Question

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

Answer 1

Answer:

Converges at -2

Step-by-step explanation:

We solve this by using integral by parts.

$\int\limits^a_b {(2x-1)/x^2 - x} \, dx$

let:

$u=2x-1$             $dv=(1/x^2)dx$

$du=2xdx$         $v=-(1/x)$

We use the following expression for integral by parts:

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

$\int\limits^a_b {(2x-1)/x^2} \, dx = -2-1/x+\int\limits {2} \, dx$

We can take the limit with x tending to -1

$\lim_{x \to \(-1} -2-1/x+x = -2+1-1=-2$

The integral converges at -2