Integral $\int\limits_{4}^9\frac{1}{\sqrt{x}}dx$

Question

Integral for $\frac{1}{\sqrt{x}}$ gives $2\sqrt{x} + c$. However, when applying the limits, $\sqrt{x}$ gives two answers ($-2$ and $2$ for $4$; $-3$ and $3$ for $9$). Which values should we use? Most solutions give an answer of $2$. Shouldn't there be four answers?

Answer 1

You must remember (for the sake of your sanity) that it is the case that when dealing with the square root function, that we have, by definition,

$$\sqrt x\ge0$$

Thus,

$$\int_4^9\frac1{\sqrt x}\ dx=2(\sqrt9-\sqrt4)=2(3-2)=2$$