We will use integration by parts

`int udv=uv-int vdu`

`int_0^infty xe^(-x/3)dx=|[u=x,dv=e^(-x/3)dx],[du=dx,v=-3e^(-x/3)]|=`

`-3xe^(-x/3)|_0^infty+3int_0^infty e^(-x/3)dx=`

`(-3xe^(-x/3)-9e^(-x/3))|_0^infty=`

`lim_(x to infty)[-3e^(-x/3)(x-3)]+3cdot0cdot e^0+9e^0=`

To calculate the above limit we will use L'Hospital's rule:

`lim_(x to c)(f(x))/(g(x))=lim_(x to c)(f'(x))/(g'(x))`

`lim_(x to infty)[-3e^(-x/3)(x-3)]=-3lim_(x to infty) (x-3)/e^(x/3)=`

Apply L'Hospital's rule.

`-3lim_(x to infty)1/e^(x/3)=0`

Let us now return to the integral.

`0+0+9=9`

**As we can see the integral converges and it has value of 9.**

The image below shows graph of the function and area under it representing the value of the integral. Looking at the image we can see that the graph approaches `x`-axis (function converges to zero) "very fast". This suggests that the integral should converge.