A simple integral with one question

Question

Question is: For $x$ equals $4$ and $9$, why is $t$ not $\pm2$ and $\pm3$ but just $2$ and $3$ ?

Answer 1

You can use the negative square root if you want; it all works out the same way in the end. If we denote the positive square root of $x$ by $\sqrt{x}$, we can substitute $t = -\sqrt{x}$ instead. We still have $t^2 = x$, so $2 t \, dt = dx $ as before. So the integral becomes \begin{align*} \int_{-2}^{-3} \frac{2 t \, dt}{-t -1} &= 2 \int_{-3}^{-2} \frac{t}{t + 1} dt \\ &= 2 \left[ \int_{-3}^{-2} dt - \int_{-3}^{-2} \frac{dt}{t+1} \right] \\ &= 2 \Big[ t - \ln |t+1| \Big]_{-3}^{-2} \\ &= 2 \Big[ -2 - (-3) - \left( \ln(1) - \ln(2) \right)\Big] \\ &= 2 + 2 \ln 2. \end{align*} The steps look slightly different, but it works out to be exactly the same result. This is an important lesson in general: there is frequently more than one possible substitution that allows you to solve an integral.

Answer 2

By convention, $\sqrt{a}$ represents the non-negative square root, or the principal square root, of $x$. Hence, the only case in which there are two opposite solutions is when you have $\pm\sqrt{a}$. (Note the extra $\pm$ sign.) Hence is important to note that $x = \sqrt{a}$ (one non-negative solution). should not be confused with $x^2 = a \iff \vert x\vert = \sqrt{a} \iff x = \pm\sqrt{a}$ (two solutions). So, for instance, $\sqrt{4} = \color{blue}{+}2$ and $\sqrt{9} = \color{blue}{+}3$.

Answer 3

Because under standard and generally accepted notation,

$$ \forall x\ge0;\sqrt x =|x^{\frac12}|$$ That is to say, the square root is used in this case to mean the positive square root.