Mister Exam

Other calculators

Integral of x*exp(3*x-5) dx

Limits of integration:

from to
v

The graph:

from to

Piecewise:

The solution

You have entered [src]
  1              
  /              
 |               
 |     3*x - 5   
 |  x*e        dx
 |               
/                
0                
01xe3x5dx\int\limits_{0}^{1} x e^{3 x - 5}\, dx
Integral(x*exp(3*x - 5), (x, 0, 1))
Detail solution
  1. There are multiple ways to do this integral.

    Method #1

    1. Rewrite the integrand:

      xe3x5=xe3xe5x e^{3 x - 5} = \frac{x e^{3 x}}{e^{5}}

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

      xe3xe5dx=xe3xdxe5\int \frac{x e^{3 x}}{e^{5}}\, dx = \frac{\int x e^{3 x}\, dx}{e^{5}}

      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)=e3x\operatorname{dv}{\left(x \right)} = e^{3 x}.

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

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

        1. Let u=3xu = 3 x.

          Then let du=3dxdu = 3 dx and substitute du3\frac{du}{3}:

          eu3du\int \frac{e^{u}}{3}\, 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: eu3\frac{e^{u}}{3}

          Now substitute uu back in:

          e3x3\frac{e^{3 x}}{3}

        Now evaluate the sub-integral.

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

        e3x3dx=e3xdx3\int \frac{e^{3 x}}{3}\, dx = \frac{\int e^{3 x}\, dx}{3}

        1. Let u=3xu = 3 x.

          Then let du=3dxdu = 3 dx and substitute du3\frac{du}{3}:

          eu3du\int \frac{e^{u}}{3}\, 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: eu3\frac{e^{u}}{3}

          Now substitute uu back in:

          e3x3\frac{e^{3 x}}{3}

        So, the result is: e3x9\frac{e^{3 x}}{9}

      So, the result is: xe3x3e3x9e5\frac{\frac{x e^{3 x}}{3} - \frac{e^{3 x}}{9}}{e^{5}}

    Method #2

    1. Rewrite the integrand:

      xe3x5=xe3xe5x e^{3 x - 5} = \frac{x e^{3 x}}{e^{5}}

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

      xe3xe5dx=xe3xdxe5\int \frac{x e^{3 x}}{e^{5}}\, dx = \frac{\int x e^{3 x}\, dx}{e^{5}}

      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)=e3x\operatorname{dv}{\left(x \right)} = e^{3 x}.

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

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

        1. Let u=3xu = 3 x.

          Then let du=3dxdu = 3 dx and substitute du3\frac{du}{3}:

          eu3du\int \frac{e^{u}}{3}\, 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: eu3\frac{e^{u}}{3}

          Now substitute uu back in:

          e3x3\frac{e^{3 x}}{3}

        Now evaluate the sub-integral.

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

        e3x3dx=e3xdx3\int \frac{e^{3 x}}{3}\, dx = \frac{\int e^{3 x}\, dx}{3}

        1. Let u=3xu = 3 x.

          Then let du=3dxdu = 3 dx and substitute du3\frac{du}{3}:

          eu3du\int \frac{e^{u}}{3}\, 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: eu3\frac{e^{u}}{3}

          Now substitute uu back in:

          e3x3\frac{e^{3 x}}{3}

        So, the result is: e3x9\frac{e^{3 x}}{9}

      So, the result is: xe3x3e3x9e5\frac{\frac{x e^{3 x}}{3} - \frac{e^{3 x}}{9}}{e^{5}}

  2. Now simplify:

    (3x1)e3x59\frac{\left(3 x - 1\right) e^{3 x - 5}}{9}

  3. Add the constant of integration:

    (3x1)e3x59+constant\frac{\left(3 x - 1\right) e^{3 x - 5}}{9}+ \mathrm{constant}


The answer is:

(3x1)e3x59+constant\frac{\left(3 x - 1\right) e^{3 x - 5}}{9}+ \mathrm{constant}

The answer (Indefinite) [src]
  /                                         
 |                     /   3*x      3*x\    
 |    3*x - 5          |  e      x*e   |  -5
 | x*e        dx = C + |- ---- + ------|*e  
 |                     \   9       3   /    
/                                           
xe3x5dx=C+xe3x3e3x9e5\int x e^{3 x - 5}\, dx = C + \frac{\frac{x e^{3 x}}{3} - \frac{e^{3 x}}{9}}{e^{5}}
The graph
0.001.000.100.200.300.400.500.600.700.800.900.2-0.1
The answer [src]
 -5      -2
e     2*e  
--- + -----
 9      9  
19e5+29e2\frac{1}{9 e^{5}} + \frac{2}{9 e^{2}}
=
=
 -5      -2
e     2*e  
--- + -----
 9      9  
19e5+29e2\frac{1}{9 e^{5}} + \frac{2}{9 e^{2}}
exp(-5)/9 + 2*exp(-2)/9
Numerical answer [src]
0.0308231681635901
0.0308231681635901

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