Daily math problems for June 8, 2026 — Calculus, Integrals

${\int }_0^3{x}^{2}dx={\displaystyle \frac{27}{3}}=9$.

${\int }_0^{\pi \mathrm{/}2}{\sin }^{2}xdx={\int }_0^{\pi \mathrm{/}2}{\displaystyle \frac{1-\cos 2x}{2}}dx={\displaystyle \frac{1}{2}}{[x-\frac{\sin 2x}{2}]}_0^{\pi \mathrm{/}2}={\displaystyle \frac{\pi }{4}}$.

Vertical: $x=\pm 1$ (two). Horizontal: $\lim f(x)=1$ (one). Total: $3$. (No slant — the rational function has equal-degree numerator and denominator.)