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x^(three)exp(-2x)
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x^(3)exp(2x)

Integral of x^(3)exp(-2x) dx

The solution

You have entered [src]

 oo            
  /            
 |             
 |   3  -2*x   
 |  x *e     dx
 |             
/              
0

$$\int\limits_{0}^{\infty} x^{3} e^{- 2 x}\, dx$$

Integral(x^3*exp(-2*x), (x, 0, oo))

Detail solution

Use integration by parts:

$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$

Let $u{\left(x \right)} = x^{3}$ and let $\operatorname{dv}{\left(x \right)} = e^{- 2 x}$ .

Then $\operatorname{du}{\left(x \right)} = 3 x^{2}$ .

To find $v{\left(x \right)}$ :
1. Let $u = - 2 x$ .
  
  Then let $du = - 2 dx$ and substitute $- \frac{du}{2}$ :
  
  $\int \left(- \frac{e^{u}}{2}\right)\, du$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\text{False}$
    1. The integral of the exponential function is itself.
      
      $\int e^{u}\, du = e^{u}$
    So, the result is: $- \frac{e^{u}}{2}$
  Now substitute $u$ back in:
  
  $- \frac{e^{- 2 x}}{2}$
Now evaluate the sub-integral.
Use integration by parts:

$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$

Let $u{\left(x \right)} = - \frac{3 x^{2}}{2}$ and let $\operatorname{dv}{\left(x \right)} = e^{- 2 x}$ .

Then $\operatorname{du}{\left(x \right)} = - 3 x$ .

To find $v{\left(x \right)}$ :
1. Let $u = - 2 x$ .
  
  Then let $du = - 2 dx$ and substitute $- \frac{du}{2}$ :
  
  $\int \left(- \frac{e^{u}}{2}\right)\, du$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\text{False}$
    1. The integral of the exponential function is itself.
      
      $\int e^{u}\, du = e^{u}$
    So, the result is: $- \frac{e^{u}}{2}$
  Now substitute $u$ back in:
  
  $- \frac{e^{- 2 x}}{2}$
Now evaluate the sub-integral.
Use integration by parts:

$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$

Let $u{\left(x \right)} = \frac{3 x}{2}$ and let $\operatorname{dv}{\left(x \right)} = e^{- 2 x}$ .

Then $\operatorname{du}{\left(x \right)} = \frac{3}{2}$ .

To find $v{\left(x \right)}$ :
1. Let $u = - 2 x$ .
  
  Then let $du = - 2 dx$ and substitute $- \frac{du}{2}$ :
  
  $\int \left(- \frac{e^{u}}{2}\right)\, du$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\text{False}$
    1. The integral of the exponential function is itself.
      
      $\int e^{u}\, du = e^{u}$
    So, the result is: $- \frac{e^{u}}{2}$
  Now substitute $u$ back in:
  
  $- \frac{e^{- 2 x}}{2}$
Now evaluate the sub-integral.
The integral of a constant times a function is the constant times the integral of the function:

$\int \left(- \frac{3 e^{- 2 x}}{4}\right)\, dx = - \frac{3 \int e^{- 2 x}\, dx}{4}$
1. Let $u = - 2 x$ .
  
  Then let $du = - 2 dx$ and substitute $- \frac{du}{2}$ :
  
  $\int \left(- \frac{e^{u}}{2}\right)\, du$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\text{False}$
    1. The integral of the exponential function is itself.
      
      $\int e^{u}\, du = e^{u}$
    So, the result is: $- \frac{e^{u}}{2}$
  Now substitute $u$ back in:
  
  $- \frac{e^{- 2 x}}{2}$
So, the result is: $\frac{3 e^{- 2 x}}{8}$
Now simplify:

$- \frac{\left(4 x^{3} + 6 x^{2} + 6 x + 3\right) e^{- 2 x}}{8}$
Add the constant of integration:

$- \frac{\left(4 x^{3} + 6 x^{2} + 6 x + 3\right) e^{- 2 x}}{8}+ \mathrm{constant}$

The answer is:

$- \frac{\left(4 x^{3} + 6 x^{2} + 6 x + 3\right) e^{- 2 x}}{8}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                                             
 |                      -2*x        -2*x      2  -2*x    3  -2*x
 |  3  -2*x          3*e       3*x*e       3*x *e       x *e    
 | x *e     dx = C - ------- - --------- - ---------- - --------
 |                      8          4           4           2    
/

$$\int x^{3} e^{- 2 x}\, dx = C - \frac{x^{3} e^{- 2 x}}{2} - \frac{3 x^{2} e^{- 2 x}}{4} - \frac{3 x e^{- 2 x}}{4} - \frac{3 e^{- 2 x}}{8}$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

3/8

$$\frac{3}{8}$$

3/8

$$\frac{3}{8}$$

3/8

Use the examples entering the upper and lower limits of integration.

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Identical expressions

Similar expressions

Integral of x^(3)exp(-2x) dx

Limits of integration:

The graph:

Piecewise:

The solution