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. 

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