# Unsure why this improper integral converges at 2.

I have the improper integral $$\int_3^\infty \frac{1}{(x-2)^{3/2}}\,dx$$

after integrating I get

$$\left.\frac{-2}{\sqrt{x-2}}\right|^b_3$$ where b is positive infinity when i evaluate though I get $$0- \frac{-2}{\pm1}$$ The book,and an integral calculator website says the answer is 2. I am not sure how to know which root to take as I know area can be negative if its under the x-axis.

Edit: fixed typo

• Hint: It makes no sense to have $\sqrt1=-1$. If we did, the idea of, say, area under a graph, would make no sense! – Simply Beautiful Art Feb 26 '17 at 0:21
• For me it diverges at $2$, by equivalence. – Bernard Feb 26 '17 at 0:22
• @SimplyBeautifulArt So, does it not make sense because technically the case $-2$ it would be the area above the part of the function that is under the x-axis? And with an integral we're looking for the area under the function? – Jude Feb 26 '17 at 0:26
• Do you mean $$\int_3^{\infty}\frac{1}{(x-2)^{3/2}}dx?$$ – Juniven Feb 26 '17 at 0:26
• @ΘΣΦGenSan oops yes – Jude Feb 26 '17 at 0:26

There is a bit of confusion, between a lot of students, about square roots. For example, $\sqrt{9}$, is $\pm3$? The answer is no. Note that $\sqrt{4}$, for example, is $2$, not $\pm 2$. The confusion comes from solving $x^2 = 4$, where $x$ could be either $2$ or $-2$. The $\pm$ sign comes from the fact that $\sqrt{x^2} = |x|$. In your case, $\sqrt{1} = 1$, not $\pm 1$. Therefore, the answer you want is $$0 - \frac{-2}{1} = 2$$.

• Oh wow what a simple mistake this makes sense lol – Jude Feb 26 '17 at 0:59