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(1-7x)e^(5x)

Integral of (1-7x)e^(5x) dx

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01(17x)e5xdx\int\limits_{0}^{1} \left(1 - 7 x\right) e^{5 x}\, dx
Integral((1 - 7*x)*E^(5*x), (x, 0, 1))
Detail solution
  1. There are multiple ways to do this integral.

    Method #1

    1. Rewrite the integrand:

      (17x)e5x=7xe5x+e5x\left(1 - 7 x\right) e^{5 x} = - 7 x e^{5 x} + e^{5 x}

    2. Integrate term-by-term:

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

        (7xe5x)dx=7xe5xdx\int \left(- 7 x e^{5 x}\right)\, dx = - 7 \int x e^{5 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)=e5x\operatorname{dv}{\left(x \right)} = e^{5 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=5xu = 5 x.

              Then let du=5dxdu = 5 dx and substitute du5\frac{du}{5}:

              eu25du\int \frac{e^{u}}{25}\, du

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

                eu5du=eudu5\int \frac{e^{u}}{5}\, du = \frac{\int e^{u}\, du}{5}

                1. The integral of the exponential function is itself.

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

                So, the result is: eu5\frac{e^{u}}{5}

              Now substitute uu back in:

              e5x5\frac{e^{5 x}}{5}

            Method #2

            1. Let u=e5xu = e^{5 x}.

              Then let du=5e5xdxdu = 5 e^{5 x} dx and substitute du5\frac{du}{5}:

              125du\int \frac{1}{25}\, du

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

                15du=1du5\int \frac{1}{5}\, du = \frac{\int 1\, du}{5}

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

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

                So, the result is: u5\frac{u}{5}

              Now substitute uu back in:

              e5x5\frac{e^{5 x}}{5}

          Now evaluate the sub-integral.

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

          e5x5dx=e5xdx5\int \frac{e^{5 x}}{5}\, dx = \frac{\int e^{5 x}\, dx}{5}

          1. Let u=5xu = 5 x.

            Then let du=5dxdu = 5 dx and substitute du5\frac{du}{5}:

            eu25du\int \frac{e^{u}}{25}\, du

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

              eu5du=eudu5\int \frac{e^{u}}{5}\, du = \frac{\int e^{u}\, du}{5}

              1. The integral of the exponential function is itself.

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

              So, the result is: eu5\frac{e^{u}}{5}

            Now substitute uu back in:

            e5x5\frac{e^{5 x}}{5}

          So, the result is: e5x25\frac{e^{5 x}}{25}

        So, the result is: 7xe5x5+7e5x25- \frac{7 x e^{5 x}}{5} + \frac{7 e^{5 x}}{25}

      1. Let u=5xu = 5 x.

        Then let du=5dxdu = 5 dx and substitute du5\frac{du}{5}:

        eu25du\int \frac{e^{u}}{25}\, du

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

          eu5du=eudu5\int \frac{e^{u}}{5}\, du = \frac{\int e^{u}\, du}{5}

          1. The integral of the exponential function is itself.

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

          So, the result is: eu5\frac{e^{u}}{5}

        Now substitute uu back in:

        e5x5\frac{e^{5 x}}{5}

      The result is: 7xe5x5+12e5x25- \frac{7 x e^{5 x}}{5} + \frac{12 e^{5 x}}{25}

    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)=17xu{\left(x \right)} = 1 - 7 x and let dv(x)=e5x\operatorname{dv}{\left(x \right)} = e^{5 x}.

      Then du(x)=7\operatorname{du}{\left(x \right)} = -7.

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

      1. Let u=5xu = 5 x.

        Then let du=5dxdu = 5 dx and substitute du5\frac{du}{5}:

        eu25du\int \frac{e^{u}}{25}\, du

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

          eu5du=eudu5\int \frac{e^{u}}{5}\, du = \frac{\int e^{u}\, du}{5}

          1. The integral of the exponential function is itself.

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

          So, the result is: eu5\frac{e^{u}}{5}

        Now substitute uu back in:

        e5x5\frac{e^{5 x}}{5}

      Now evaluate the sub-integral.

