\hfill \thepage} %} \input{tcilatex} \begin{document} \section{\protect\vspace{1pt}Ma 115 Lecture 11/1/99} \begin{center} \vspace{1pt} \end{center} \section{Antiderivatives} \vspace{0in} \subsection{Antiderivatives---Indefinite Integrals} \vspace{0in}\vspace{1pt} \begin{quotation} Given a function $f(x)$, if $F(x)$ is any function for which $F^{\prime }(x)=f(x)$, then $F(x)$ is called an \emph{antiderivative\/} of $f(x)$. \end{quotation} \vspace{1pt} Notice in the definition above that $F(x)$ is called \emph{an\/} antiderivative of $f(x)$. The reason is that there are many antiderivatives of the same function. For example, if $f(x)=x^{2}$, then $F(x)=\frac{1}{3}% x^{3}$ is an antiderivative (note: $F^{\prime }(x)=x^{2}=f(x)$), but so are $% G(x)=\frac{1}{3}x^{2}+1$ and $H(x)=\frac{1}{3}x^{3}+7$. \ There are more. Indeed, $F(x)=\frac{1}{3}x^{3}+C$, where $C$ is any constant, is an antiderivative. \vspace{1pt} In general, \begin{quotation} \qquad \qquad \qquad if $F(x)$ \emph{is an antiderivative of\/} $f(x)$, \emph{so also is\/} $F(x)+C$. \end{quotation} \vspace{1pt} YOU\ CAN\ ALWAYS check if you have found a correct antiderivative by differentiating. \begin{center} \FRAME{dtbpF}{4.1943in}{0.0623in}{0pt}{}{}{maroon.wmf}{\special{language "Scientific Word";type "GRAPHIC";display "PICT";valid_file "F";width 4.1943in;height 0.0623in;depth 0pt;original-width 290.5625pt;original-height 1pt;cropleft "0";croptop "0.6633";cropright "1";cropbottom "0.6621";filename 'graphics/maroon.wmf';file-properties "XNPEU";}} \end{center} \vspace{0in} \subsection{Antiderivatives---The Basic Theorem} \vspace{0in}\vspace{1pt} At this point two questions arise: \begin{enumerate} \item How does one find these antiderivatives? \item Given an antiderivative $F(x)$ of $f(x)$, are there others than those produced by adding constants? \end{enumerate} \noindent The first question will be answered partly in this chapter and more fully in the next chapters. As to the second question, we can say, happily, that, once a particular antiderivative has been found, all others can be determined by adding on constants. Formally, we state this as a theorem. \begin{quotation} \textbf{Theorem}~~~If $F_{1}(x)$ and $F_{2}(x)$ are antiderivatives of $f(x)$% , then the difference $F_{1}(x)-F_{2}(x)$ is a constant function. That is, there is a constant $C$ such that, for all $x$ in the domain of $f(x)$, \[ F_{1}(x)-F_{2}(x)=C \] \end{quotation} \vspace{1pt} \begin{center} \FRAME{dtbpF}{4.1943in}{0.0623in}{0pt}{}{}{maroon.wmf}{\special{language "Scientific Word";type "GRAPHIC";display "PICT";valid_file "F";width 4.1943in;height 0.0623in;depth 0pt;original-width 290.5625pt;original-height 1pt;cropleft "0";croptop "0.6633";cropright "1";cropbottom "0.6621";filename 'graphics/maroon.wmf';file-properties "XNPEU";}} \end{center} \subsection{\protect\vspace{1pt}The Integral} The general antiderivative warrants a special name and symbol. \begin{quotation} If $F(x)$ is an antiderivative of $f(x)$, then the function $F(x)+C,C$ any constant, is called the \emph{indefinite integral\/} of $f(x)$, and it is denoted by \[ \int f(x)dx=F(x)+C \] Here $f(x)$ is called the \emph{integrand\/}. The process of