Let's first calculate indefinite integral by using integration by parts

`int11xe^(-x)dx=11int x e^(-x)dx=|(u=x, dv=e^(-x)dx),(du=dx, v=-e^(-x))|`

`=11(-xe^(-x)-int-e^(-x)dx)=11(-xe^(-x)+e^(-x))`

Now we can calculate our improper integral.

`int_0^infty11xe^(-x)dx=lim_(R->infty)int_0^R11xe^(-x)dx=`

Now we use the what we have calculated before.

`lim_(R->infty)11(-xe^(-x)-e^(-x))|_0^R=`

`11(lim_(R->infty)(-Re^(-R)-e^(-R))+0+1)=11(0+0+1)=11`

In the line above `lim_(R->infty)(-Re^(-R)-e^(-R))=0` because

`lim_(R->infty)-e^(-R)=0` which is easily seen. And by L'Hospital rule

`lim_(R->infty)-Re^(-R)=lim_(R->infty)-e^(-R)=0`.

**So your solution is** `int_0^infty11xe^(-x)dx=11.`