<latex>

  \frac{3}{4 \pi}   \sqrt{4 \cdot x^2   12}\\
  \lim_{n \to \infty}
  \sum_{k=1}^n \frac{1}{k^2} = \frac{\pi^2}{6}
  e^{i \pi} + 1 = 0 \\

</latex>

$$ \begin{align*} \int x^2 dx & = \frac{1}{3}x^3 \therefore\quad\int_0^1 x^2 dx &= \frac{1}{3} \end{align*} $$