finding integrals is called \emph{integration\/}, and $C$ is called the \emph{% constant of integration\/}. \end{quotation} \vspace{1pt} The expression $\int f(x)dx$ is read ``the indefinite integral of $f(x)$ with respect to $x$,'' and, though it may look rather contrived, it has special significance, as we will see later. \vspace{1pt} From the previous section, we have \begin{example} Using our new notation, we have $\int x\,dx=\dfrac{1}{2}x^{2}+C$ \end{example} \begin{example} Using our new notation, we have $\int x^{4}dx=\allowbreak \frac{1}{5}x^{5}+C$ \end{example} \vspace{1pt} \FRAME{dtbpF}{4.1943in}{0.0623in}{0pt}{}{}{maroon.wmf}{\special{language "Scientific Word";type "GRAPHIC";display "PICT";valid_file "F";width 4.1943in;height 0.0623in;depth 0pt;original-width 290.5625pt;original-height 1pt;cropleft "0";croptop "0.6633";cropright "1";cropbottom "0.6621";filename 'graphics/maroon.wmf';file-properties "XNPEU";}} \vspace{1pt}\vspace{0in} \subsection{Antiderivatives---Examples} \vspace{0in}\vspace{1pt} \begin{example} Find $\int x^{2/3}\,dx$. That is, find the indefinite integral of $% f(x)=x^{2/3}$. \end{example} \noindent \emph{Solution}~~The function $F(x)=\frac{3}{5}x^{5/3}$ is an antiderivative of $x^{2/3}$. (Check it by differentiating.) So, \[ \int x^{2/3}dx=\frac{3}{5}x^{5/3}+C \] \begin{example} Find $\int x^{-3}\ dx$. \end{example} \noindent \emph{Solution}~~The function $F(x)=-\frac{1}{2}x^{-2}$ is an antiderivative of $x^{-3}$, as can be verified by differentiation. So, \[ \int x^{-3}\ dx=-\frac{1}{2}x^{-2}+C \] \vspace{1pt}\FRAME{dtbpF}{4.1943in}{0.0623in}{0pt}{}{}{maroon.wmf}{\special% {language "Scientific Word";type "GRAPHIC";display "PICT";valid_file "F";width 4.1943in;height 0.0623in;depth 0pt;original-width 290.5625pt;original-height 1pt;cropleft "0";croptop "0.6633";cropright "1";cropbottom "0.6621";filename 'graphics/maroon.wmf';file-properties "XNPEU";}} \begin{center} \qquad \vspace{0in} \end{center} \subsection{Antiderivatives---Examples} \vspace{0in}\vspace{1pt} \begin{example} Find $\int \sin x\,dx$. That is, find the indefinite integral of $f(x)=\sin x $. \end{example} \noindent \emph{Solution}~~The function $F(x)=-\cos x$ \ is an antiderivative of $\sin x$. (Check it by differentiating.) So, \[ \int \sin x\,dx=-\cos x+C \] \begin{example} Find $\int \cos \ dx$. \end{example} \noindent \emph{Solution}~~The function $F(x)=\sin x$ is an antiderivative of $\cos x$, as can be verified by differentiation. So, \[ \int \cos x\ dx=\allowbreak \sin x+C \] $\ $ \begin{example} Find $\int \sin 3x\,dx$ \end{example} \emph{Solution}. \ This time there is an extra multiplying factor in the antidertivative. We have \[ \int \sin 3x\,dx=\allowbreak -\frac{1}{3}\cos 3x+C \] \FRAME{dtbpF}{4.1943in}{0.0623in}{0pt}{}{}{maroon.wmf}{\special{language "Scientific Word";type "GRAPHIC";display "PICT";valid_file "F";width 4.1943in;height 0.0623in;depth 0pt;original-width 290.5625pt;original-height 1pt;cropleft "0";croptop "0.6633";cropright "1";cropbottom "0.6621";filename 'graphics/maroon.wmf';file-properties "XNPEU";}} \begin{remark} For each example, we added the extra constant $C.