate $dx$ in the previous equation

$\frac{du}{-2}=dx$
18

Substituting $u$ and $dx$ in the integral

$\int x\frac{e^u}{-2}dx-\frac{1}{2}x^2e^{-2x}$
19

Taking the constant out of the integral

$\int x\frac{e^u}{-2}dx-\frac{1}{2}x^2e^{-2x}$
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 $a \neq 1$

$\int x\frac{e^u}{-2}dx-\frac{1}{2}x^2e^{-2x}$
21

Substitute $u$ back for it's value, $-2x$

$\int x\frac{e^u}{-2}dx-\frac{1}{2}x^2e^{-2x}$
22

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

$\frac{1}{2}\int e^{-2x}dx-\frac{1}{2}xe^{-2x}-\frac{1}{2}x^2e^{-2x}$
23

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

$\begin{matrix}u=-2x \\ du=-2dx\end{matrix}$
24

Isolate $dx$ in the previous equation

$\frac{du}{-2}=dx$
25

Substituting $u$ and $dx$ in the integral

$\frac{1}{2}\int\frac{e^u}{-2}du-\frac{1}{2}xe^{-2x}-\frac{1}{2}x^2e^{-2x}$
26

Taking the constant out of the integral

$\frac{1}{2}\left(-\frac{1}{2}\right)\int e^udu-\frac{1}{2}xe^{-2x}-\frac{1}{2}x^2e^{-2x}$
27

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

$-\frac{1}{2}x^2e^{-2x}-\frac{1}{2}xe^{-2x}-\frac{1}{4}\int e^udu$
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 $a \neq 1$

$-\frac{1}{2}x^2e^{-2x}-\frac{1}{2}xe^{-2x}-\frac{1}{4}e^u$
29

Substitute $u$ back for it's value, $-2x$

$-\frac{1}{2}x^2e^{-2x}-\frac{1}{2}xe^{-2x}-\frac{1}{4}e^{-2x}$
30

Add the constant of integration

$-\frac{1}{2}xe^{-2x}-\frac{1}{4}e^{-2x}-\frac{1}{2}x^2e^{-2x}+C_0$

Answer

$-\frac{1}{2}xe^{-2x}-\frac{1}{4}e^{-2x}-\frac{1}{2}x^2e^{-2x}+C_0$

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Problem Analysis

Main topic:

Integration by parts

Time to solve it:

0.27 seconds

Views:

131