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(two y^2+y)exp^(-2y)
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(two y squared plus y) exponent of to the power of ( minus 2y)
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Similar expressions
(2y^2-y)exp^(-2y)
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Integral of (2y^2+y)exp^(-2y) dx

The solution

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  1                    
  /                    
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 |  /   2    \  -2*y   
 |  \2*y  + y/*E     dy
 |                     
/                      
0

$$\int\limits_{0}^{1} e^{- 2 y} \left(2 y^{2} + y\right)\, dy$$

Integral((2*y^2 + y)*E^(-2*y), (y, 0, 1))

Detail solution

Rewrite the integrand:

$e^{- 2 y} \left(2 y^{2} + y\right) = 2 y^{2} e^{- 2 y} + y e^{- 2 y}$
Integrate term-by-term:
1. The integral of a constant times a function is the constant times the integral of the function:
  
  $\int 2 y^{2} e^{- 2 y}\, dy = 2 \int y^{2} e^{- 2 y}\, dy$
  1. Use integration by parts:
    
    $\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$
    
    Let $u{\left(y \right)} = y^{2}$ and let $\operatorname{dv}{\left(y \right)} = e^{- 2 y}$ .
    
    Then $\operatorname{du}{\left(y \right)} = 2 y$ .
    
    To find $v{\left(y \right)}$ :
    1. Let $u = - 2 y$ .
      
      Then let $du = - 2 dy$ 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:
        
        $\text{False}$
        
        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:
      
      $- \frac{e^{- 2 y}}{2}$
    Now evaluate the sub-integral.
  2. Use integration by parts:
    
    $\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$
    
    Let $u{\left(y \right)} = - y$ and let $\operatorname{dv}{\left(y \right)} = e^{- 2 y}$ .
    
    Then $\operatorname{du}{\left(y \right)} = -1$ .
    
    To find $v{\left(y \right)}$ :
    1. Let $u = - 2 y$ .
      
      Then let $du = - 2 dy$ 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:
        
        $\text{False}$
        
        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:
      
      $- \frac{e^{- 2 y}}{2}$
    Now evaluate the sub-integral.
  3. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \frac{e^{- 2 y}}{2}\, dy = \frac{\int e^{- 2 y}\, dy}{2}$
    1. Let $u = - 2 y$ .
      
      Then let $du = - 2 dy$ 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:
        
        $\text{False}$
        
        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:
      
      $- \frac{e^{- 2 y}}{2}$
    So, the result is: $- \frac{e^{- 2 y}}{4}$
  So, the result is: $- y^{2} e^{- 2 y} - y e^{- 2 y} - \frac{e^{- 2 y}}{2}$
1. Use integration by parts:
  
  $\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$
  
  Let $u{\left(y \right)} = y$ and let $\operatorname{dv}{\left(y \right)} = e^{- 2 y}$ .
  
  Then $\operatorname{du}{\left(y \right)} = 1$ .
  
  To find $v{\left(y \right)}$ :
  1. Let $u = - 2 y$ .
    
    Then let $du = - 2 dy$ 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:
    
    $- \frac{e^{- 2 y}}{2}$
  Now evaluate the sub-integral.
2. The integral of a constant times a function is the constant times the integral of the function:
  
  $\int \left(- \frac{e^{- 2 y}}{2}\right)\, dy = - \frac{\int e^{- 2 y}\, dy}{2}$
  1. Let $u = - 2 y$ .
    
    Then let $du = - 2 dy$ 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:
    
    $- \frac{e^{- 2 y}}{2}$
  So, the result is: $\frac{e^{- 2 y}}{4}$
The result is: $- y^{2} e^{- 2 y} - \frac{3 y e^{- 2 y}}{2} - \frac{3 e^{- 2 y}}{4}$
Now simplify:

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

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

The answer is:

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

The answer (Indefinite) [src]

  /                                                        
 |                              -2*y                   -2*y
 | /   2    \  -2*y          3*e        2  -2*y   3*y*e    
 | \2*y  + y/*E     dy = C - ------- - y *e     - ---------
 |                              4                     2    
/

$$\int e^{- 2 y} \left(2 y^{2} + y\right)\, dy = C - y^{2} e^{- 2 y} - \frac{3 y e^{- 2 y}}{2} - \frac{3 e^{- 2 y}}{4}$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

        -2
3   13*e  
- - ------
4     4

$$\frac{3}{4} - \frac{13}{4 e^{2}}$$

        -2
3   13*e  
- - ------
4     4

$$\frac{3}{4} - \frac{13}{4 e^{2}}$$

3/4 - 13*exp(-2)/4

Numerical answer [src]

0.310160329481009

0.310160329481009

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

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Integral of (2y^2+y)exp^(-2y) dx

Limits of integration:

The graph:

Piecewise:

The solution