$ \ Don't forget this as it will be very important later on. \end{remark} \FRAME{dtbpF}{4.1943in}{0.0623in}{0pt}{}{}{maroon.wmf}{\special{language "Scientific Word";type "GRAPHIC";display "PICT";valid_file "F";width 4.1943in;height 0.0623in;depth 0pt;original-width 290.5625pt;original-height 1pt;cropleft "0";croptop "0.6633";cropright "1";cropbottom "0.6621";filename 'graphics/maroon.wmf';file-properties "XNPEU";}}\vspace{0in} \subsection{Antiderivatives---Basic Rules} \vspace{0in}\vspace{1pt} {\Large The basic rules }for antiderivatives are simply the ordinary rules for derivatives stated backwards. \subsubsection{The first set of rules:} \begin{enumerate} \item Power rule: \[ \int x^{r}\ dx=\left\{ \begin{array}{lll} \dfrac{x^{r+1}}{r+1}+C & \text{if} & r\neq -1 \\ \ln \,\left| x\right| +C & \text{if} & \;r=1% \end{array}% \right. \] \item[2.] Constant Multiplier Rule:\ If $k$ is any constant, \[ \int kf(x)dx=k\int f(x)dx \] \item[3.] Sum Rule \[ \int [f(x)\pm g(x)]dx=\int f(x)dx\pm \int g(x)dx \] \end{enumerate} \begin{center} \FRAME{dtbpF}{4.1943in}{0.0623in}{0pt}{}{}{maroon.wmf}{\special{language "Scientific Word";type "GRAPHIC";display "PICT";valid_file "F";width 4.1943in;height 0.0623in;depth 0pt;original-width 290.5625pt;original-height 1pt;cropleft "0";croptop "0.6633";cropright "1";cropbottom "0.6621";filename 'graphics/maroon.wmf';file-properties "XNPEU";}} \end{center} \vspace{1pt} From the previous section, we have \begin{example} Applying the power rule, we have $\int x^{10}\,\,dx=\dfrac{x^{10+1}}{10+1}+C=% \dfrac{x^{11}}{11}\allowbreak +C\medskip $ \end{example} \begin{example} Applying the power rule, we have $\int x^{-2}\,dx=\dfrac{x^{-2+1}}{-2+1}% \allowbreak +C=-x^{-1}\allowbreak +C.$ \ \ Also, applying the power rule, we have $\int 4x^{-1}\,dx=\allowbreak 4\ln \,\left| x\right| \allowbreak +C\medskip $ \end{example} \begin{example} Applying the power rule, and constant multiplier rule we have \[ \int 5x^{3}\,dx=5\int x^{3}\,dx=5\dfrac{x^{3+1}}{3+1}+C=\dfrac{5x^{4}}{4}+C \] \medskip \end{example} \begin{example} Applying all three rules we have \begin{eqnarray*} \int \left( 4x+\dfrac{7}{x^{3}}\right) \,dx &=&\int 4x\,dx+\int \dfrac{7}{% x^{3}\,}dx=4\dfrac{x^{2}}{2}+7\dfrac{x^{-3+1}}{-3+1}+C \\ &=&2x^{2}-\frac{7}{2x^{2}}+C \end{eqnarray*} \end{example} \begin{remark} In the above example, why don't we get two constants of integration? \ \dotfill \CustomNote[\hyperref{\TCIIcon{BITMAPSETAnswer}{0.1609in}{0.1487in}{0in}}{}{% }{}]{Margin Hint}{% The answer is easy. \ Suppose we do. \ For example \begin{eqnarray*} \int \left( x+x^{2}\right) dx=\allowbreak \frac{1}{2}x^{2}+C_{1}+\frac{1}{3}% x^{3}+C_{2}\allowbreak \\ =\frac{1}{2}x^{2}+\frac{1}{3}x^{3}+C_{1}+C_{2}{}_{{}} \end{eqnarray*} But $C_{1}+C_{2}$ is just another constant. \ Call it $C.$ \ Thus \[ \int \left( x+x^{2}\right) dx=\allowbreak \frac{1}{2}x^{2}\ +\frac{1}{3}% x^{3}+C \]% \par \vspace{1pt} \par \emph{''A constant is a constant.''}}\medskip \end{remark} \begin{example} Find $\int \dfrac{x+1}{x}\,dx.$ \end{example} \emph{Solution}. \ Split the fraction into two fractions, \ $\dfrac{x+1}{x}% =\allowbreak 1+\dfrac{1}{x}.