finding bounds for $\int_0^X\lfloor x^2\rfloor \, dx$ I am trying to find bounds for:
$$I=\int_0^X\lfloor x^2\rfloor \, dx$$ and through integrating two different ways I ended up with the sums:
$$I=(X^2+1)^{3/2}-\sum_{i=1}^{X^2+1}\sqrt{i}$$
$$I=X^3-\sum_{i=1}^{X^2}\sqrt{i}$$
I can't verify if either of these are correct firstly, and also is it possible for me to use these to create bounds? Thanks
 A: $\newcommand{\d}{\text{d}}$Assuming $X$ is an integer, we have :
$\begin{align}
I(X)&=\int_0^X\lfloor x^2\rfloor\d x\\
&=\sum_{n=0}^{X^2-1}\int_{\sqrt{n}}^{\sqrt{n+1}}\lfloor x^2\rfloor\d x\\
&=\sum_{n=0}^{X^2-1}(\sqrt{n+1}-\sqrt{n})n\\
&=\sum_{n=0}^{X^2-1}n\sqrt{n+1}-\sum_{n=0}^{X^2-1}n\sqrt{n}\\
&=\sum_{n=1}^{X^2}(n-1)\sqrt{n}-\sum_{n=1}^{X^2-1}n\sqrt{n}\\
&=X(X^2-1)-\sum_{n=1}^{X^2-1}\sqrt{n}.
\end{align}$
Now, according to Is there any way to approximate a sum of square roots
, we have : $$\frac{2}{3}N^{3/2}\leqslant\sum_{n=1}^N\sqrt{n}\leqslant\frac{2}{3}\left[(N+1)^{3/2}-1\right],$$
which can be used with $N=X^2-1$ to provide : $$X(X^2-1)-\frac{2}{3}(X^3-1)\leqslant I(X)\leqslant X(X^2-1)-\frac{2}{3}(X^2-1)^{3/2}.$$
Now, in the case where $X$ is not necessarily an integer, by assuming that $X\geqslant0$, we have $\lfloor X\rfloor\leqslant X<\lfloor X\rfloor+1$, from which we obtain : $$I(\lfloor X\rfloor)\leqslant I(X)<I(\lfloor X\rfloor+1).$$
In the end, the previous bounds can be used to provide : $$\lfloor X\rfloor(\lfloor X\rfloor^2-1)-\frac{2}{3}(\lfloor X\rfloor^3-1)\leqslant I(X)\leqslant (\lfloor X\rfloor+1)\left[(\lfloor X\rfloor+1)^2-1\right]-\frac{2}{3}\left[(\lfloor X\rfloor+1)^2-1\right]^{3/2}.$$
A: If the two are equal then
$(X^2+1)^{3/2}-\sum_{i=1}^{X^2+1}\sqrt{i}
=X^3-\sum_{i=1}^{X^2}\sqrt{i}
$
so
$(X^2+1)^{3/2}-X^3
=\sqrt{X^2+1}
$
which is true
only at $X = 0$.
For example,
for $X=1$ this is
$1
=\sqrt{2}^3-\sqrt{2}
=\sqrt{2}
$.
