# 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?

• what makes you believe you are wrong? Sep 27, 2015 at 17:21
• Why do you think your answer is wrong? Sep 27, 2015 at 17:21
• On quora, in green, it is written that $(\sqrt{2})^2=2^2$ from where the extra $2$ comes. This is clearly wrong. You are correct. Sep 27, 2015 at 17:24
• Having an admittedly very very brief look at the link you provided I'm under the impression this is not a place you should consult for advice regarding mathematical questions.... Sep 27, 2015 at 17:24
• Or you can note that $$I=\int_0^{\sqrt2}u\,du=\frac12\left[u^2\right]_0^{\sqrt2}=1\;,$$ making the substitution $u=\sqrt{x}$. Sep 27, 2015 at 17:24

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

$$\int_0^2 \sqrt{x} d\sqrt{x}$$

$$\int_{\color{brown}{\sqrt{x}=0}}^{\color{brown}{\sqrt{x}=2}} \sqrt{x} d\color{brown}{\sqrt{x}}$$

$$\sqrt{x}=t \implies \frac{d\sqrt{x}}{dt}=1 \implies d\sqrt{x}=dt$$

$$\bbox[2pt, border: 2pt green solid]{\sqrt{x}=t=2, \sqrt{x}=t=0}$$

$$=\int_{t=0}^{t=2} t dt=\frac{t^2}{2}\Big]_0^2=2$$

• math.unl.edu/~jorr1/math826/RiemannStieltjes.pdf See theorem 6.7.8
– user223391
Sep 28, 2015 at 4:12
• What you have written does not make sense. How does \begin{align}\left[\frac{\left(\sqrt{x}\right)^2}{2}\right]_0^2=\left[\frac{2^2}{2}\right]-\left[\frac{0^2}{2}\right]=2?\end{align} This should instead be \begin{align}\left[\frac{\left(\sqrt{x}\right)^2}{2}\right]_0^2=\left[\frac{x}{2}\right]_0^2=\frac{2}{2}-\frac{0}{2}=1.\end{align} Sep 28, 2015 at 4:38
• When computing Stieltjes integrals you DO NOT change limits of integration. This is blatantly false.
– user223391
Sep 28, 2015 at 4:40
• @avid19 when have I changed? Sep 28, 2015 at 4:54
• @TheArtist See the theorem in the paper I linked you. It's not $\sqrt{x}$ going from $0$ to $2$, it is $x$ that is going from $0$ to $2$.
– user223391
Sep 28, 2015 at 4:56

http://www.math.unl.edu/~jorr1/math826/RiemannStieltjes.pdf (theorem 6.7.8)

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:

$$\int_0^2 \operatorname{id}(\sqrt{x})d(\operatorname{id}(\sqrt{x}))=\int_0^{\sqrt{2}}\operatorname{id}(u)d(\operatorname{id}(u))=\int_0^{\sqrt{2}}u du=1$$

• Thank you for all your hard work, this all makes complete sense to me. And for some reason the link wouldn't work, but I think it should be wolframalpha.com/input/… Sep 28, 2015 at 5:04
• (+1) for writing up the essentials of stieltjes integration that even a physicist understands it! Sep 28, 2015 at 15:14
• Thanks for the edit! This, again, makes a lot of sense. Sep 28, 2015 at 16:00
• @jm324354 Of course. On Quora (and I guess here as well) there is a lot of misinformation. Always check academic sources first, and try to understand things yourself.
– user223391
Sep 28, 2015 at 16:06

You need to change the limits of integration - if $\sqrt x$ goes from 0 to 2 then $x$ goes from 0 to 4.

• math.unl.edu/~jorr1/math826/RiemannStieltjes.pdf Consult theorem 6.7.8
– user223391
Sep 28, 2015 at 1:50
• You DO NOT change limits of integration with computing these. Can you find an academic source, or better yet give me a proof using the definition of Stieltjes integral?
– user223391
Sep 28, 2015 at 4:54