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xe^(-x/2)
Solving integrals/ xe^(-x/2)

Integral of xe^(-x/2) dx

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The solution

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∫01xe(−1)x2 dx\int\limits_{0}^{1} x e^{\frac{\left(-1\right) x}{2}}\, dx 
Integral(x*E^(-x/2), (x, 0, 1))
Detail solution

  1. Use integration by parts:

    ∫u⁡dv⁡=u⁡v⁡−∫v⁡du⁡\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}

    Let u(x)=xu{\left(x \right)} = x and let dv⁡(x)=e−x2\operatorname{dv}{\left(x \right)} = e^{- \frac{x}{2}}.

    Then du⁡(x)=1\operatorname{du}{\left(x \right)} = 1.

    To find v(x)v{\left(x \right)}:

    1. There are multiple ways to do this integral.

      Method #1

      1. Let u=−x2u = - \frac{x}{2}.

        Then let du=−dx2du = - \frac{dx}{2} and substitute −2du- 2 du:

        ∫4eu du\int 4 e^{u}\, du

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

          ∫(−2eu) du=−2∫eu du\int \left(- 2 e^{u}\right)\, du = - 2 \int e^{u}\, du

          1. The integral of the exponential function is itself.

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

          So, the result is: −2eu- 2 e^{u}

        Now substitute uu back in:

        −2e−x2- 2 e^{- \frac{x}{2}}

      Method #2

      1. Let u=e−x2u = e^{- \frac{x}{2}}.

        Then let du=−e−x2dx2du = - \frac{e^{- \frac{x}{2}} dx}{2} and substitute −2du- 2 du:

        ∫4 du\int 4\, du

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

          ∫(−2) du=−2∫1 du\int \left(-2\right)\, du = - 2 \int 1\, du

          1. The integral of a constant is the constant times the variable of integration:

            ∫1 du=u\int 1\, du = u

          So, the result is: −2u- 2 u

        Now substitute uu back in:

        −2e−x2- 2 e^{- \frac{x}{2}}

    Now evaluate the sub-integral.

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

    ∫(−2e−x2) dx=−2∫e−x2 dx\int \left(- 2 e^{- \frac{x}{2}}\right)\, dx = - 2 \int e^{- \frac{x}{2}}\, dx

    1. Let u=−x2u = - \frac{x}{2}.

      Then let du=−dx2du = - \frac{dx}{2} and substitute −2du- 2 du:

      ∫4eu du\int 4 e^{u}\, du

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

        ∫(−2eu) du=−2∫eu du\int \left(- 2 e^{u}\right)\, du = - 2 \int e^{u}\, du

        1. The integral of the exponential function is itself.

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

        So, the result is: −2eu- 2 e^{u}

      Now substitute uu back in:

      −2e−x2- 2 e^{- \frac{x}{2}}

    So, the result is: 4e−x24 e^{- \frac{x}{2}}

  3. Now simplify:

    2(−x−2)e−x22 \left(- x - 2\right) e^{- \frac{x}{2}}

  4. Add the constant of integration:

    2(−x−2)e−x2+constant2 \left(- x - 2\right) e^{- \frac{x}{2}}+ \mathrm{constant}

The answer is:

2(−x−2)e−x2+constant2 \left(- x - 2\right) e^{- \frac{x}{2}}+ \mathrm{constant}

The answer (Indefinite) [src] 
  /                                 
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 |     2               2          2 
 | x*e    dx = C - 4*e    - 2*x*e   
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(−2 x−4) e−x2\left(-2\,x-4\right)\,e^ {- {{x}\over{2}} }
Simplify
The graph
0.001.000.100.200.300.400.500.600.700.800.905-5
Plot the graph f(x)
The answer [src] 
       -1/2
4 - 6*e
4−6e4-{{6}\over{\sqrt{e}}} 
=
= 
       -1/2
4 - 6*e
−6e12+4- \frac{6}{e^{\frac{1}{2}}} + 4
Simplify
Numerical answer [src] 
0.360816041724199
0.360816041724199
The graph
Integral of xe^(-x/2) dx

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

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