Dropping signs while doing integration In high school textbooks, we see that while taking square roots, the negative sign is not taken in calculus problems. For example, when asked to find the area bounded by $y^2=-x$ between $-2$ to $-1,$ we do it like this: $$\int_{-2}^{-1} \sqrt{-x} \,\mathrm dx.$$ Why don't we take the $y=-\sqrt{-x}\,?$ In all calculus problems, why is it okay to only the take $\sqrt{a^2}=a$ instead of $\sqrt{a^2}=|a|\,?$
 A: 
when asked to find the area bounded by $y^2=-x$ between $-2$ to $-1,$ we do it like this: $$\int_{-2}^{-1} \sqrt{-x} \,\mathrm dx.\tag1$$ Why don't we take the $y=-\sqrt{-x}\,?$

Perhaps you are showing just part of the working, since the required area in this example is obtained by multiplying expression $(1)$ by two, due to the figure's symmetry.

On the other hand, the signed area bounded by $y^2=-x$ between $-2$ to $-1$ does involve the $y=-\sqrt{-x}$ portion, and equals $$\int_{-2}^{-1} \sqrt{-x} \,\mathrm dx+\int_{-2}^{-1} -\sqrt{-x} \,\mathrm dx,$$ which equals $0.$

In all calculus problems, why is it okay to only the take $\sqrt{a^2}=a$ instead of $\sqrt{a^2}=|a|\,?$

This premise is false: the workings above are based on geometry rather than some generic calculus rule.
A: $y = -\sqrt{x}$ is complex on (-2,-1). you would get $y = -\sqrt{-1}$. What you want is $y=\sqrt{1}$. What you can do is let $z = -x$ then $dz = -dx$ and you get:
$$-\int_2^1 \sqrt{z} dz = \int_1^2 \sqrt{x} dx.$$
For your second question $\sqrt{(-1)^2} = \sqrt{1}=1 = |-1| \neq -1$.