$ \ Thus \[ \int \dfrac{x+1}{x}\,dx=\int \allowbreak 1+\dfrac{1}{x}\,dx=x+\ln \left| x\right| +C \] \begin{center} \vspace{1pt}\FRAME{dtbpF}{4.1943in}{0.0623in}{0pt}{}{}{maroon.wmf}{\special% {language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio TRUE;display "PICT";valid_file "F";width 4.1943in;height 0.0623in;depth 0pt;original-width 290.5625pt;original-height 1pt;cropleft "0";croptop "0.9405";cropright "1";cropbottom "0.3849";filename 'graphics/maroon__1.wmf';file-properties "XNPEU";}} \end{center} Test yourself: \ Determine the following: \begin{itemize} \item \begin{itemize} \item $\int x^{6}\,dx\qquad $\dotfill \CustomNote[\hyperref{\TCIIcon{BITMAPSETAnswer}{0.1609in}{0.1487in}{0in}}{}{% }{}]{Margin Hint}{$\int x^{6}\,dx=\allowbreak \frac{1}{7}x^{7}+C$} \item $\int \left( 3x-4x^{7}\right) \,dx\qquad $\dotfill \CustomNote[\hyperref{\TCIIcon{BITMAPSETAnswer}{0.1609in}{0.1487in}{0in}}{}{% }{}]{Margin Hint}{$\int \left( 3x-4x^{7}\right) \,dx=\allowbreak \dfrac{3}{2}% x^{2}-\dfrac{1}{2}x^{8}+C$} \item $\int 3x^{-1}\,dx\qquad $\dotfill \CustomNote[\hyperref{\TCIIcon{BITMAPSETAnswer}{0.1609in}{0.1487in}{0in}}{}{% }{}]{Margin Hint}{$\int 3x^{-1}\,dx=\allowbreak 3\ln \left| x\right| +C$} \item $\int \dfrac{x^{2}+1}{3x}\,dx$ $\qquad $\dotfill \CustomNote[\hyperref{\TCIIcon{BITMAPSETAnswer}{0.1609in}{0.1487in}{0in}}{}{% }{}]{Margin Hint}{$\int \dfrac{x^{2}+1}{3x}\,dx=\int $ $\left( \dfrac{1}{3}x+% \dfrac{1}{3x}\right) dx=\allowbreak \dfrac{1}{6}x^{2}+\dfrac{1}{3}\ln \left| x\right| $ $+C$% \par See, we split the fraction into two parts, and applied the power rule to each part and then the sum rule.} \end{itemize} \end{itemize} $\qquad $\FRAME{dtbpF}{4.1943in}{0.0623in}{0pt}{}{}{maroon.wmf}{\special% {language "Scientific Word";type "GRAPHIC";display "PICT";valid_file "F";width 4.1943in;height 0.0623in;depth 0pt;original-width 290.5625pt;original-height 1pt;cropleft "0";croptop "0.6633";cropright "1";cropbottom "0.6621";filename 'graphics/maroon.wmf';file-properties "XNPEU";}} \vspace{1pt}On the next page you will find the very important rules for $% \sin x$, $\cos x$ and exponential funtions. \vspace{0in} \subsection{Antiderivatives---Basic Rules} \vspace{0in}\vspace{1pt} {\Large The basic rules }for antiderivatives are simply the ordinary rules for derivatives stated backwards. \subsubsection{The second set of rules:} \begin{enumerate} \item[4.] \qquad Trigonometric functions: \begin{eqnarray*} \int \sin ax\ dx &=&-\dfrac{1}{a}\cos ax+C \\ \int \cos ax\,dx &=&\dfrac{1}{a}\sin ax+C\qquad \end{eqnarray*} \begin{enumerate} \item[5.] Exponential functions: \[ \int e^{ax}dx=\dfrac{1}{a}e^{ax}+C \] \item[6.] Inverse trigonometric functions: \begin{eqnarray*} \int \dfrac{1}{\sqrt{1-x^{2}}}\,dx &=&\sin ^{-1}x+C \\ \int \dfrac{-1}{\sqrt{1-x^{2}}}\,dx &=&\cos ^{-1}x+C \end{eqnarray*} \end{enumerate} \end{enumerate} \begin{center} \FRAME{dtbpF}{4.1943in}{0.0623in}{0pt}{}{}{maroon.wmf}{\special{language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio TRUE;display "PICT";valid_file "F";width 4.1943in;height 0.0623in;depth 0pt;original-width 290.5625pt;original-height 1pt;cropleft "0";croptop "0.7183";cropright "1";cropbottom "0.6071";filename 'graphics/maroon__1.wmf';file-properties "XNPEU";}} \end{center} We included, but will not use, the basic form for the integrals of inverse trig functions here. \ They will play a greater role in later sections. \begin{center} \CustomNote[{\fbox{\underline{Where are the rules for the other trig functions?