How do you determine if the improper integral converges or diverges #int_0^oo 1/ (x-2)^2 dx #?

Question

Scarlett Allison · Accepted Answer

This integral converges to #-1/2#.

#int_0^∞ dx/(x-2)^2# =#-1/(∞-2)+1/(0-2)# =#0-1/2# =#-1/2#

Christopher Agresta · Answer

The integral is not convergent.

The function: #f(x) = 1/(x-2)^2# is not continuous in the interval of integration so we must split the integral as: #int_0^oo 1/(x-2)^2dx = int_0^2 1/(x-2)^2dx+ int_2^A 1/(x-2)^2dx + int_A^oo 1/(x-2)^2dx# with #A > 2# Now: #int_0^2 1/(x-2)^2dx = lim_(t->2^-) int_0^t 1/(x-2)^2dx# #int_0^2 1/(x-2)^2dx = lim_(t->2^-) [-1/(x-2)]_0^t# #int_0^2 1/(x-2)^2dx = lim_(t->2^-) (1/(2-t)-1/2) = +oo# and: #int_2^A 1/(x-2)^2dx = lim_(t->2^+) int_t^A 1/(x-2)^2dx# #int_2^A 1/(x-2)^2dx = lim_(t->2^+) [-1/(x-2)]_t^A# #int_2^A 1/(x-2)^2dx = lim_(t->2^+) (-1/(A-2)+1/(t-2)) = +oo# Finally: #int_A^oo 1/(x-2)^2dx = lim_(t->oo) int_A^t 1/(x-2)^2dx# #int_A^oo 1/(x-2)^2dx = lim_(t->oo) [-1/(x-2)]_A^t# #int_A^oo 1/(x-2)^2dx = lim_(t->oo) (-1/(t-2) +1/(A-2)) = 1/(A-2)# Thus the integral is not convergent.

Emilia Aguilar · Answer

undefined To determine if the improper integral $ \int_0^\infty \frac{1}{(x-2)^2} \, dx $ converges or diverges:

1. Evaluate the integral from $ x = 0 $ to $ x = M $, where $ M $ is a large positive number.
2. Then, take the limit as $ M $ approaches infinity.

If the limit exists and is a finite number, the integral converges. If the limit does not exist or is infinite, the integral diverges.

Evelyn Agard · Answer

undefined To determine if the improper integral $\int_0^\infty \frac{1}{{(x-2)^2}} \, dx$ converges or diverges, we need to evaluate its convergence using the properties of improper integrals.

We can break down this improper integral into two parts:

1. $\int_0^2 \frac{1}{{(x-2)^2}} \, dx$
2. $\int_2^\infty \frac{1}{{(x-2)^2}} \, dx$

First, let's consider the integral from 0 to 2. At $x = 2$, the denominator becomes $0$, resulting in an infinite value. Thus, the integral from 0 to 2 diverges.

Next, let's consider the integral from 2 to $\infty$. We can evaluate this integral using the limit:

$$
\lim_{{a \to \infty}} \int_2^a \frac{1}{{(x-2)^2}} \, dx
$$

By integrating the function, we get:

$$
\lim_{{a \to \infty}} \left[-\frac{1}{{x-2}}\right]_2^a = \lim_{{a \to \infty}} \left(-\frac{1}{{a-2}} + \frac{1}{0}\right)
$$

This limit diverges to $\infty$.

Since one part of the improper integral diverges and the other part also diverges, the entire improper integral $\int_0^\infty \frac{1}{{(x-2)^2}} \, dx$ diverges.