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

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

Limits of integration:

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Piecewise:

The solution

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  2            
  /            
 |             
 |   2*x - 1   
 |  E        dx
 |             
/              
1
∫12e2x−1 dx\int\limits_{1}^{2} e^{2 x - 1}\, dx 
Integral(E^(2*x - 1), (x, 1, 2))
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}:

      ∫eu2 du\int \frac{e^{u}}{2}\, du

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

        False\text{False}

        1. The integral of the exponential function is itself.

          ∫eu du=eu\int e^{u}\, du = e^{u}

        So, the result is: eu2\frac{e^{u}}{2}

      Now substitute uu back in:

      e2x−12\frac{e^{2 x - 1}}{2}

    Method #2

    1. Rewrite the integrand:

      e2x−1=e2xee^{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:

      ∫e2xe dx=∫e2x dxe\int \frac{e^{2 x}}{e}\, dx = \frac{\int e^{2 x}\, dx}{e}

      1. Let u=2xu = 2 x.

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

        ∫eu2 du\int \frac{e^{u}}{2}\, du

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

          False\text{False}

          1. The integral of the exponential function is itself.

            ∫eu du=eu\int e^{u}\, du = e^{u}

          So, the result is: eu2\frac{e^{u}}{2}

        Now substitute uu back in:

        e2x2\frac{e^{2 x}}{2}

      So, the result is: e2x2e\frac{e^{2 x}}{2 e}

    Method #3

    1. Rewrite the integrand:

      e2x−1=e2xee^{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:

      ∫e2xe dx=∫e2x dxe\int \frac{e^{2 x}}{e}\, dx = \frac{\int e^{2 x}\, dx}{e}

      1. Let u=2xu = 2 x.

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

        ∫eu2 du\int \frac{e^{u}}{2}\, du

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

          False\text{False}

          1. The integral of the exponential function is itself.

            ∫eu du=eu\int e^{u}\, du = e^{u}

          So, the result is: eu2\frac{e^{u}}{2}

        Now substitute uu back in:

        e2x2\frac{e^{2 x}}{2}

      So, the result is: e2x2e\frac{e^{2 x}}{2 e}

  2. Now simplify:

    e2x−12\frac{e^{2 x - 1}}{2}

  3. Add the constant of integration:

    e2x−12+constant\frac{e^{2 x - 1}}{2}+ \mathrm{constant}

The answer is:

e2x−12+constant\frac{e^{2 x - 1}}{2}+ \mathrm{constant}

The answer (Indefinite) [src] 
  /                          
 |                    2*x - 1
 |  2*x - 1          e       
 | E        dx = C + --------
 |                      2    
/
∫e2x−1 dx=C+e2x−12\int e^{2 x - 1}\, dx = C + \frac{e^{2 x - 1}}{2}
Simplify
The graph
1.002.001.101.201.301.401.501.601.701.801.90040
Plot the graph f(x)
The answer [src] 
 3    
e    E
-- - -
2    2
−e2+e32- \frac{e}{2} + \frac{e^{3}}{2} 
=
= 
 3    
e    E
-- - -
2    2
−e2+e32- \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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