Question

I need help in limits and integration. I'll post the question on the comment below

Answer 1

\[\int\limits_{1}^{4}\frac{ 1 }{ (x-2)^2 }dx = \lim_{a \rightarrow 2-}\int\limits_{1}^{a}\frac{ 1 }{ (x-2)^2 }dx + \lim_{a \rightarrow 2+}\int\limits_{b}^{4}\frac{ 1 }{ (x-2)^2 }dx\]
This is what my TA gave me

Answer 2

and she said that the answer does not exist. Why ?

Answer 3

sorry, it's suppose to be "b to 2+" not "a to 2+"

Answer 4

For starters, we note that \(\dfrac{1}{(x-2)^2}\) is undefined when \(x=2\); hence this is an improper integral when evaluated over the interval \((1,4)\). So, we first see that
\[\large \int_1^4\frac{1}{(x-2)^2}\,dx = \lim_{a\to 2^{-}}\int_1^a \frac{1}{(x-2)^2}\,dx + \lim_{b\to 2^+}\int_b^4\frac{1}{(x-2)^2}\,dx.\]If you proceed with the integration, we see that we get \[\large \begin{aligned} &\phantom{=} \lim_{a\to 2^-}\left.\left(-\frac{1}{x-2}\right)\right|_1^a + \lim_{b\to 2^+}\left.\left(-\frac{1}{x-2}\right)\right|_b^4 \\ &= \lim_{a\to 2^-}\left(-\frac{1}{a-2}\right) -1 -\frac{1}{2} + \lim_{b\to 2^+}\frac{1}{b-2}\end{aligned}\]However, recall that \(\large\displaystyle\lim_{a\to 2^-}\left(-\frac{1}{a-2}\right) = \infty\) and \(\large\displaystyle\lim_{b\to 2^+}\left(\frac{1}{b-2}\right) = \infty\). Hence \[\large \int_1^4\frac{1}{(x-2)^2}\,dx = \infty - 1 - \frac{1}{2} + \infty = \infty.\]Therefore, the integral diverges (to infinity).

Does this clarify things? :-)

Answer 5

Thank you so much for writing all that down, Chris! (I hope u don't mind me calling u Chris)
It does clarify a lot of things. One question though, how do u differentiate infinity and no answer, because what I know is that if your limit from the left and from the right is different, the answer doesn't exist.

Answer 6

It's cool - I'm fine with you calling me Chris. :-)

Hmmm...I'll have to think about that; it's been a while since I thought about that. XD

Typically, whenever an integral went to infinity or didn't give me something that was concrete and finite, I just said that the integral diverged instead of saying no answer.

Answer 7

ok, thanks anyways Chris, much appreciated