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Say we have $y = x^{0.5}$ and we want to find the area under the curve between the values of $0$ and $1$.

$\int_0^1 y dx = \left(\frac{2}{3} x^{1.5}\right)_0^1$

Now $x^{1.5} = x\cdot x^{0.5}$

If we plot the curve it is obvious that the area is positive, but what's to say, in the equation $\frac{2}{3} x^{1.5}$ that the answer is positive, a possible root of $x = -1$ so the area $= \pm \frac{2}{3}$

Is there an explanation or do we simply use the positive root because we know it has to be the positive root? Is there any case when you instead take the negative root?

Ian Miller
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Edward Garemo
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2 Answers2

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Because most people in the world do not know that $\sqrt{x^2}=|x|$.

In another hand, if we write $x^{0.5}$ then by definition $x>0$.

Michael Rozenberg
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  • I understand that x must be greater than 0, but root x doesn't, root x as I said in my example could be negative one xx^0.5 = 1+-1 = +-1. Unless there's something I'm missing you didn't actually answer my question, why do we assume x^0.5 is positive when we integrate? When you're first taught quadratics the idea that you can't simply ignore the negative root is thoroughly reinforced since it is of course of great importance. – Edward Garemo May 16 '17 at 07:15
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By definition the $\sqrt{}$ symbol (and hence $x^{0.5}$) means the positive root. If you are asked to graph $y=\sqrt{x}$ then your graph is the positive root and has domain and range of $[0,\infty)$. This is different to the graph of $y^2=x$ which is a parabola rotated by $90^\circ$. In the case of $y^2=x$ you get two solution when doing the inverse operation of square and write $y=\pm\sqrt{x}$.

Ian Miller
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  • All good and well but, unless there's something I'm missing you didn't actually answer my question; why do we assume x^0.5 is positive when we integrate? When you're first taught quadratics the idea that you can't simply ignore the negative root is thoroughly reinforced since it is of course of great importance. Why doesn't the same apply here? – Edward Garemo May 16 '17 at 07:18
  • When you first learn square root you only learnt the positive value which is the definition. It was later on that you tackled problems like $x^2=4$ which has two solutions. The two solutions are $\pm\sqrt{4}$ where $\sqrt{4}$ is equal to positive $2$ by definition. (It is always positive, not just when you integrate.) – Ian Miller May 16 '17 at 07:19
  • The questions "What is the square root of 4?" and "What is the solution to $x^2=4$?" are not the same. The first has answer of only $2$. The second has two answers, $-2$ and $2$. – Ian Miller May 16 '17 at 07:24