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x*exp(-x)

Integral of x*exp(-x) dx

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The solution

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01xexdx\int\limits_{0}^{1} x e^{- x}\, dx
Integral(x*exp(-x), (x, 0, 1))
Detail solution
  1. There are multiple ways to do this integral.

    Method #1

    1. Let u=xu = - x.

      Then let du=dxdu = - dx and substitute dudu:

      ueudu\int u e^{u}\, du

      1. Use integration by parts:

        udv=uvvdu\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}

        Let u(u)=uu{\left(u \right)} = u and let dv(u)=eu\operatorname{dv}{\left(u \right)} = e^{u}.

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

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

        1. The integral of the exponential function is itself.

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

        Now evaluate the sub-integral.

      2. The integral of the exponential function is itself.

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

      Now substitute uu back in:

      xexex- x e^{- x} - e^{- x}

    Method #2

    1. Use integration by parts:

      udv=uvvdu\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)=ex\operatorname{dv}{\left(x \right)} = e^{- x}.

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

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

      1. Let u=xu = - x.

        Then let du=dxdu = - dx and substitute du- du:

        (eu)du\int \left(- e^{u}\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.

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

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

        Now substitute uu back in:

        ex- e^{- x}

      Now evaluate the sub-integral.

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

      (ex)dx=exdx\int \left(- e^{- x}\right)\, dx = - \int e^{- x}\, dx

      1. Let u=xu = - x.

        Then let du=dxdu = - dx and substitute du- du:

        (eu)du\int \left(- e^{u}\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.

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

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

        Now substitute uu back in:

        ex- e^{- x}

      So, the result is: exe^{- x}

  2. Now simplify:

    (x+1)ex- \left(x + 1\right) e^{- x}

  3. Add the constant of integration:

    (x+1)ex+constant- \left(x + 1\right) e^{- x}+ \mathrm{constant}


The answer is:

(x+1)ex+constant- \left(x + 1\right) e^{- x}+ \mathrm{constant}

The answer (Indefinite) [src]
  /                          
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 | x*e   dx = C - e   - x*e  
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xexdx=Cxexex\int x e^{- x}\, dx = C - x e^{- x} - e^{- x}
The graph
0.001.000.100.200.300.400.500.600.700.800.901-2
The answer [src]
       -1
1 - 2*e  
12e1 - \frac{2}{e}
=
=
       -1
1 - 2*e  
12e1 - \frac{2}{e}
1 - 2*exp(-1)
Numerical answer [src]
0.264241117657115
0.264241117657115
The graph
Integral of x*exp(-x) dx

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