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Identical expressions
(x- three)e^(-2x)
(x minus 3)e to the power of ( minus 2x)
(x minus three)e to the power of ( minus 2x)
(x-3)e(-2x)
x-3e-2x
x-3e^-2x
(x-3)e^(-2x)dx
Similar expressions
(x+3)e^(-2x)
(x-3)e^(2x)

Solving integrals/ (x-3)e^(-2x)

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

The solution

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  1                 
  /                 
 |                  
 |           -2*x   
 |  (x - 3)*e     dx
 |                  
/                   
0

$$\int\limits_{0}^{1} \left(x - 3\right) e^{- 2 x}\, dx$$

Integral((x - 1*3)/E^(2*x), (x, 0, 1))

Detail solution

There are multiple ways to do this integral.
Method #1
1. 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 $\operatorname{du}{\left(x \right)} = 1$ .
  
  To find $v{\left(x \right)}$ :
  1. There are multiple ways to do this integral.
    Method #1
    
    Let $u = - 2 x$ .
    
    Then let $du = - 2 dx$ and substitute $- \frac{du}{2}$ :
    
    $\int \frac{e^{u}}{4}\, du$
    
    The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- \frac{e^{u}}{2}\right)\, du = - \frac{\int e^{u}\, du}{2}$
    
    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:
    
    $- \frac{e^{- 2 x}}{2}$
    Method #2
    
    Let $u = e^{- 2 x}$ .
    
    Then let $du = - 2 e^{- 2 x} dx$ and substitute $- \frac{du}{2}$ :
    
    $\int \frac{1}{4}\, du$
    
    The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- \frac{1}{2}\right)\, du = - \frac{\int 1\, du}{2}$
    
    The integral of a constant is the constant times the variable of integration:
    
    $\int 1\, du = u$
    
    So, the result is: $- \frac{u}{2}$
    
    Now substitute $u$ back in:
    
    $- \frac{e^{- 2 x}}{2}$
  Now evaluate the sub-integral.
2. The integral of a constant times a function is the constant times the integral of the function:
  
  $\int \left(- \frac{e^{- 2 x}}{2}\right)\, dx = - \frac{\int e^{- 2 x}\, dx}{2}$
  1. Let $u = - 2 x$ .
    
    Then let $du = - 2 dx$ and substitute $- \frac{du}{2}$ :
    
    $\int \frac{e^{u}}{4}\, du$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{e^{u}}{2}\right)\, du = - \frac{\int e^{u}\, du}{2}$
      
      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: $\frac{e^{- 2 x}}{4}$
Method #2
1. Rewrite the integrand:
  
  $\left(x - 3\right) e^{- 2 x} = x e^{- 2 x} - 3 e^{- 2 x}$
2. Integrate term-by-term:
  1. Use integration by parts:
    
    $\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$
    
    Let $u{\left(x \right)} = x$ and let $\operatorname{dv}{\left(x \right)} = e^{- 2 x}$ .
    
    Then $\operatorname{du}{\left(x \right)} = 1$ .
    
    To find $v{\left(x \right)}$ :
    1. Let $u = - 2 x$ .
      
      Then let $du = - 2 dx$ and substitute $- \frac{du}{2}$ :
      
      $\int \frac{e^{u}}{4}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{e^{u}}{2}\right)\, du = - \frac{\int e^{u}\, du}{2}$
      
      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.
  2. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- \frac{e^{- 2 x}}{2}\right)\, dx = - \frac{\int e^{- 2 x}\, dx}{2}$
    1. Let $u = - 2 x$ .
      
      Then let $du = - 2 dx$ and substitute $- \frac{du}{2}$ :
      
      $\int \frac{e^{u}}{4}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{e^{u}}{2}\right)\, du = - \frac{\int e^{u}\, du}{2}$
      
      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: $\frac{e^{- 2 x}}{4}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- 3 e^{- 2 x}\right)\, dx = - 3 \int e^{- 2 x}\, dx$
    1. Let $u = - 2 x$ .
      
      Then let $du = - 2 dx$ and substitute $- \frac{du}{2}$ :
      
      $\int \frac{e^{u}}{4}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{e^{u}}{2}\right)\, du = - \frac{\int e^{u}\, du}{2}$
      
      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: $\frac{3 e^{- 2 x}}{2}$
  The result is: $- \frac{x e^{- 2 x}}{2} + \frac{5 e^{- 2 x}}{4}$
Now simplify:

$\frac{\left(5 - 2 x\right) e^{- 2 x}}{4}$
Add the constant of integration:

$\frac{\left(5 - 2 x\right) e^{- 2 x}}{4}+ \mathrm{constant}$

The answer is:

$\frac{\left(5 - 2 x\right) e^{- 2 x}}{4}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                             
 |                         -2*x             -2*x
 |          -2*x          e       (-3 + x)*e    
 | (x - 3)*e     dx = C - ----- - --------------
 |                          4           2       
/

$$\int \left(x - 3\right) e^{- 2 x}\, dx = C - \frac{\left(x - 3\right) e^{- 2 x}}{2} - \frac{e^{- 2 x}}{4}$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

         -2
  5   3*e  
- - + -----
  4     4

$$- \frac{5}{4} + \frac{3}{4 e^{2}}$$

         -2
  5   3*e  
- - + -----
  4     4

$$- \frac{5}{4} + \frac{3}{4 e^{2}}$$

Упростить

Numerical answer [src]

-1.14849853757254

-1.14849853757254

The graph

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

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Identical expressions

Similar expressions

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

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #1

Method #2

Method #2