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Solving integrals/ 2^(2*x-1)

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

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The solution

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  1            
  /            
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 |   2*x - 1   
 |  2        dx
 |             
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0
∫0122x−1 dx\int\limits_{0}^{1} 2^{2 x - 1}\, dx 
Integral(2^(2*x - 1), (x, 0, 1))
Detail solution

  1. There are multiple ways to do this integral.

    Method #1

    1. Let u=2x−1u = 2 x - 1.

      Then let du=2dxdu = 2 dx and substitute du2\frac{du}{2}:

      ∫2u2 du\int \frac{2^{u}}{2}\, du

      1. The integral of a constant times a function is the constant times the integral of the function:

        ∫2u du=∫2u du2\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.

          ∫2u du=2ulog⁡(2)\int 2^{u}\, du = \frac{2^{u}}{\log{\left(2 \right)}}

        So, the result is: 2u2log⁡(2)\frac{2^{u}}{2 \log{\left(2 \right)}}

      Now substitute uu back in:

      22x−12log⁡(2)\frac{2^{2 x - 1}}{2 \log{\left(2 \right)}}

    Method #2

    1. Rewrite the integrand:

      22x−1=22x22^{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:

      ∫22x2 dx=∫22x dx2\int \frac{2^{2 x}}{2}\, dx = \frac{\int 2^{2 x}\, dx}{2}

      1. Let u=2xu = 2 x.

        Then let du=2dxdu = 2 dx and substitute du2\frac{du}{2}:

        ∫2u2 du\int \frac{2^{u}}{2}\, du

        1. The integral of a constant times a function is the constant times the integral of the function:

          ∫2u du=∫2u du2\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.

            ∫2u du=2ulog⁡(2)\int 2^{u}\, du = \frac{2^{u}}{\log{\left(2 \right)}}

          So, the result is: 2u2log⁡(2)\frac{2^{u}}{2 \log{\left(2 \right)}}

        Now substitute uu back in:

        22x2log⁡(2)\frac{2^{2 x}}{2 \log{\left(2 \right)}}

      So, the result is: 22x4log⁡(2)\frac{2^{2 x}}{4 \log{\left(2 \right)}}

    Method #3

    1. Rewrite the integrand:

      22x−1=22x22^{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:

      ∫22x2 dx=∫22x dx2\int \frac{2^{2 x}}{2}\, dx = \frac{\int 2^{2 x}\, dx}{2}

      1. Let u=2xu = 2 x.

        Then let du=2dxdu = 2 dx and substitute du2\frac{du}{2}:

        ∫2u2 du\int \frac{2^{u}}{2}\, du

        1. The integral of a constant times a function is the constant times the integral of the function:

          ∫2u du=∫2u du2\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.

            ∫2u du=2ulog⁡(2)\int 2^{u}\, du = \frac{2^{u}}{\log{\left(2 \right)}}

          So, the result is: 2u2log⁡(2)\frac{2^{u}}{2 \log{\left(2 \right)}}

        Now substitute uu back in:

        22x2log⁡(2)\frac{2^{2 x}}{2 \log{\left(2 \right)}}

      So, the result is: 22x4log⁡(2)\frac{2^{2 x}}{4 \log{\left(2 \right)}}

  2. Now simplify:

    22x−2log⁡(2)\frac{2^{2 x - 2}}{\log{\left(2 \right)}}

  3. Add the constant of integration:

    22x−2log⁡(2)+constant\frac{2^{2 x - 2}}{\log{\left(2 \right)}}+ \mathrm{constant}

The answer is:

22x−2log⁡(2)+constant\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)
/
∫22x−1 dx=22x−12log⁡(2)+C\int 2^{2 x - 1}\, dx = \frac{2^{2 x - 1}}{2 \log{\left(2 \right)}} + C
Simplify
The graph
0.001.000.100.200.300.400.500.600.700.800.9004
Plot the graph f(x)
The answer [src] 
   3    
--------
4*log(2)
34log⁡(2)\frac{3}{4 \log{\left(2 \right)}} 
=
= 
   3    
--------
4*log(2)
34log⁡(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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