Integral of (1-6x)×e^(2x) dx

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$$\int\limits_{0}^{1} \left(1 - 6 x\right) e^{2 x}\, dx$$

Integral((1 - 6*x)*E^(2*x), (x, 0, 1))

Detail solution

There are multiple ways to do this integral.
Method #1
1. Rewrite the integrand:
  
  $\left(1 - 6 x\right) e^{2 x} = - 6 x e^{2 x} + e^{2 x}$
2. Integrate term-by-term:
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- 6 x e^{2 x}\right)\, dx = - 6 \int x e^{2 x}\, dx$
    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 $\operatorname{dv}{\left(x \right)} = e^{2 x}$ .
      
      Then $\operatorname{du}{\left(x \right)} = 1$ .
      
      To find $v{\left(x \right)}$ :
      
      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:
      
      $\int \frac{e^{u}}{2}\, du = \frac{\int e^{u}\, du}{2}$
      
      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:
      
      $\int \frac{1}{2}\, du = \frac{\int 1\, du}{2}$
      
      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:
      
      $\int \frac{e^{2 x}}{2}\, dx = \frac{\int e^{2 x}\, dx}{2}$
      
      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:
      
      $\int \frac{e^{u}}{2}\, du = \frac{\int e^{u}\, du}{2}$
      
      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: $\frac{e^{2 x}}{4}$
    So, the result is: $- 3 x e^{2 x} + \frac{3 e^{2 x}}{2}$
  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:
      
      $\int \frac{e^{u}}{2}\, du = \frac{\int e^{u}\, du}{2}$
      
      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:
    
    $\frac{e^{2 x}}{2}$
  The result is: $- 3 x e^{2 x} + 2 e^{2 x}$
Method #2
1. Use integration by parts:
  
  $\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$
  
  Let $u{\left(x \right)} = 1 - 6 x$ and let $\operatorname{dv}{\left(x \right)} = e^{2 x}$ .
  
  Then $\operatorname{du}{\left(x \right)} = -6$ .
  
  To find $v{\left(x \right)}$ :
  1. Let $u = 2 x$ .
    
    Then let $du = 2 dx$ and substitute $\frac{du}{2}$ :
    
    $\int \frac{e^{u}}{4}\, du$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{e^{u}}{2}\, du = \frac{\int e^{u}\, du}{2}$
      
      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:
  
  $\int \left(- 3 e^{2 x}\right)\, dx = - 3 \int e^{2 x}\, dx$
  1. Let $u = 2 x$ .
    
    Then let $du = 2 dx$ and substitute $\frac{du}{2}$ :
    
    $\int \frac{e^{u}}{4}\, du$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{e^{u}}{2}\, du = \frac{\int e^{u}\, du}{2}$
      
      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: $- \frac{3 e^{2 x}}{2}$
Method #3
1. Rewrite the integrand:
  
  $\left(1 - 6 x\right) e^{2 x} = - 6 x e^{2 x} + e^{2 x}$
2. Integrate term-by-term:
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- 6 x e^{2 x}\right)\, dx = - 6 \int x e^{2 x}\, dx$
    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 $\operatorname{dv}{\left(x \right)} = e^{2 x}$ .
      
      Then $\operatorname{du}{\left(x \right)} = 1$ .
      
      To find $v{\left(x \right)}$ :
      
      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:
      
      $\int \frac{e^{u}}{2}\, du = \frac{\int e^{u}\, du}{2}$
      
      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:
      
      $\int \frac{e^{2 x}}{2}\, dx = \frac{\int e^{2 x}\, dx}{2}$
      
      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:
      
      $\int \frac{e^{u}}{2}\, du = \frac{\int e^{u}\, du}{2}$
      
      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: $\frac{e^{2 x}}{4}$
    So, the result is: $- 3 x e^{2 x} + \frac{3 e^{2 x}}{2}$
  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:
      
      $\int \frac{e^{u}}{2}\, du = \frac{\int e^{u}\, du}{2}$
      
      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:
    
    $\frac{e^{2 x}}{2}$
  The result is: $- 3 x e^{2 x} + 2 e^{2 x}$
Now simplify:

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

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

The answer is:

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

The answer (Indefinite) [src]

  /                                         
 |                                          
 |            2*x             2*x        2*x
 | (1 - 6*x)*e    dx = C + 2*e    - 3*x*e   
 |                                          
/

$${{e^{2\,x}}\over{2}}-{{3\,\left(2\,x-1\right)\,e^{2\,x}}\over{2}}$$

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The answer [src]

      2
-2 - e

$$-e^2-2$$

=

      2
-2 - e

$$- e^{2} - 2$$

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Numerical answer [src]

-9.38905609893065

-9.38905609893065

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Integral of (1-6x)×e^(2x) dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #1

Method #2

Method #2

Method #3