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Solving integrals/ (x^2+2)×e^(x/2)

Integral of (x^2+2)×e^(x/2) dx

The solution

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  1               
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 |            x   
 |            -   
 |  / 2    \  2   
 |  \x  + 2/*e  dx
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0

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

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

Detail solution

There are multiple ways to do this integral.
Method #1
1. Rewrite the integrand:
  
  $\left(x^{2} + 2\right) e^{\frac{x}{2}} = x^{2} e^{\frac{x}{2}} + 2 e^{\frac{x}{2}}$
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 $u{\left(x \right)} = x^{2}$ and let $\operatorname{dv}{\left(x \right)} = e^{\frac{x}{2}}$ .
    
    Then $\operatorname{du}{\left(x \right)} = 2 x$ .
    
    To find $v{\left(x \right)}$ :
    1. There are multiple ways to do this integral.
      
      Method #1
      
      Let $u = \frac{x}{2}$ .
      
      Then let $du = \frac{dx}{2}$ and substitute $2 du$ :
      
      $\int 4 e^{u}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int 2 e^{u}\, du = 2 \int e^{u}\, du$
      
      The integral of the exponential function is itself.
      
      $\int e^{u}\, du = e^{u}$
      
      So, the result is: $2 e^{u}$
      
      Now substitute $u$ back in:
      
      $2 e^{\frac{x}{2}}$
      
      Method #2
      
      Let $u = e^{\frac{x}{2}}$ .
      
      Then let $du = \frac{e^{\frac{x}{2}} dx}{2}$ and substitute $2 du$ :
      
      $\int 4\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int 2\, du = 2 \int 1\, du$
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int 1\, du = u$
      
      So, the result is: $2 u$
      
      Now substitute $u$ back in:
      
      $2 e^{\frac{x}{2}}$
    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 $u{\left(x \right)} = 4 x$ and let $\operatorname{dv}{\left(x \right)} = e^{\frac{x}{2}}$ .
    
    Then $\operatorname{du}{\left(x \right)} = 4$ .
    
    To find $v{\left(x \right)}$ :
    1. Let $u = \frac{x}{2}$ .
      
      Then let $du = \frac{dx}{2}$ and substitute $2 du$ :
      
      $\int 4 e^{u}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int 2 e^{u}\, du = 2 \int e^{u}\, du$
      
      The integral of the exponential function is itself.
      
      $\int e^{u}\, du = e^{u}$
      
      So, the result is: $2 e^{u}$
      
      Now substitute $u$ back in:
      
      $2 e^{\frac{x}{2}}$
    Now evaluate the sub-integral.
  3. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int 8 e^{\frac{x}{2}}\, dx = 8 \int e^{\frac{x}{2}}\, dx$
    1. Let $u = \frac{x}{2}$ .
      
      Then let $du = \frac{dx}{2}$ and substitute $2 du$ :
      
      $\int 4 e^{u}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int 2 e^{u}\, du = 2 \int e^{u}\, du$
      
      The integral of the exponential function is itself.
      
      $\int e^{u}\, du = e^{u}$
      
      So, the result is: $2 e^{u}$
      
      Now substitute $u$ back in:
      
      $2 e^{\frac{x}{2}}$
    So, the result is: $16 e^{\frac{x}{2}}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int 2 e^{\frac{x}{2}}\, dx = 2 \int e^{\frac{x}{2}}\, dx$
    1. Let $u = \frac{x}{2}$ .
      
      Then let $du = \frac{dx}{2}$ and substitute $2 du$ :
      
      $\int 4 e^{u}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int 2 e^{u}\, du = 2 \int e^{u}\, du$
      
      The integral of the exponential function is itself.
      
      $\int e^{u}\, du = e^{u}$
      
      So, the result is: $2 e^{u}$
      
      Now substitute $u$ back in:
      
      $2 e^{\frac{x}{2}}$
    So, the result is: $4 e^{\frac{x}{2}}$
  The result is: $2 x^{2} e^{\frac{x}{2}} - 8 x e^{\frac{x}{2}} + 20 e^{\frac{x}{2}}$
Method #2
1. Use integration by parts:
  
  $\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$
  
  Let $u{\left(x \right)} = x^{2} + 2$ and let $\operatorname{dv}{\left(x \right)} = e^{\frac{x}{2}}$ .
  
  Then $\operatorname{du}{\left(x \right)} = 2 x$ .
  
  To find $v{\left(x \right)}$ :
  1. Let $u = \frac{x}{2}$ .
    
    Then let $du = \frac{dx}{2}$ and substitute $2 du$ :
    
    $\int 4 e^{u}\, du$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int 2 e^{u}\, du = 2 \int e^{u}\, du$
      
      The integral of the exponential function is itself.
      
      $\int e^{u}\, du = e^{u}$
      
      So, the result is: $2 e^{u}$
    Now substitute $u$ back in:
    
    $2 e^{\frac{x}{2}}$
  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 $u{\left(x \right)} = 4 x$ and let $\operatorname{dv}{\left(x \right)} = e^{\frac{x}{2}}$ .
  
  Then $\operatorname{du}{\left(x \right)} = 4$ .
  
  To find $v{\left(x \right)}$ :
  1. Let $u = \frac{x}{2}$ .
    
    Then let $du = \frac{dx}{2}$ and substitute $2 du$ :
    
    $\int 4 e^{u}\, du$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int 2 e^{u}\, du = 2 \int e^{u}\, du$
      
      The integral of the exponential function is itself.
      
