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Solving integrals/ x^(3)exp(-2x)

Integral of x^(3)exp(-2x) dx

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The solution

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∫0∞x3e−2x dx\int\limits_{0}^{\infty} x^{3} e^{- 2 x}\, dx 
Integral(x^3*exp(-2*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−2x\operatorname{dv}{\left(x \right)} = e^{- 2 x}.

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

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

    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}

    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)=−3x22u{\left(x \right)} = - \frac{3 x^{2}}{2} and let dv⁡(x)=e−2x\operatorname{dv}{\left(x \right)} = e^{- 2 x}.

    Then du⁡(x)=−3x\operatorname{du}{\left(x \right)} = - 3 x.

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

    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}

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

    Then du⁡(x)=32\operatorname{du}{\left(x \right)} = \frac{3}{2}.

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

    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}

    Now evaluate the sub-integral.

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

    ∫(−3e−2x4) dx=−3∫e−2x dx4\int \left(- \frac{3 e^{- 2 x}}{4}\right)\, dx = - \frac{3 \int e^{- 2 x}\, dx}{4}

    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: 3e−2x8\frac{3 e^{- 2 x}}{8}

  5. Now simplify:

    −(4x3+6x2+6x+3)e−2x8- \frac{\left(4 x^{3} + 6 x^{2} + 6 x + 3\right) e^{- 2 x}}{8}

  6. Add the constant of integration:

    −(4x3+6x2+6x+3)e−2x8+constant- \frac{\left(4 x^{3} + 6 x^{2} + 6 x + 3\right) e^{- 2 x}}{8}+ \mathrm{constant}

The answer is:

−(4x3+6x2+6x+3)e−2x8+constant- \frac{\left(4 x^{3} + 6 x^{2} + 6 x + 3\right) e^{- 2 x}}{8}+ \mathrm{constant}

The answer (Indefinite) [src] 
  /                                                             
 |                      -2*x        -2*x      2  -2*x    3  -2*x
 |  3  -2*x          3*e       3*x*e       3*x *e       x *e    
 | x *e     dx = C - ------- - --------- - ---------- - --------
 |                      8          4           4           2    
/
∫x3e−2x dx=C−x3e−2x2−3x2e−2x4−3xe−2x4−3e−2x8\int x^{3} e^{- 2 x}\, dx = C - \frac{x^{3} e^{- 2 x}}{2} - \frac{3 x^{2} e^{- 2 x}}{4} - \frac{3 x e^{- 2 x}}{4} - \frac{3 e^{- 2 x}}{8}
Simplify
The graph
0.001.000.100.200.300.400.500.600.700.800.900.5-0.5
Plot the graph f(x)
The answer [src] 
3/8
38\frac{3}{8} 
=
= 
3/8
38\frac{3}{8} 
3/8

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

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