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____ [38]
4 years ago
11

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

Mathematics
1 answer:
dem82 [27]4 years ago
3 0

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

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