      $\int e^{u}\, du = e^{u}$
      
      So, the result is: $2 e^{u}$
    Now substitute $u$ back in:
    
    $2 e^{\frac{x}{2}}$
  Now evaluate the sub-integral.
3. The integral of a constant times a function is the constant times the integral of the function:
  
  $\int 8 e^{\frac{x}{2}}\, dx = 8 \int e^{\frac{x}{2}}\, dx$
  1. Let $u = \frac{x}{2}$ .
    
    Then let $du = \frac{dx}{2}$ and substitute $2 du$ :
    
    $\int 4 e^{u}\, du$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int 2 e^{u}\, du = 2 \int e^{u}\, du$
      
      The integral of the exponential function is itself.
      
      $\int e^{u}\, du = e^{u}$
      
      So, the result is: $2 e^{u}$
    Now substitute $u$ back in:
    
    $2 e^{\frac{x}{2}}$
  So, the result is: $16 e^{\frac{x}{2}}$
Method #3
1. Rewrite the integrand:
  
  $\left(x^{2} + 2\right) e^{\frac{x}{2}} = x^{2} e^{\frac{x}{2}} + 2 e^{\frac{x}{2}}$
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 $u{\left(x \right)} = x^{2}$ and let $\operatorname{dv}{\left(x \right)} = e^{\frac{x}{2}}$ .
    
    Then $\operatorname{du}{\left(x \right)} = 2 x$ .
    
    To find $v{\left(x \right)}$ :
    1. Let $u = \frac{x}{2}$ .
      
      Then let $du = \frac{dx}{2}$ and substitute $2 du$ :
      
      $\int 4 e^{u}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int 2 e^{u}\, du = 2 \int e^{u}\, du$
      
      The integral of the exponential function is itself.
      
      $\int e^{u}\, du = e^{u}$
      
      So, the result is: $2 e^{u}$
      
      Now substitute $u$ back in:
      
      $2 e^{\frac{x}{2}}$
    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 $u{\left(x \right)} = 4 x$ and let $\operatorname{dv}{\left(x \right)} = e^{\frac{x}{2}}$ .
    
    Then $\operatorname{du}{\left(x \right)} = 4$ .
    
    To find $v{\left(x \right)}$ :
    1. Let $u = \frac{x}{2}$ .
      
      Then let $du = \frac{dx}{2}$ and substitute $2 du$ :
      
      $\int 4 e^{u}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int 2 e^{u}\, du = 2 \int e^{u}\, du$
      
      The integral of the exponential function is itself.
      
      $\int e^{u}\, du = e^{u}$
      
      So, the result is: $2 e^{u}$
      
      Now substitute $u$ back in:
      
      $2 e^{\frac{x}{2}}$
    Now evaluate the sub-integral.
  3. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int 8 e^{\frac{x}{2}}\, dx = 8 \int e^{\frac{x}{2}}\, dx$
    1. Let $u = \frac{x}{2}$ .
      
      Then let $du = \frac{dx}{2}$ and substitute $2 du$ :
      
      $\int 4 e^{u}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int 2 e^{u}\, du = 2 \int e^{u}\, du$
      
      The integral of the exponential function is itself.
      
      $\int e^{u}\, du = e^{u}$
      
      So, the result is: $2 e^{u}$
      
      Now substitute $u$ back in:
      
      $2 e^{\frac{x}{2}}$
    So, the result is: $16 e^{\frac{x}{2}}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int 2 e^{\frac{x}{2}}\, dx = 2 \int e^{\frac{x}{2}}\, dx$
    1. Let $u = \frac{x}{2}$ .
      
      Then let $du = \frac{dx}{2}$ and substitute $2 du$ :
      
      $\int 4 e^{u}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int 2 e^{u}\, du = 2 \int e^{u}\, du$
      
      The integral of the exponential function is itself.
      
      $\int e^{u}\, du = e^{u}$
      
      So, the result is: $2 e^{u}$
      
      Now substitute $u$ back in:
      
      $2 e^{\frac{x}{2}}$
    So, the result is: $4 e^{\frac{x}{2}}$
  The result is: $2 x^{2} e^{\frac{x}{2}} - 8 x e^{\frac{x}{2}} + 20 e^{\frac{x}{2}}$
Now simplify:

$2 \left(x^{2} - 4 x + 10\right) e^{\frac{x}{2}}$
Add the constant of integration:

$2 \left(x^{2} - 4 x + 10\right) e^{\frac{x}{2}}+ \mathrm{constant}$

The answer is:

$2 \left(x^{2} - 4 x + 10\right) e^{\frac{x}{2}}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                             
 |                                              
 |           x              x        x         x
 |           -              -        -         -
 | / 2    \  2              2        2      2  2
 | \x  + 2/*e  dx = C + 20*e  - 8*x*e  + 2*x *e 
 |                                              
/

$$\int \left(x^{2} + 2\right) e^{\frac{x}{2}}\, dx = C + 2 x^{2} e^{\frac{x}{2}} - 8 x e^{\frac{x}{2}} + 20 e^{\frac{x}{2}}$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

          1/2
-20 + 14*e

$$-20 + 14 e^{\frac{1}{2}}$$

          1/2
-20 + 14*e

$$-20 + 14 e^{\frac{1}{2}}$$

Упростить

Numerical answer [src]

3.08209778980179

3.08209778980179

The graph

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

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Identical expressions

Similar expressions

Integral of (x^2+2)×e^(x/2) dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #1

Method #2

Method #2

Method #3