Question

Determine whether the improper integral converges or diverges, and find the value of each that converges.

Answer 1

Answer:

Converges at -1

Step-by-step explanation:

The integral converges if the limit exists, if the limit does not exist or if the limit is infinity it diverges.

We will make use of integral by parts to determine:

let:

$u=x$             $dv=e^(2x)\cdot{dx}$

$du=dx$         $v=2\cdot{e^(2x)}$

$\int\limits^a_b {u} \, dv = uv -\int\limits^a_b {v} \, du$

$\int\limits^a_b {x\cdot{e^2^x} \, dx =2xe^2^x- \int\limits^a_b {2e^2^x} \, dx$

$\int\limits^a_b {xe^2^x} \, dx = 2xe^2^x-2e^2^x-C$

We can therefore determine that if x tends to 0 the limit is -1

$\lim_{x \to \0} 2xe^2^x-2e^2^x=0-1=-1$