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\begin{document}
	
	\mbox{}
	\title{Math1B: Quiz 1}
	\maketitle
	
	Name:\underline{\phantom{This is a blank space to place the student's name}} \phantom{spacespace space} Section: 209

	\vspace{10pt}
%	%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%	%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%	%\section{Introduction} \label{sec:intro}
%	\noindent {\footnotesize Note: Most of the problems were taken from the textbook} \cite{jS15}.
		
	\begin{problem}
		Solve the integral
		\[
			\int \cos^5 t \sin^6 t \, dt.
		\]
	\end{problem}
	
	\vspace{8cm}
		
	\begin{problem}
		Solve the integral
		\[
			\int x^2 5^x \, dx.
		\]
	\end{problem}
	
	\newpage
	
	\vspace{20pt}
	
	\begin{problem}
		Evaluate the integral
		\[
			\bigintsss_0^2 \frac{dt}{\sqrt{4 + t^2}}.
		\]
	\end{problem}
	
	\vspace{10cm}
	
	\begin{problem}
		Solve the integral
		\[
			\bigintsss \frac{x^3 - 3x^2 + 9x + 14}{x^2 - 4x + 13} \, dx.
		\]
	\end{problem}
	
%	\noindent {\it Solution:} We integrate by parts. Taking $u = z$ and $dv = e^{2z}$, we find that
%	\[
%		du = dz \ \text{ and } \ v = \int e^{2z} \, dz = \frac12 e^{2z}.
%	\]
%	Therefore
%	\[
%		\int_0^1 ze^{2z} \, dz = \frac{ze^{2z}}{2}\bigg{|}_0^1 - \frac12 \int_0^1 e^{2z} \, dz = \frac{e^2}{2} - \frac{e^{2z}}4\bigg{|}_0^1 = \frac{e^2}4 + \frac14.
%	\]
%	\hfill \qed
	
%	\begin{thebibliography}{20}
%		
%		\bibitem{jS15} J.~Stewart: \emph{Single Variable Calculus} 8th Edition, Cengage Learning, Boston 2015.
%		
%	\end{thebibliography}
	
\end{document}