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

      (7e5x5)dx=7e5xdx5\int \left(- \frac{7 e^{5 x}}{5}\right)\, dx = - \frac{7 \int e^{5 x}\, dx}{5}

      1. Let u=5xu = 5 x.

        Then let du=5dxdu = 5 dx and substitute du5\frac{du}{5}:

        eu25du\int \frac{e^{u}}{25}\, du

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

          eu5du=eudu5\int \frac{e^{u}}{5}\, du = \frac{\int e^{u}\, du}{5}

          1. The integral of the exponential function is itself.

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

          So, the result is: eu5\frac{e^{u}}{5}

        Now substitute uu back in:

        e5x5\frac{e^{5 x}}{5}

      So, the result is: 7e5x25- \frac{7 e^{5 x}}{25}

    Method #3

    1. Rewrite the integrand:

      (17x)e5x=7xe5x+e5x\left(1 - 7 x\right) e^{5 x} = - 7 x e^{5 x} + e^{5 x}

    2. Integrate term-by-term:

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

        (7xe5x)dx=7xe5xdx\int \left(- 7 x e^{5 x}\right)\, dx = - 7 \int x e^{5 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)=e5x\operatorname{dv}{\left(x \right)} = e^{5 x}.

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

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

          1. Let u=5xu = 5 x.

            Then let du=5dxdu = 5 dx and substitute du5\frac{du}{5}:

            eu25du\int \frac{e^{u}}{25}\, du

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

              eu5du=eudu5\int \frac{e^{u}}{5}\, du = \frac{\int e^{u}\, du}{5}

              1. The integral of the exponential function is itself.

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

              So, the result is: eu5\frac{e^{u}}{5}

            Now substitute uu back in:

            e5x5\frac{e^{5 x}}{5}

          Now evaluate the sub-integral.

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

          e5x5dx=e5xdx5\int \frac{e^{5 x}}{5}\, dx = \frac{\int e^{5 x}\, dx}{5}

          1. Let u=5xu = 5 x.

            Then let du=5dxdu = 5 dx and substitute du5\frac{du}{5}:

            eu25du\int \frac{e^{u}}{25}\, du

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

              eu5du=eudu5\int \frac{e^{u}}{5}\, du = \frac{\int e^{u}\, du}{5}

              1. The integral of the exponential function is itself.

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

              So, the result is: eu5\frac{e^{u}}{5}

            Now substitute uu back in:

            e5x5\frac{e^{5 x}}{5}

          So, the result is: e5x25\frac{e^{5 x}}{25}

        So, the result is: 7xe5x5+7e5x25- \frac{7 x e^{5 x}}{5} + \frac{7 e^{5 x}}{25}

      1. Let u=5xu = 5 x.

        Then let du=5dxdu = 5 dx and substitute du5\frac{du}{5}:

        eu25du\int \frac{e^{u}}{25}\, du

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

          eu5du=eudu5\int \frac{e^{u}}{5}\, du = \frac{\int e^{u}\, du}{5}

          1. The integral of the exponential function is itself.

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

          So, the result is: eu5\frac{e^{u}}{5}

        Now substitute uu back in:

        e5x5\frac{e^{5 x}}{5}

      The result is: 7xe5x5+12e5x25- \frac{7 x e^{5 x}}{5} + \frac{12 e^{5 x}}{25}

  2. Now simplify:

    (1235x)e5x25\frac{\left(12 - 35 x\right) e^{5 x}}{25}

  3. Add the constant of integration:

    (1235x)e5x25+constant\frac{\left(12 - 35 x\right) e^{5 x}}{25}+ \mathrm{constant}


The answer is:

(1235x)e5x25+constant\frac{\left(12 - 35 x\right) e^{5 x}}{25}+ \mathrm{constant}

The answer (Indefinite) [src]
  /                                          
 |                             5*x        5*x
 |            5*x          12*e      7*x*e   
 | (1 - 7*x)*e    dx = C + ------- - --------
 |                            25        5    
/                                            
(17x)e5xdx=C7xe5x5+12e5x25\int \left(1 - 7 x\right) e^{5 x}\, dx = C - \frac{7 x e^{5 x}}{5} + \frac{12 e^{5 x}}{25}
The graph
0.001.000.100.200.300.400.500.600.700.800.90-10001000
The answer [src]
           5
  12   23*e 
- -- - -----
  25     25 
23e5251225- \frac{23 e^{5}}{25} - \frac{12}{25}
=
=
           5
  12   23*e 
- -- - -----
  25     25 
23e5251225- \frac{23 e^{5}}{25} - \frac{12}{25}
Numerical answer [src]
-137.02010637437
-137.02010637437
The graph
Integral of (1-7x)e^(5x) dx

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