[Algebra]
Problem: Let $T=20 \sqrt{7}$. Let $w>0$ be a real number such that $T$ is the area of the region above the $x$-axis, below the graph of $y=\lceil x\rceil^{2}$, and between the lines $x=0$ and $x=w$. Compute $\lceil 2 w\rceil$.

Solution: Write $w=k+\alpha$, where $k$ is an integer, and $0 \leq \alpha<1$. Then

$$
T=1^{2}+2^{2}+\cdots+k^{2}+(k+1)^{2} \cdot \alpha .
$$

Computing $\lceil 2 w\rceil$ requires computing $w$ to the nearest half-integer. First obtain the integer $k$. As $\sqrt{7}>2$, with $T=20 \sqrt{7}$, one obtains $T>40$. As $1^{2}+2^{2}+3^{2}+4^{2}=30$, it follows that $k \geq 4$. To obtain an upper bound for $k$, note that $700<729$, so $10 \sqrt{7}<27$, and $T=20 \sqrt{7}<54$. As $1^{2}+2^{2}+3^{2}+4^{2}+5^{2}=55$, it follows that $4<w<5$, and hence $k=4$.

It now suffices to determine whether or not $\alpha>0.5$. To this end, one must determine whether $T>1^{2}+2^{2}+3^{2}+4^{2}+5^{2} / 2=42.5$. Indeed, note that $2.5^{2}=6.25<7$, so $T>(20)(2.5)=50$. It follows that $\alpha>0.5$, so $4.5<w<5$. Thus $9<2 w<10$, and $\lceil 2 w\rceil=\mathbf{1 0}$.
Once it has been determined that $4<w<5$, the formula for $T$ yields $1+4+9+16+25 \cdot \alpha=20 \sqrt{7}$, hence $\alpha=\frac{4 \sqrt{7}-6}{5}$. Thus $2 \alpha=\frac{8 \sqrt{7}-12}{5}=\frac{\sqrt{448}-12}{5}>\frac{21-12}{5}=1.8$. Because $2 w=2 k+2 \alpha$, it follows that $\lceil 2 w\rceil=\lceil 8+2 \alpha\rceil=\mathbf{1 0}$, because $1.8<2 \alpha<2$.

Final answer: 10