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Identical expressions
(one -5x)*e^(-3x)
(1 minus 5x) multiply by e to the power of ( minus 3x)
(one minus 5x) multiply by e to the power of ( minus 3x)
(1-5x)*e(-3x)
1-5x*e-3x
(1-5x)e^(-3x)
(1-5x)e(-3x)
1-5xe-3x
1-5xe^-3x
(1-5x)*e^(-3x)dx
Similar expressions
(1-5x)*e^(3x)
(1+5x)*e^(-3x)

Integral of (1-5x)*e^(-3x) dx

The solution

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  1                   
  /                   
 |                    
 |             -3*x   
 |  (1 - 5*x)*E     dx
 |                    
/                     
0

$$\int\limits_{0}^{1} e^{- 3 x} \left(1 - 5 x\right)\, dx$$

Integral((1 - 5*x)*E^(-3*x), (x, 0, 1))

Detail solution

There are multiple ways to do this integral.
Method #1
1. Rewrite the integrand:
  
  $e^{- 3 x} \left(1 - 5 x\right) = - \left(5 x - 1\right) e^{- 3 x}$
2. The integral of a constant times a function is the constant times the integral of the function:
  
  $\int \left(- \left(5 x - 1\right) e^{- 3 x}\right)\, dx = - \int \left(5 x - 1\right) e^{- 3 x}\, dx$
  1. Use integration by parts:
    
    $\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$
    
    Let $u{\left(x \right)} = 5 x - 1$ and let $\operatorname{dv}{\left(x \right)} = e^{- 3 x}$ .
    
    Then $\operatorname{du}{\left(x \right)} = 5$ .
    
    To find $v{\left(x \right)}$ :
    1. Let $u = - 3 x$ .
      
      Then let $du = - 3 dx$ and substitute $- \frac{du}{3}$ :
      
      $\int \left(- \frac{e^{u}}{3}\right)\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\text{False}$
      
      The integral of the exponential function is itself.
      
      $\int e^{u}\, du = e^{u}$
      
      So, the result is: $- \frac{e^{u}}{3}$
      
      Now substitute $u$ back in:
      
      $- \frac{e^{- 3 x}}{3}$
    Now evaluate the sub-integral.
  2. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- \frac{5 e^{- 3 x}}{3}\right)\, dx = - \frac{5 \int e^{- 3 x}\, dx}{3}$
    1. Let $u = - 3 x$ .
      
      Then let $du = - 3 dx$ and substitute $- \frac{du}{3}$ :
      
      $\int \left(- \frac{e^{u}}{3}\right)\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\text{False}$
      
      The integral of the exponential function is itself.
      
      $\int e^{u}\, du = e^{u}$
      
      So, the result is: $- \frac{e^{u}}{3}$
      
      Now substitute $u$ back in:
      
      $- \frac{e^{- 3 x}}{3}$
    So, the result is: $\frac{5 e^{- 3 x}}{9}$
  So, the result is: $\frac{\left(5 x - 1\right) e^{- 3 x}}{3} + \frac{5 e^{- 3 x}}{9}$
Method #2
1. Rewrite the integrand:
  
  $e^{- 3 x} \left(1 - 5 x\right) = - 5 x e^{- 3 x} + e^{- 3 x}$
2. Integrate term-by-term:
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- 5 x e^{- 3 x}\right)\, dx = - 5 \int x e^{- 3 x}\, dx$
    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 $\operatorname{dv}{\left(x \right)} = e^{- 3 x}$ .
      
      Then $\operatorname{du}{\left(x \right)} = 1$ .
      
      To find $v{\left(x \right)}$ :
      
      Let $u = - 3 x$ .
      
      Then let $du = - 3 dx$ and substitute $- \frac{du}{3}$ :
      
      $\int \left(- \frac{e^{u}}{3}\right)\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\text{False}$
      
      The integral of the exponential function is itself.
      
      $\int e^{u}\, du = e^{u}$
      
      So, the result is: $- \frac{e^{u}}{3}$
      
      Now substitute $u$ back in:
      
      $- \frac{e^{- 3 x}}{3}$
      
      Now evaluate the sub-integral.
    2. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{e^{- 3 x}}{3}\right)\, dx = - \frac{\int e^{- 3 x}\, dx}{3}$
      
      Let $u = - 3 x$ .
      
      Then let $du = - 3 dx$ and substitute $- \frac{du}{3}$ :
      
      $\int \left(- \frac{e^{u}}{3}\right)\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\text{False}$
      
      The integral of the exponential function is itself.
      
      $\int e^{u}\, du = e^{u}$
      
      So, the result is: $- \frac{e^{u}}{3}$
      
      Now substitute $u$ back in:
      
      $- \frac{e^{- 3 x}}{3}$
      
      So, the result is: $\frac{e^{- 3 x}}{9}$
    So, the result is: $\frac{5 x e^{- 3 x}}{3} + \frac{5 e^{- 3 x}}{9}$
  1. Let $u = - 3 x$ .
    
    Then let $du = - 3 dx$ and substitute $- \frac{du}{3}$ :
    
    $\int \left(- \frac{e^{u}}{3}\right)\, du$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\text{False}$
      
      The integral of the exponential function is itself.
      
      $\int e^{u}\, du = e^{u}$
      
      So, the result is: $- \frac{e^{u}}{3}$
    Now substitute $u$ back in:
    
    $- \frac{e^{- 3 x}}{3}$
  The result is: $\frac{5 x e^{- 3 x}}{3} + \frac{2 e^{- 3 x}}{9}$
Now simplify:

$\frac{\left(15 x + 2\right) e^{- 3 x}}{9}$
Add the constant of integration:

$\frac{\left(15 x + 2\right) e^{- 3 x}}{9}+ \mathrm{constant}$

The answer is:

$\frac{\left(15 x + 2\right) e^{- 3 x}}{9}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                                   
 |                             -3*x               -3*x
 |            -3*x          5*e       (-1 + 5*x)*e    
 | (1 - 5*x)*E     dx = C + ------- + ----------------
 |                             9             3        
/

$$\int e^{- 3 x} \left(1 - 5 x\right)\, dx = C + \frac{\left(5 x - 1\right) e^{- 3 x}}{3} + \frac{5 e^{- 3 x}}{9}$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

          -3
  2   17*e  
- - + ------
  9     9

$$- \frac{2}{9} + \frac{17}{9 e^{3}}$$

          -3
  2   17*e  
- - + ------
  9     9

$$- \frac{2}{9} + \frac{17}{9 e^{3}}$$

-2/9 + 17*exp(-3)/9

Numerical answer [src]

-0.128179981971813

-0.128179981971813

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

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Identical expressions

Similar expressions

Integral of (1-5x)*e^(-3x) dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2