Skip to main content\(\newcommand{\N}{\mathbb N} \newcommand{\Z}{\mathbb Z} \newcommand{\Q}{\mathbb Q} \newcommand{\R}{\mathbb R}
\newcommand{\lt}{<}
\newcommand{\gt}{>}
\newcommand{\amp}{&}
\definecolor{fillinmathshade}{gray}{0.9}
\newcommand{\fillinmath}[1]{\mathchoice{\colorbox{fillinmathshade}{$\displaystyle \phantom{\,#1\,}$}}{\colorbox{fillinmathshade}{$\textstyle \phantom{\,#1\,}$}}{\colorbox{fillinmathshade}{$\scriptstyle \phantom{\,#1\,}$}}{\colorbox{fillinmathshade}{$\scriptscriptstyle\phantom{\,#1\,}$}}}
\)
Section 1.1 Section Title
So I think the integral \(\int_0^5 f(x) \, dx \)
\begin{equation*}
\frac{d}{dx} \int_1^x \frac{1}{t}\, dt = \frac{1}{x}
\end{equation*}
\begin{equation*}
\int_a^b f(x)\, dx = F(b) - F(a)
\end{equation*}
Checkpoint 1.1.1.
