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e^(5-2x)
Solving integrals/ e^(5-2x)

Integral of e^(5-2x) dx

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The solution

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 |   5 - 2*x   
 |  E        dx
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0
∫01e5−2x dx\int\limits_{0}^{1} e^{5 - 2 x}\, dx 
Integral(E^(5 - 2*x), (x, 0, 1))
Detail solution

  1. There are multiple ways to do this integral.

    Method #1

    1. Let u=5−2xu = 5 - 2 x.

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

      ∫(−eu2) du\int \left(- \frac{e^{u}}{2}\right)\, 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:

      −e5−2x2- \frac{e^{5 - 2 x}}{2}

    Method #2

    1. Rewrite the integrand:

      e5−2x=e5e−2xe^{5 - 2 x} = e^{5} e^{- 2 x}

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

      ∫e5e−2x dx=e5∫e−2x dx\int e^{5} e^{- 2 x}\, dx = e^{5} \int e^{- 2 x}\, dx

      1. Let u=−2xu = - 2 x.

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

        ∫(−eu2) du\int \left(- \frac{e^{u}}{2}\right)\, 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:

        −e−2x2- \frac{e^{- 2 x}}{2}

      So, the result is: −e5e−2x2- \frac{e^{5} e^{- 2 x}}{2}

    Method #3

    1. Rewrite the integrand:

      e5−2x=e5e−2xe^{5 - 2 x} = e^{5} e^{- 2 x}

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

      ∫e5e−2x dx=e5∫e−2x dx\int e^{5} e^{- 2 x}\, dx = e^{5} \int e^{- 2 x}\, dx

      1. Let u=−2xu = - 2 x.

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

        ∫(−eu2) du\int \left(- \frac{e^{u}}{2}\right)\, 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:

        −e−2x2- \frac{e^{- 2 x}}{2}

      So, the result is: −e5e−2x2- \frac{e^{5} e^{- 2 x}}{2}

  2. Add the constant of integration:

    −e5−2x2+constant- \frac{e^{5 - 2 x}}{2}+ \mathrm{constant}

The answer is:

−e5−2x2+constant- \frac{e^{5 - 2 x}}{2}+ \mathrm{constant}

The answer (Indefinite) [src] 
  /                          
 |                    5 - 2*x
 |  5 - 2*x          e       
 | E        dx = C - --------
 |                      2    
/
∫e5−2x dx=C−e5−2x2\int e^{5 - 2 x}\, dx = C - \frac{e^{5 - 2 x}}{2}
Simplify
The graph
0.001.000.100.200.300.400.500.600.700.800.90-250250
Plot the graph f(x)
The answer [src] 
 5    3
e    e 
-- - --
2    2
−e32+e52- \frac{e^{3}}{2} + \frac{e^{5}}{2} 
=
= 
 5    3
e    e 
-- - --
2    2
−e32+e52- \frac{e^{3}}{2} + \frac{e^{5}}{2} 
exp(5)/2 - exp(3)/2
Numerical answer [src] 
64.1638110896945
64.1638110896945
The graph
Integral of e^(5-2x) dx

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

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