To find the integral $$I= \int_1^a \lfloor x^2\rfloor f'(x) dx$$ where $a > 1$ is a positive real number and $ \lfloor x\rfloor \ $ is the greatest integer function.
I tried splitting the limits as $1 $ to $ \sqrt 2$ and so on. But I couldn't get a general solution for this problem.
Integration by parts using the Riemann-Stieltjes integral gives $$ \begin{align} \int_1^a\left\lfloor x^2\right\rfloor f'(x)\,\mathrm{d}x &=f(a)\left\lfloor a^2\right\rfloor-\int_1^a f(x)\,\mathrm{d}\!\left\lfloor x^2\right\rfloor\\ &=\bbox[5px,border:2px solid #C0A000]{f(a)\left\lfloor a^2\right\rfloor-\sum_{k=1}^{\left\lfloor a^2\right\rfloor}f\!\left(\sqrt{k}\right)}\tag1 \end{align} $$ Note that this is a continuous function regardless of the floor function and the summation if $f$ is smooth.
We can verify $(1)$ by looking at its derivative. For $n\lt a^2\lt n+1$, the derivative of both sides is $$ n\,f'(a)\tag2 $$ Since $f$ is continuous at $\sqrt{n}$, $$ \forall\epsilon\gt0,\exists\delta\in(0,1):\sqrt{n-\delta}\le x,y\le\sqrt{n+\delta}\implies|f(x)-f(y)|\le\frac\epsilon{n+1}\tag3 $$ Thus, $$ \begin{align} &\left|\left[f\!\left(\sqrt{n+\delta}\right)\left\lfloor n+\delta\right\rfloor-\sum_{k=1}^{\left\lfloor n+\delta\right\rfloor}f\!\left(\sqrt{k}\right)\right]-\left[f\!\left(\sqrt{n-\delta}\right)\left\lfloor n-\delta\right\rfloor-\sum_{k=1}^{\left\lfloor n-\delta\right\rfloor}f\!\left(\sqrt{k}\right)\right]\right|\\[6pt] &=\left|n\left[f\!\left(\sqrt{n+\delta}\right)-f\!\left(\sqrt{n-\delta}\right)\right]+f\!\left(\sqrt{n-\delta}\right)-f\!\left(\sqrt{n}\right)\right|\\[9pt] &\le\epsilon\tag4 \end{align} $$ That is, the right side of $(1)$ is continuous at $\sqrt{n}$.
Since the derivatives match when $a^2\not\in\mathbb{Z}$, both are continuous when $a^2\in\mathbb{Z}$, and both are $0$ when $a=1$, the left and right sides of $(1)$ are equal.