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Similar expressions
2^(2*x+1)

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

The solution

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  1            
  /            
 |             
 |   2*x - 1   
 |  2        dx
 |             
/              
0

$$\int\limits_{0}^{1} 2^{2 x - 1}\, dx$$

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

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 $\frac{du}{2}$ :
  
  $\int \frac{2^{u}}{2}\, du$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int 2^{u}\, du = \frac{\int 2^{u}\, du}{2}$
    1. The integral of an exponential function is itself divided by the natural logarithm of the base.
      
      $\int 2^{u}\, du = \frac{2^{u}}{\log{\left(2 \right)}}$
    So, the result is: $\frac{2^{u}}{2 \log{\left(2 \right)}}$
  Now substitute $u$ back in:
  
  $\frac{2^{2 x - 1}}{2 \log{\left(2 \right)}}$
Method #2
1. Rewrite the integrand:
  
  $2^{2 x - 1} = \frac{2^{2 x}}{2}$
2. The integral of a constant times a function is the constant times the integral of the function:
  
  $\int \frac{2^{2 x}}{2}\, dx = \frac{\int 2^{2 x}\, dx}{2}$
  1. Let $u = 2 x$ .
    
    Then let $du = 2 dx$ and substitute $\frac{du}{2}$ :
    
    $\int \frac{2^{u}}{2}\, du$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int 2^{u}\, du = \frac{\int 2^{u}\, du}{2}$
      
      The integral of an exponential function is itself divided by the natural logarithm of the base.
      
      $\int 2^{u}\, du = \frac{2^{u}}{\log{\left(2 \right)}}$
      
      So, the result is: $\frac{2^{u}}{2 \log{\left(2 \right)}}$
    Now substitute $u$ back in:
    
    $\frac{2^{2 x}}{2 \log{\left(2 \right)}}$
  So, the result is: $\frac{2^{2 x}}{4 \log{\left(2 \right)}}$
Method #3
1. Rewrite the integrand:
  
  $2^{2 x - 1} = \frac{2^{2 x}}{2}$
2. The integral of a constant times a function is the constant times the integral of the function:
  
  $\int \frac{2^{2 x}}{2}\, dx = \frac{\int 2^{2 x}\, dx}{2}$
  1. Let $u = 2 x$ .
    
    Then let $du = 2 dx$ and substitute $\frac{du}{2}$ :
    
    $\int \frac{2^{u}}{2}\, du$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int 2^{u}\, du = \frac{\int 2^{u}\, du}{2}$
      
      The integral of an exponential function is itself divided by the natural logarithm of the base.
      
      $\int 2^{u}\, du = \frac{2^{u}}{\log{\left(2 \right)}}$
      
      So, the result is: $\frac{2^{u}}{2 \log{\left(2 \right)}}$
    Now substitute $u$ back in:
    
    $\frac{2^{2 x}}{2 \log{\left(2 \right)}}$
  So, the result is: $\frac{2^{2 x}}{4 \log{\left(2 \right)}}$
Now simplify:

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

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

The answer is:

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

The answer (Indefinite) [src]

  /                          
 |                    2*x - 1
 |  2*x - 1          2       
 | 2        dx = C + --------
 |                   2*log(2)
/

$$\int 2^{2 x - 1}\, dx = \frac{2^{2 x - 1}}{2 \log{\left(2 \right)}} + C$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

   3    
--------
4*log(2)

$$\frac{3}{4 \log{\left(2 \right)}}$$

   3    
--------
4*log(2)

$$\frac{3}{4 \log{\left(2 \right)}}$$

3/(4*log(2))

Numerical answer [src]

1.08202128066672

1.08202128066672

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

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Identical expressions

Similar expressions

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

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #3