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\title[]{MA140-Engineering Calculus}
\author[ ]{\large{} 
\and %\\ \small{Supervisor: Dr. John Burns}
\\ 
 \includegraphics[scale=0.4]{Figures/lec14.jpg}}
%\date[June 2-4, 2016]{June 2-4, 2016}
%\institute[Uppsala University]{ }


\begin{document}



\section{MA140}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{frame}
An
{\bf improper integral}
is a definite integral 
$$
\int_a^b f(x) dx
$$
where either
\begin{enumerate}
\item
one, or both limits is $\infty$ or $-\infty$, or
\item
the function $f(x)$ is $\infty$ or $-\infty$ in some region of the domain
\end{enumerate}
\begin{description}
\item[Case 1: One of the limits is $\pm\infty$]
If $a = -\infty$, calculate $\int_a^b f(x) dx$ as normal (your answer will be a 
function of $a$), then take $\lim_{a\to -\infty}$.
If $b = \infty$, calculate $\int_a^b f(x) dx$ as normal (your answer will be a 
function of $b$), then take $\lim_{b\to \infty}$.
\item[Case 2: The function is $\pm\infty$ at some point $c$]
Break up the integral $\int_a^b f(x) dx = \int_a^d f(x) dx + \int_d^b f(x) dx$.
Then take the limit of the first integral as $ d\to c$ from the left, and of the
second integral as $ d \to c$ from the right.
\end{description}

\end{frame}
%\begin{frame}
%\begin{definition}
%The volume of a solid of known integrable cross-sectional area $A(x)$ from $x=a$ to $x=b$ is the integral of $A$ from $a$ to $b$, $$V=\int_{a}^{b} A(x) dx$$
%\end{definition}
%\begin{theorem}
%{\textcolor{red}{Calculating the volume of solids:}}
%\begin{itemize}
%\item[(1)] Sketch the solid and a typical cross-section.
%\item[(2)] Find a formula for $A(x)$, the area of a typical cross-section.
%\item[(3)] Find the limits of integration.
%\item[(4)] Integrate $A(x)$ using the Fundamental Theorem.  
%\end{itemize}
%\end{theorem}
%\end{frame}
%\begin{frame}
%\begin{example}
%A pyramid $3$ m high has a square base that is $3$ m on a side. Find
%the volume of the pyramid.
%\end{example}
%We draw the pyramid: 
%\begin{center}
%			\includegraphics[scale=0.28]{Figures/lec603.pdf}
%		\end{center}
%\end{frame}
%\begin{frame}
%The cross-section at $x$ is a square $x$ meters on a side, so its area is: $$A(x)=x^{2}$$ 
%The squares lie on the planes from $x=0$ to $x=b$. So the volume is $$V=\int_{0}^{3} A(x) dx=\int_{0}^{3} x^{2} dx=\frac{x^{3}}{3} \Big ]_{0}^{3}=9$$ 
%\end{frame}





\end{document}