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Identical expressions
x^ three *e^(four *x)
x cubed multiply by e to the power of (4 multiply by x)
x to the power of three multiply by e to the power of (four multiply by x)
x3*e(4*x)
x3*e4*x
x³*e^(4*x)
x to the power of 3*e to the power of (4*x)
x^3e^(4x)
x3e(4x)
x3e4x
x^3e^4x
x^3*e^(4*x)dx
What you mean?
x^3*e^(4*x)
x^((3*e)^(4*x))
x^((3*e)^(4*x))

You entered:

x^3*e^(4*x)

What you mean?

x^3*e^(4*x)

x^((3*e)^(4*x))

x^((3*e)^(4*x))

Integral of x^3e^(4x) dx

The solution

You have entered [src]

  1           
  /           
 |            
 |   3  4*x   
 |  x *e    dx
 |            
/             
0

$$\int\limits_{0}^{1} x^{3} e^{4 x}\, dx$$

Integral(x^3*E^(4*x), (x, 0, 1))

Detail solution

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 $\operatorname{dv}{\left(x \right)} = e^{4 x}$ .

Then $\operatorname{du}{\left(x \right)} = 3 x^{2}$ .

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

$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$

Let $u{\left(x \right)} = \frac{3 x^{2}}{4}$ and let $\operatorname{dv}{\left(x \right)} = e^{4 x}$ .

Then $\operatorname{du}{\left(x \right)} = \frac{3 x}{2}$ .

To find $v{\left(x \right)}$ :
1. Let $u = 4 x$ .
  
  Then let $du = 4 dx$ and substitute $\frac{du}{4}$ :
  
  $\int \frac{e^{u}}{16}\, du$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \frac{e^{u}}{4}\, du = \frac{\int e^{u}\, du}{4}$
    1. The integral of the exponential function is itself.
      
      $\int e^{u}\, du = e^{u}$
    So, the result is: $\frac{e^{u}}{4}$
  Now substitute $u$ back in:
  
  $\frac{e^{4 x}}{4}$
Now evaluate the sub-integral.
Use integration by parts:

$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$

Let $u{\left(x \right)} = \frac{3 x}{8}$ and let $\operatorname{dv}{\left(x \right)} = e^{4 x}$ .

Then $\operatorname{du}{\left(x \right)} = \frac{3}{8}$ .

To find $v{\left(x \right)}$ :
1. Let $u = 4 x$ .
  
  Then let $du = 4 dx$ and substitute $\frac{du}{4}$ :
  
  $\int \frac{e^{u}}{16}\, du$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \frac{e^{u}}{4}\, du = \frac{\int e^{u}\, du}{4}$
    1. The integral of the exponential function is itself.
      
      $\int e^{u}\, du = e^{u}$
    So, the result is: $\frac{e^{u}}{4}$
  Now substitute $u$ back in:
  
  $\frac{e^{4 x}}{4}$
Now evaluate the sub-integral.
The integral of a constant times a function is the constant times the integral of the function:

$\int \frac{3 e^{4 x}}{32}\, dx = \frac{3 \int e^{4 x}\, dx}{32}$
1. Let $u = 4 x$ .
  
  Then let $du = 4 dx$ and substitute $\frac{du}{4}$ :
  
  $\int \frac{e^{u}}{16}\, du$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \frac{e^{u}}{4}\, du = \frac{\int e^{u}\, du}{4}$
    1. The integral of the exponential function is itself.
      
      $\int e^{u}\, du = e^{u}$
    So, the result is: $\frac{e^{u}}{4}$
  Now substitute $u$ back in:
  
  $\frac{e^{4 x}}{4}$
So, the result is: $\frac{3 e^{4 x}}{128}$
Now simplify:

$\frac{\left(32 x^{3} - 24 x^{2} + 12 x - 3\right) e^{4 x}}{128}$
Add the constant of integration:

$\frac{\left(32 x^{3} - 24 x^{2} + 12 x - 3\right) e^{4 x}}{128}+ \mathrm{constant}$

The answer is:

$\frac{\left(32 x^{3} - 24 x^{2} + 12 x - 3\right) e^{4 x}}{128}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                                        
 |                     4*x      2  4*x    3  4*x        4*x
 |  3  4*x          3*e      3*x *e      x *e      3*x*e   
 | x *e    dx = C - ------ - --------- + ------- + --------
 |                   128         16         4         32   
/

$${{\left(32\,x^3-24\,x^2+12\,x-3\right)\,e^{4\,x}}\over{128}}$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

          4
 3    17*e 
--- + -----
128    128

$${{17\,e^4}\over{128}}+{{3}\over{128}}$$

          4
 3    17*e 
--- + -----
128    128

$$\frac{3}{128} + \frac{17 e^{4}}{128}$$

Упростить

Numerical answer [src]

7.27475430127697

7.27475430127697

The graph

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

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How to use it?

Identical expressions

What you mean?

You entered:

What you mean?

Integral of x^3e^(4x) dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Other calculators

How to use it?

Identical expressions

What you mean?

You entered:

What you mean?

Integral of x^3*e^(4*x) dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Integral of x^3e^(4x) dx