`int_0^infty11xe^(-x)dx` Evaluate the improper integral. Give your answer using the constant e, or round to six decimal places. Enter diverges if the integral diverges.

Expert Answers

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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))|`


Now we can calculate our improper integral. 


Now we use the what we have calculated before.



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


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

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