Improper Integrals converge or not?

[flaren5]

Question:Determine whether the following improper integral converges or not. If it does converge, find the value. Use proper limit notation.

. . . . .\(\displaystyle {\large \displaystyle\int_1^2\, \dfrac{dx}{(x\, -\, 2)^2}}\)

I have come to determine that the above improper integral diverges to infinity. However, the way I have figured it out is quite messy and very hard to follow, plus I'm not sure if I'm even correct. If anyone has a "proper" way, or an eaiser way to follow, to determine the improper integral, it would be greatly appreciated for some insight. Thank you.

 

[daon2]

The standard antiderivative is \(\displaystyle F(x) = \dfrac{-1}{x-2}\).

What you need to calculate is \(\displaystyle \displaystyle \lim_{b\to 2} \,\left(F(b)-F(1)\right)\)

 

[flaren5]

The standard antiderivative is \(\displaystyle F(x) = \dfrac{-1}{x-2}\).

What you need to calculate is \(\displaystyle \displaystyle \lim_{b\to 2} \,\left(F(b)-F(1)\right)\)

I did as you suggested, and I got a different result as my value. I initially thought that it was divergent because I ended up with infinity, but now I have a value of -1, does this still make it divergent because it is not within the limit from 1 to 2? Does how I calculated look like I'm on the right track?

 

[DrPhil]

Question:Determine whether the following improper integral converges or not. If it does converge, find the value. Use proper limit notation.

. . . . .\(\displaystyle {\large \displaystyle\int_1^2\, \dfrac{dx}{(x\, -\, 2)^2}}\)

I have come to determine that the above improper integral diverges to infinity. However, the way I have figured it out is quite messy and very hard to follow, plus I'm not sure if I'm even correct. If anyone has a "proper" way, or an eaiser way to follow, to determine the improper integral, it would be greatly appreciated for some insight. Thank you.

After you make the u-substitution, the limits of integration MUST be converted to limits of u.

Let u = x - 2,.....du = dx,.....x=1-->u=-1,.....x=2->u=0

\(\displaystyle \displaystyle \int_{-1}^0 \dfrac{du}{u^2} = \left. -\dfrac{1}{u}\right|_{-1}^0\)

What you did is equivalent after re-substituting back to x. But it is the UPPER limit of the integral that makes it improper. Your error was to say the limit of 1/u is 0.

To evaluate the upper limit, take the limit as u-->0- (or as x-->2-).