How to properly find the antiderivative of $\sqrt{x}$ at $x=0$?

Question

I used to write it this way: $$\int \sqrt{x}dx=\int x^{\frac{1}{2}}dx=\frac{x^{\frac{3}{2}}}{\frac{3}{2}}+C=\frac{2}{3}x^{\frac{3}{2}}+C \tag{1}$$ because of the rule: $$\int x^ndx=\frac{x^{n+1}}{n+1}+C, n \neq -1.\tag{2}$$ But now I remember a rule that says I can write $x^{\frac{m}{n}}=\sqrt[n]{x^m}$ only when $x>0$. So in fact (1) is just finding the antiderivative on $(0,+\infty)$.

What about the antiderivative at $x=0$? I know that it should still be $\frac{2}{3}x^{\frac{3}{2}}+C$, or more precisely, $\frac{2}{3}\sqrt{x^3}+C$ (the same $C$) by taking its derivative but I don't know how to do it the right way.

Answer 1

So you've done most of the job. If we take a look at the domain of $f(x)=\sqrt{x}$, it is $x\in[0,+\infty)$. So that means that in that set there exists an antiderivative and you found it correctly, it is $\frac{2}{3}\sqrt{x^3}+C, C\in\mathbb{R}$. If we plug in $x=0$ we find that an antiderivative at that point is $C$.

Answer 2

Here is my View Point which might resolve your Issue :

Let $f(x)=\sqrt{x}=x^{1/2}=\sqrt[2]{x^1}$
Let the Integral be $g(x)=(2/3)x^{3/2}+C=(2/3)\sqrt[2]{x^3}+C=(2/3){(\sqrt[2]{x^1})}^3+C$

We then have $g(x)=(2/3)[f(x)]^3+C$ , thus :
[1] Both $f(x)$ & $g(x)$ have the "SAME" Domain
[2] Both use the "Power of the Square root"
[3] Both "interfere" with that given rule @ $x=0$ !

If we DO allow $x=0$ for $f(x)$ & we take $f(0)=0$ , then the same applies to $g(x)$ to allow $g(0)=0+C$.

If we DO NOT allow $x=0$ for $f(x)$ & we take Domain to be Positive $x$ , then the same applies to $g(x)$ to take Domain to be Positive $x$.

Both Cases , we DO have Consistency & we DO NOT get Contradiction.

Answer 3

Your thinking is a bit befuddled, since $\sqrt{x}:=x^{\frac{1}{2}}$. They're just different symbols, one modern, one older. They both mean $x=y^2$ for some $y$ given $x$. The integral $$\int \sqrt{x}dx$$ means, for $x=y^2,$ $$\int ydx=\int ydy^2\\=2\int y^2 dy=\frac{2y^3}{3}+C.$$ You can see how being over- or hypertechnical is actually contrary to the truth itself, since it doesn't leave any room for insight. As an exercise to the reader, be a bit lazy when it comes to domains and ranges, and see if you can't invent some new math.