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e^(-x)*x^3
Solving integrals/ e^(-x)*x^3

Integral of e^(-x)*x^3 dx

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The solution

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∫0∞x3e−x dx\int\limits_{0}^{\infty} x^{3} e^{- x}\, dx 
Integral(x^3/E^x, (x, 0, oo))
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)=x3u{\left(x \right)} = x^{3} and let dv⁡(x)=e−x\operatorname{dv}{\left(x \right)} = e^{- x}.

    Then du⁡(x)=3x2\operatorname{du}{\left(x \right)} = 3 x^{2}.

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

    1. There are multiple ways to do this integral.

      Method #1

      1. Let u=−xu = - x.

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

        ∫eu du\int e^{u}\, du

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

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

        Now substitute uu back in:

        −e−x- e^{- x}

      Method #2

      1. Let u=e−xu = e^{- x}.

        Then let du=−e−xdxdu = - e^{- x} dx and substitute −du- du:

        ∫1 du\int 1\, du

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

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

        Now substitute uu back in:

        −e−x- e^{- x}

    Now evaluate the sub-integral.

  2. 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)=−3x2u{\left(x \right)} = - 3 x^{2} and let dv⁡(x)=e−x\operatorname{dv}{\left(x \right)} = e^{- x}.

    Then du⁡(x)=−6x\operatorname{du}{\left(x \right)} = - 6 x.

    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 e^{u}\, du

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

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

      Now substitute uu back in:

      −e−x- e^{- x}

    Now evaluate the sub-integral.

  3. 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)=6xu{\left(x \right)} = 6 x and let dv⁡(x)=e−x\operatorname{dv}{\left(x \right)} = e^{- x}.

    Then du⁡(x)=6\operatorname{du}{\left(x \right)} = 6.

    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 e^{u}\, du

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

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

      Now substitute uu back in:

      −e−x- e^{- x}

    Now evaluate the sub-integral.

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

    ∫(−6e−x) dx=−6∫e−x dx\int \left(- 6 e^{- x}\right)\, dx = - 6 \int e^{- x}\, dx

    1. Let u=−xu = - x.

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

      ∫eu du\int e^{u}\, du

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

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

      Now substitute uu back in:

      −e−x- e^{- x}

    So, the result is: 6e−x6 e^{- x}

  5. Now simplify:

    (−x3−3x2−6x−6)e−x\left(- x^{3} - 3 x^{2} - 6 x - 6\right) e^{- x}

  6. Add the constant of integration:

    (−x3−3x2−6x−6)e−x+constant\left(- x^{3} - 3 x^{2} - 6 x - 6\right) e^{- x}+ \mathrm{constant}

The answer is:

(−x3−3x2−6x−6)e−x+constant\left(- x^{3} - 3 x^{2} - 6 x - 6\right) e^{- x}+ \mathrm{constant}

The answer (Indefinite) [src] 
  /                                                   
 |                                                    
 |  -x  3             -x    3  -x        -x      2  -x
 | e  *x  dx = C - 6*e   - x *e   - 6*x*e   - 3*x *e  
 |                                                    
/
∫x3e−x dx=C−x3e−x−3x2e−x−6xe−x−6e−x\int x^{3} e^{- x}\, dx = C - x^{3} e^{- x} - 3 x^{2} e^{- x} - 6 x e^{- x} - 6 e^{- x}
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The answer [src] 
6
66 
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= 
6
66
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Integral of e^(-x)*x^3 dx

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

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