Solve $\int_0^2\sqrt{x} \, d\sqrt{x}\overset{?}{=}$ I've recently come across this question on Quora, and I don't know which answer is right, or why my own might be wrong.
What is the value of
\begin{align}I=\int_0^2\sqrt{x}\:d\sqrt{x}.\end{align}
I believe it is $=1$ since, first and foremost,
\begin{align}\int_0^2\sqrt{x}\:d\sqrt{x}&\overset{?}{=}\int_0^2 \frac{x^{1/2}x^{-1/2}}{2}\:dx\\&=\frac{1}{2}\int_0^2\:dx\\&=\frac{x}{2}\Bigg|_0^2\\&=1-0=\color{red}{1}.\end{align}
This is also since $d\alpha\left(x\right)=\alpha'\left(x\right)\:dx$, and further since graphing the set $\left\{\left(\sqrt{t},\sqrt{t}\right):t\in\mathbb{R}_0^{+}\right\}$ results in a line very similar to $y=x$ in the plane, therefore making the area from $\left[0,2\right]$ equal to $\frac{\sqrt{2}\sqrt{2}}{2}-\frac{\sqrt{0}\sqrt{0}}{2}=1$.
Where am I going wrong? Is there something I'm overlooking?
 A: Method 1
$$\int_0^2 \sqrt{x} d\sqrt{x}$$
$$=\int_{\color{brown}{\sqrt{x}=0}}^{\color{brown}{\sqrt{x}=2}} \sqrt{x} d\color{brown}{\sqrt{x}}$$
$$= \left[\frac{(\sqrt{x})^2}{2}\right]_0^2$$
$$=\left[\frac{(2)^2}{2}\right]-\left[\frac{(0)^2}{2}\right]=2$$

Method 2
$$\int_0^2 \sqrt{x} d\sqrt{x}$$
$$\int_{\color{brown}{\sqrt{x}=0}}^{\color{brown}{\sqrt{x}=2}} \sqrt{x} d\color{brown}{\sqrt{x}}$$
$$\sqrt{x}=t \implies \frac{d\sqrt{x}}{dt}=1 \implies d\sqrt{x}=dt$$
$$\bbox[2pt, border: 2pt green solid]{\sqrt{x}=t=2, \sqrt{x}=t=0}$$
$$=\int_{t=0}^{t=2} t dt=\frac{t^2}{2}\Big]_0^2=2$$
A: Using a few academic sources:
http://www.math.unl.edu/~jorr1/math826/RiemannStieltjes.pdf (theorem 6.7.8)
http://ocw.nctu.edu.tw/upload/classbfs1209122139184046.pdf (theorem E6)
These both state that:
$$\int_a^b f(x) d\alpha(x)=\int_a^b f(x) \alpha'(x) dx$$
no changing of limits of integration, agreeing with your answer of $1$. You DO NOT change limits with a Stieljes integral.
However lets go to the definition of Stieltjes integral
Lets compute
$$\begin{align*}
\lim\limits_{n\to\infty} \sum\limits_{i=0}^{n-1} \sqrt{\frac{2i}{n}}\left(\sqrt{\frac{2(i+1)}{n}}-\sqrt{\frac{2i}{n}}\right)&=\lim\limits_{n\to\infty} \frac{2}{n}\sum\limits_{i=0}^{n-1} \sqrt{i^2+i}-i \\
\end{align*}$$
I tried to find a clever way of summing this (I couldn't), but pick a large number, say $n=1,000,001$ and using Wolfram Alpha 
http://www.wolframalpha.com/input/?i=sum+sqrt%28k%5E2%2Bk%29-k+from+0+to+1000000
We get the approximation 
$$\approx\frac{2}{1,000,001}\times 499,999$$
This is needless to say, pretty close to $1$.
Also, another argument look at the first source, theorem 6.7.6 integration by parts. This says that
$$\int_a^b f dg +\int_a^b g df=g(b)f(b)-g(a)f(a)$$
Since $f=g$, we get that:
$$2\int_0^2 \sqrt{x} d\sqrt{x}=\sqrt{2}\sqrt{2}-\sqrt{0}\sqrt{0}=2$$
That is,
$$\int_0^2 \sqrt{x} d\sqrt{x}=1$$
Edit: I wanted to add why there is the confusion about changing limits. I want to reiterate. With a Stieltjes integral:
$$\int_a^b f(x) d\alpha(x)=\int_a^b f(x) \alpha'(x) dx\neq \int_{\alpha^{-1}(a)}^{\alpha^{-1}(b)} f(x) \alpha'(x) dx$$
And in both the left and middle terms, you are integrating with limits of $x$. Not $\alpha(x)$, but in fact $x$. This is why things like $\int_{\sqrt{x}=0}^{\sqrt{x}=2}\sqrt{x}d\sqrt{x}$ in the other answer are nonsense.
The issue here is that this looks like u substitution from freshman calculus. $u$ substitution has the following form:
$$\int_{a}^{b} f(\alpha(x))\alpha'(x)dx=\int_{\alpha(a)}^{\alpha(b)}f(u) du$$
However, with Stieltjes integration there is an analogous formula (check theorem E9 in the second reference).   
$$\int_a^b f(g(x))d\alpha(g(x))=\int_{g(a)}^{g(b)} f(u) d\alpha(u)$$
So in this case, let $f(x)=\operatorname{id}(x)=\alpha(x)$, $g(x)=\sqrt{x}$. Then:
$$\int_0^2 \operatorname{id}(\sqrt{x})d(\operatorname{id}(\sqrt{x}))=\int_0^{\sqrt{2}}\operatorname{id}(u)d(\operatorname{id}(u))=\int_0^{\sqrt{2}}u du=1$$
A: You need to change the limits of integration - if $\sqrt x$ goes from 0 to 2 then $x$ goes from 0 to 4.
