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
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 $$0 - \frac{-2}{1} = 2$$.