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4*x*e^(2*x)

Integral of 4*x*e^(2*x) dx

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The solution

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014xe2xdx\int\limits_{0}^{1} 4 x e^{2 x}\, dx
Integral(4*x*E^(2*x), (x, 0, 1))
Detail solution
  1. The integral of a constant times a function is the constant times the integral of the function:

    4xe2xdx=4xe2xdx\int 4 x e^{2 x}\, dx = 4 \int x e^{2 x}\, dx

    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)=e2x\operatorname{dv}{\left(x \right)} = e^{2 x}.

      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=2xu = 2 x.

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

          eu4du\int \frac{e^{u}}{4}\, du

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

            eu2du=eudu2\int \frac{e^{u}}{2}\, du = \frac{\int e^{u}\, du}{2}

            1. The integral of the exponential function is itself.

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

            So, the result is: eu2\frac{e^{u}}{2}

          Now substitute uu back in:

          e2x2\frac{e^{2 x}}{2}

        Method #2

        1. Let u=e2xu = e^{2 x}.

          Then let du=2e2xdxdu = 2 e^{2 x} dx and substitute du2\frac{du}{2}:

          14du\int \frac{1}{4}\, du

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

            12du=1du2\int \frac{1}{2}\, du = \frac{\int 1\, du}{2}

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

              1du=u\int 1\, du = u

            So, the result is: u2\frac{u}{2}

          Now substitute uu back in:

          e2x2\frac{e^{2 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:

      e2x2dx=e2xdx2\int \frac{e^{2 x}}{2}\, dx = \frac{\int e^{2 x}\, dx}{2}

      1. Let u=2xu = 2 x.

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

        eu4du\int \frac{e^{u}}{4}\, du

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

          eu2du=eudu2\int \frac{e^{u}}{2}\, du = \frac{\int e^{u}\, du}{2}

          1. The integral of the exponential function is itself.

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

          So, the result is: eu2\frac{e^{u}}{2}

        Now substitute uu back in:

        e2x2\frac{e^{2 x}}{2}

      So, the result is: e2x4\frac{e^{2 x}}{4}

    So, the result is: 2xe2xe2x2 x e^{2 x} - e^{2 x}

  2. Now simplify:

    (2x1)e2x\left(2 x - 1\right) e^{2 x}

  3. Add the constant of integration:

    (2x1)e2x+constant\left(2 x - 1\right) e^{2 x}+ \mathrm{constant}


The answer is:

(2x1)e2x+constant\left(2 x - 1\right) e^{2 x}+ \mathrm{constant}

The answer (Indefinite) [src]
  /                                 
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 |      2*x           2*x        2*x
 | 4*x*e    dx = C - e    + 2*x*e   
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(2x1)e2x\left(2\,x-1\right)\,e^{2\,x}
The graph
0.001.000.100.200.300.400.500.600.700.800.90-5050
The answer [src]
     2
1 + e 
4(e24+14)4\,\left({{e^2}\over{4}}+{{1}\over{4}}\right)
=
=
     2
1 + e 
1+e21 + e^{2}
Numerical answer [src]
8.38905609893065
8.38905609893065
The graph
Integral of 4*x*e^(2*x) dx

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