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How can I evaluate the integral $$\int_1^2x\,d[x^2]$$

I think this meaningless, because the area below separate points is not defined.

Note: $[.]$ is the floor function.

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    I edited my answer. Apologies for the misunderstanding. – 2017-02-12
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    @User8128 My answer is $\sqrt{2}+\sqrt{3}+\sqrt{4}$ but I don't know is true or false. – 2017-02-12
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    It's incorrect. The correct answer is $\sqrt 2$. – 2017-02-12
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    This joke is not it? What makes the floor function under a differential sign? – 2017-02-13
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    @MyGlasses I'm sorry, I completely botched this one. Your answer is correct – 2017-02-13
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    Usually, the $\texttt{floor}$ function is denoted by, for example, $\left\lfloor x^{2}\right\rfloor$. $\LaTeX$ code is $\verb@\left\lfloor x^{2}\right\rfloor@$ – 2017-02-19

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Compute this as Riemann-Stieltjes integral: $$\int\limits_a^b f(x)\text{d}g(x)=\int\limits_a^b f(x)g'(x)\text{d}x$$ under suitable regularity asssumptions (which are in this case satisfied).

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    They are not, really. – 2017-02-12
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    @MarianoSuárez-Álvarez, you are right, I have treated $[x^2]$ as the parenthesis. Nevertheless, Riemann-Stieltjes integral does exist, because $x$ is continuous and $[x^2]$ has a bounded variation. But, of course, more careful analysis is needed. It was given by User8128 above. Thanks for pointing my mistake. – 2017-02-12
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This is just the Riemann-Stieltjes integral. If you have monotone function $g$ which is continuously differentiable (and the range of integration is a compact interval $[a,b]$), then $$\int_a^b f(x) dg(x) = \int^b_a f(x)g'(x) dx.$$ Here we have $$\int^2_1 x d[x^2] = \int^2_1 2x^2 dx = \frac 2 3(2^3 - 1^3) = \frac {14} 3$$

For more info: https://en.wikipedia.org/wiki/Riemann%E2%80%93Stieltjes_integral

EDIT: I see now that we are using $g(x) = [x^2]$ as the integrator where $[\cdot]$ is the floor function. Generally, if $f$ is continuous on $[a,b]$ and $g$ is piecewise constant with jump discontinuities at $a_1, a_2, \ldots, a_\ell$ in the interval (but $g$ has one-sided limits $g(a_j^+)$ and $g(a_j^-)$ on each side of the jumps), then a similar argument will give $$\int^b_a f(x) dg(x) = \sum_{j=1}^\ell f(a_j) (g(a_j^+) - g(a_j^-)).$$ This gives $$\int^2_1 x d[x^2] = \sqrt 2 + \sqrt 3 + \sqrt 4.$$

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    Doesn't the bounds change? Because we go from $x^2=1$ to $x^2=2$ – 2017-02-12
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    No. There is no substitution being made here, so the bounds wont change – 2017-02-12
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    As I said, we aren't really making a substitution here. It's more like we are integrating on the interval $[1,2]$ with respect to a different measure than usual (here our measure is $dg(x)$ rather than the usual $dx$). – 2017-02-12
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    Are you using the greatest integer function in the differential term? – 2017-02-12
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    I assume we are putting $g(x) = x^2$ and doing the integral $$\int^2_1 x dg(x).$$ Is it actually $g(x) = [x^2]$ (as in the floor of $x^2$)? This would be an unusual question since $[x^2]$ is not smooth. You would need to revert back to the definition of the Riemann-Stieltjes integral using upper and lower Riemann sums and taking limits. – 2017-02-12
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    I think you have to take care of all $1\leq x\leq 2$ where $[x^2]$ jumps, see my answer. – 2017-02-13
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    Oh of course, I completely botched this one. Thanks for the correction – 2017-02-13
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One has $$ \int_1^2 x\,d[x^2]=\sqrt{2}+\sqrt{3}+\sqrt{4}. $$

Argument 1

We use

Theorem (9-11 in Apostol's Real Analysis) Let $\alpha$ be a step function defined on $[a,b]$ with jump $\alpha_k$ at $x_k$, where $a\leq x_1

If we take $f(x)=x$ and $\alpha(x)=[x^2]$ the conditions are fulfilled, and with $x_1=\sqrt{2}$, $x_2=\sqrt{3}$ and $x_3=\sqrt{4}$, we find that $$ \int_1^2x\,d[x^2]=\sqrt{2}(2-1)+\sqrt{3}(3-2)+\sqrt{4}(4-2)=\sqrt{2}+\sqrt{3}+\sqrt{4}. $$

Argument 2

We can use integration by parts. The general formula reads $$ \int_a^b f(x)\,d\alpha(x)+\int_a^b \alpha(x)\,df(x)=f(b)\alpha(b)-f(a)\alpha(a). $$ With $f(x)=x$ and $\alpha(x)=[x^2]$, we have $$ \begin{aligned} \int_1^2 x\,d[x^2]&=-\int_1^2 [x^2]\,dx+2[2^2]-1[1^2]\\ &=-\bigl\{(\sqrt{2}-1)+2(\sqrt{3}-\sqrt{2})+3(\sqrt{4}-\sqrt{3})\bigr\}+8-1\\ &=\sqrt{2}+\sqrt{3}+\sqrt{4}. \end{aligned} $$

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    why is the point $1$ not included in the point of discontinuity – 2017-02-13
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    Because the function $x\mapsto [x^2]$, defined on $[1,2]$, does not jump there. – 2017-02-13
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    How did you get $(\sqrt 2-1)+...$ ? I mean how did you compute $\int_1^2[x^2]dx$ don't see it :( (in argument 2) – 2017-06-10