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Graphing y =:
e^(2*x-1)
Derivative of:
e^(2*x-1)
Identical expressions
e^(two *x- one)
e to the power of (2 multiply by x minus 1)
e to the power of (two multiply by x minus one)
e(2*x-1)
e2*x-1
e^(2x-1)
e(2x-1)
e2x-1
e^2x-1
e^(2*x-1)dx
Similar expressions
e^(2*x+1)
(e^(2*x)-1)/e^x

Integral of e^(2*x-1) dx

The solution

You have entered [src]

  2            
  /            
 |             
 |   2*x - 1   
 |  E        dx
 |             
/              
1

$$\int\limits_{1}^{2} e^{2 x - 1}\, dx$$

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

Detail solution

There are multiple ways to do this integral.
Method #1
1. Let $u = 2 x - 1$ .
  
  Then let $du = 2 dx$ and substitute $\frac{du}{2}$ :
  
  $\int \frac{e^{u}}{2}\, du$
  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:
  
  $\frac{e^{2 x - 1}}{2}$
Method #2
1. Rewrite the integrand:
  
  $e^{2 x - 1} = \frac{e^{2 x}}{e}$
2. The integral of a constant times a function is the constant times the integral of the function:
  
  $\int \frac{e^{2 x}}{e}\, dx = \frac{\int e^{2 x}\, dx}{e}$
  1. Let $u = 2 x$ .
    
    Then let $du = 2 dx$ and substitute $\frac{du}{2}$ :
    
    $\int \frac{e^{u}}{2}\, du$
    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:
    
    $\frac{e^{2 x}}{2}$
  So, the result is: $\frac{e^{2 x}}{2 e}$
Method #3
1. Rewrite the integrand:
  
  $e^{2 x - 1} = \frac{e^{2 x}}{e}$
2. The integral of a constant times a function is the constant times the integral of the function:
  
  $\int \frac{e^{2 x}}{e}\, dx = \frac{\int e^{2 x}\, dx}{e}$
  1. Let $u = 2 x$ .
    
    Then let $du = 2 dx$ and substitute $\frac{du}{2}$ :
    
    $\int \frac{e^{u}}{2}\, du$
    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:
    
    $\frac{e^{2 x}}{2}$
  So, the result is: $\frac{e^{2 x}}{2 e}$
Now simplify:

$\frac{e^{2 x - 1}}{2}$
Add the constant of integration:

$\frac{e^{2 x - 1}}{2}+ \mathrm{constant}$

The answer is:

$\frac{e^{2 x - 1}}{2}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                          
 |                    2*x - 1
 |  2*x - 1          e       
 | E        dx = C + --------
 |                      2    
/

$$\int e^{2 x - 1}\, dx = C + \frac{e^{2 x - 1}}{2}$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

 3    
e    E
-- - -
2    2

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

 3    
e    E
-- - -
2    2

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

exp(3)/2 - E/2

Numerical answer [src]

8.68362754736431

8.68362754736431

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

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Identical expressions

Similar expressions

Integral of e^(2*x-1) dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #3