Integrate x^2e^(-1x)

\int x^2 e^{\left(-1\right)\cdot x}dx

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Answer

$-x^2e^{-x}-2e^{-x}-2xe^{-x}+C_0$

Step by step solution

Problem

$\int x^2 e^{\left(-1\right)\cdot x}dx$
1

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

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

Solve the integral

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

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

$\begin{matrix}u=-x \\ du=-1dx\end{matrix}$
6

Isolate $dx$ in the previous equation

$\frac{du}{-1}=dx$
7

Substituting $u$ and $dx$ in the integral

$\int-e^udu$
8

Taking the constant out of the integral

$-\int e^udu$
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 $a \neq 1$

$-e^u$
10

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

$-e^{-x}$
11

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

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

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

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

Solve the integral

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

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

$\begin{matrix}u=-x \\ du=-1dx\end{matrix}$
17

Isolate $dx$ in the previous equation

$\frac{du}{-1}=dx$
18

Substituting $u$ and $dx$ in the integral

$2\int x-e^udx-x^2e^{-x}$
19

Taking the constant out of the integral

$2\int-xe^udx-x^2e^{-x}$
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 $a \neq 1$

$2\int-xe^udx-x^2e^{-x}$
21

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

$2\int-xe^udx-x^2e^{-x}$
22

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

$2\left(\int e^{-x}dx-xe^{-x}\right)-x^2e^{-x}$
23

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

$\begin{matrix}u=-x \\ du=-1dx\end{matrix}$
24

Isolate $dx$ in the previous equation

$\frac{du}{-1}=dx$
25

Substituting $u$ and $dx$ in the integral

$2\left(\int-e^udu-xe^{-x}\right)-x^2e^{-x}$
26

Taking the constant out of the integral

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

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 $a \neq 1$

$2\left(-xe^{-x}-e^u\right)-x^2e^{-x}$
28

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

$2\left(-xe^{-x}-e^{-x}\right)-x^2e^{-x}$
29

Multiply $\left(-xe^{-x}+-e^{-x}\right)$ by $2$

$-x^2e^{-x}-2e^{-x}-2xe^{-x}$
30

Add the constant of integration

$-x^2e^{-x}-2e^{-x}-2xe^{-x}+C_0$

Answer

$-x^2e^{-x}-2e^{-x}-2xe^{-x}+C_0$

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Problem Analysis

Main topic:

Integration by parts

Time to solve it:

0.32 seconds

Views:

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