}}}]{Margin Hint}{% These rules will come in the next chapters as applications of powerful methods for finding antiderivatives. \par But if you're curious, we have \par $\int \sec x\,dx=\allowbreak \ln \left| \sec x+\tan x\right| +C$% \par $\int \tan x\,dx=\allowbreak -\ln \left| \cos x\right| +C$% \par with similar rules for $\int \csc x\,dx$ and $\int \cot x\,dx.$}$% \longleftarrow $Click here\FRAME{dtbpF}{4.1943in}{0.0623in}{0pt}{}{}{% maroon.wmf}{\special{language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio TRUE;display "PICT";valid_file "F";width 4.1943in;height 0.0623in;depth 0pt;original-width 290.5625pt;original-height 1pt;cropleft "0";croptop "0.9405";cropright "1";cropbottom "0.3849";filename 'graphics/maroon__1.wmf';file-properties "XNPEU";}} \end{center} \begin{example} Applying rule 4, we have $\int \sin 3x\,\,dx=-\dfrac{1}{3}\cos x+C$ \end{example} \begin{example} Applying rule 5, we have $\int e^{4x}\,dx=\dfrac{1}{4}e^{4x}+C$ \end{example} \begin{example} Applying rule 4 and the \hyperref{constant multiplier rule}{}{}{% ./santi0301.tex#cmr} we have \[ \int 5\cos 2x\,dx=\allowbreak \dfrac{5}{2}\sin 2x+C \] \end{example} \begin{example} Applying rule 4, the \hyperref{constant multiplier rule}{}{}{% ./santi0301.tex#cmr} and the \hyperref{sum rule}{}{}{./santi0301.tex#sumrule} we have \[ \int \left( 4e^{3x}-7e^{-5x}\right) \,dx=\dfrac{4}{3}e^{3x}+\dfrac{7}{5}% e^{-5x}+C \] \end{example} \vspace{1pt}\textsl{Question:} On the last example do you understand why the minus $\left( -\right) $ sign changed to a plus $\left( +\right) $ sign?\dotfill \CustomNote[\hyperref{\TCIIcon{BITMAPSETAnswer}{0.1609in}{0.1487in}{0in}}{}{% }{}]{Margin Hint}{% The reason is that when you compute the antiderivative of $e^{-5x}$you get $-% \dfrac{1}{5}e^{-5x}.$ \ Thus the minus in the integrand and the minus from the antiderivative multiply to a plus.} \vspace{1pt} Test yourself: \ Determine the following: \begin{itemize} \item \begin{itemize} \item $\int 3\sin 5x\,dx\qquad $\dotfill \CustomNote[\hyperref{\TCIIcon{BITMAPSETAnswer}{0.1609in}{0.1487in}{0in}}{}{% }{}]{Margin Hint}{$\int 3\sin 5x\,dx=\allowbreak -\frac{3}{5}\cos 5x+C$} \item $\int 6e^{-x}\,dx$ $\qquad $\dotfill \CustomNote[\hyperref{\TCIIcon{BITMAPSETAnswer}{0.1609in}{0.1487in}{0in}}{}{% }{}]{Margin Hint}{$\int 6e^{-x}\,dx=\allowbreak -6e^{-x}+C$} \item $\int \left( e^{3x}+2\cos 4x-6\sin \left( -2x\right) \right) \,dx\qquad $\dotfill \CustomNote[\hyperref{\TCIIcon{BITMAPSETAnswer}{0.1609in}{0.1487in}{0in}}{}{% }{}]{Margin Hint}{$\int \left( e^{3x}+2\cos 4x-6\sin \left( -2x\right) \right) \,dx=\allowbreak \frac{1}{3}e^{3x}+\frac{1}{2}\sin 4x-3\cos 2x+C$% \par Can you explain the minus sign preceeding the term $3\cos 2x$ ? \ To answer requires you to explain why the integral has a $\cos 2x$ term rather than $% \cos \left( -2x\right) .$} \end{itemize} \end{itemize} $\qquad $\FRAME{dtbpF}{4.1943in}{0.0623in}{0pt}{}{}{maroon.wmf}{\special% {language "Scientific Word";type "GRAPHIC";display "PICT";valid_file "F";width 4.1943in;height 0.0623in;depth 0pt;original-width 290.5625pt;original-height 1pt;cropleft "0";croptop "0.6633";cropright "1";cropbottom "0.6621";filename 'graphics/maroon.wmf';file-properties "XNPEU";}} \vspace{1pt}\vspace{0in} \subsection{Antiderivatives---Applications} \vspace{0in}\vspace{1pt} {\Large What is that constant of integration all about?} \ In short, since we get a constant of integration, this allows us to specify that the antiderivative should satisfy some condition. \ That is we can solve the problem: \vspace{1pt} \begin{quotation} Problem: \ \ Given a function $f\,\left( x\right) .$ \ Find an antiderivative $F\left( x\right) $ of $f\,\left( x\right) $ such that $% F\left( x_{0}\right) =y_{0}$. \end{quotation} \vspace{1pt} \ See the following example. \begin{example} Find an antiderivative $F(x)$ of the function $f\,(x)=\dfrac{(x+1)}{x}$ for which $F(1)=~2$. \end{example} \noindent \emph{Solution}~~By the Sum and Power Rules, \begin{eqnarray*} F(x)=\int \frac{x+1}{x}dx &=&\int \left( \frac{x}{x}+\frac{1}{x}\right) dx \\ &=&\int 1\ dx+\int \frac{dx}{x} \\ &=&x+\ln |x|+C \end{eqnarray*} Since $F(1)=2$, \[ F(1)=1+\ln 1+C=2 \] and, recalling that $\ln 1=0$, we obtain \[ C=1 \] Hence, \[ F(x)=x+\ln |x|+1 \] Test yourself: \begin{itemize} \item \begin{itemize} \item Find the antiderivative $F\left( x\right) $ \ of $f\left( x\right) =x^{2}$ \ for which $F\left( 2\right) =4$ \ \dotfill \CustomNote[\hyperref{\TCIIcon{BITMAPSETAnswer}{0.1609in}{0.1487in}{0in}}{}{% }{}]{Margin Hint}{% The general antiderivative of $f\,\left( x\right) =x^{2}$ \ is \ $F\left( x\right) =$ $\dfrac{1}{3}x^{3}+C$ . \par If $F\left( 2\right) =4,$ \ then we must have \[ F\left( 2\right) =\dfrac{1}{3}2^{3}+C=4 \] Solving for $C$ give \[ C=4-\dfrac{1}{3}2^{3}=\dfrac{4}{3} \]% } \item Find the antiderivative $F\left( x\right) $ \ of $f\left( x\right) =\sin 2x$ for which $F\left( \dfrac{\pi }{6}\right) =1$ \ \dotfill \CustomNote[\hyperref{\TCIIcon{BITMAPSETAnswer}{0.1609in}{0.1487in}{0in}}{}{% }{}]{Margin Hint}{% The antiderivative of $\sin 2x$ \ is $F\left( x\right) =\dfrac{1}{2}\cos 2x+C.$ \ If $F\left( \dfrac{\pi }{6}\right) =1,$ then we must have \[ F\left( \dfrac{\pi }{6}\right) =\dfrac{1}{2}\cos \left( 2\dfrac{\pi }{6}% \right) +C=1 \] Solving for $C$ \ gives \begin{eqnarray*} \dfrac{1}{2}\cos \left( 2\dfrac{\pi }{6}\right) +C=1 \\ \dfrac{1}{2}\cos \dfrac{\pi }{3}+C=1 \\ \dfrac{1}{2}\left( \dfrac{1}{2}\right) +C=1 \\ C=\dfrac{4}{3} \end{eqnarray*}% \par {} \par $\dfrac{1}{2}\cos \left( 2\dfrac{\pi }{6}\right) +C=1$, Solution is : $% \left\{ C=\frac{3}{4}\right\} $} \end{itemize} \end{itemize} \begin{center} \FRAME{dtbpF}{4.1943in}{0.0623in}{0pt}{}{}{maroon.wmf}{\special{language "Scientific Word";type "GRAPHIC";display "PICT";valid_file "F";width 4.1943in;height 0.0623in;depth 0pt;original-width 290.5625pt;original-height 1pt;cropleft "0";croptop "0.6633";cropright "1";cropbottom "0.6621";filename 'graphics/maroon.wmf';file-properties "XNPEU";}} \end{center} \vspace{1pt} \begin{example} The \CustomNote[{\fbox{\underline{marginal cost}}}]{Margin Hint}{% Recall, the \emph{marginal cost} of a cost function $C\left( x\right) $ is just its derivative. So, \[ C_{M}\left( x\right) =C^{\prime }\left( x\right) \]% } $C_{M}(x)$ of producing $x$ kilograms of penicillin is given by \[ C_{M}(x)=2500+10x^{3/2}\;\text{dollars/kilogram} \]% If the start-up manufacturing cost of \$3500, what is the cost of manufacturing $x$ kilograms? \end{example} \noindent \emph{Solution}~~The derivative of the cost function $C(x)$ is the marginal cost $C_{M}(x)$ ; therefore, the cost function is an antiderivative of the marginal cost function. So we have \begin{eqnarray*} C(x)=\int C_{M}(x)dx &=&\int (2500+10x^{3/2})\,dx \\ &=&2500\int 1\ dx+10\int x^{3/2}\ dx \\ &=&2500x+10x^{5/2}/\left( \frac{5}{2}\right) +C \\ &=&2500x+4x^{5/2}+C \end{eqnarray*} The cost $C(0)$ is specified as \$3500, so \[ C(0)=2500\cdot 0+4\cdot 0^{5/2}+C=3500 \] and \[ C=3500 \] The cost $C(x)$ is thus \[ C(x)=2500x+4x^{5/2}+3500\text{ \ kilograms} \] $\qquad $\FRAME{dtbpF}{4.1943in}{0.0623in}{0pt}{}{}{maroon.wmf}{\special% {language "Scientific Word";type "GRAPHIC";display "PICT";valid_file "F";width 4.1943in;height 0.0623in;depth 0pt;original-width 290.5625pt;original-height 1pt;cropleft "0";croptop "0.6633";cropright "1";cropbottom "0.6621";filename 'graphics/maroon.wmf';file-properties "XNPEU";}} \vspace{0in} \subsection{Antiderivatives---Applications} \vspace{0in}\vspace{1pt} {\Large What is that constant of integration all about?} Another example \begin{center} \FRAME{dtbpF}{4.1943in}{0.0623in}{0pt}{}{}{maroon.wmf}{\special{language "Scientific Word";type "GRAPHIC";display "PICT";valid_file "F";width 4.1943in;height 0.0623in;depth 0pt;original-width 290.5625pt;original-height 1pt;cropleft "0";croptop "0.6633";cropright "1";cropbottom "0.6621";filename 'graphics/maroon.wmf';file-properties "XNPEU";}} \end{center} \vspace{1pt} \begin{example} Suppose the \CustomNote[{\fbox{\underline{velocity}}}]{Margin Hint}{% Recall, if $s\left( t\right) $ \ is the position of an object at time $t,$ \ then its velocity is its derivative. \ So, \[ v\left( t\right) =s^{\prime }\left( t\right) \] Alternatively, the antiderivative of \emph{velocity }is \emph{position}. \ \par \vspace{1pt} \par Here, it is easy to see the \underline{WHY} of the constant $C.$ \ Knowing only the velocity of an object, it is impossible to its determine the position with some reference, or starting point. \ This is just what the constant of integration $C$ is.} (in feet per second) of a rocket fired straight up is $v(t)=1500-32t$ after $t$ seconds. If the rocket is shot at $% t=0$ from a mountaintop 5000 feet high, what is its height after $t$ seconds? \end{example} \noindent \emph{Solution}~~The height $s(t)$ of the rocket is related to its velocity by $s^{\prime }(t)=v(t)$. So $s(t)$ is an antiderivative of $v(t)$. \ By \hyperref{rules 1, 2, and 3}{}{}{./santi0301.tex#derrules}, \begin{eqnarray*} s(t)=\int v(t)dt &=&\int (1500-32t)dt \\ &=&\int 1500\ dt-\int 32t\ dt \\ &=&1500\int 1\ dt-32\int t\ dt \\ &=&1500t-16t^{2}+C \end{eqnarray*} The question specifies $s(0)=5000$, so \[ s(0)=1500\cdot 0-16\cdot 0^{2}+C=5000 \] and this equation determines that \[ C=5000 \] Hence, the answer is \[ s(t)=5000+1500t-16t^{2}\text{ \ feet} \] $\qquad $\FRAME{dtbpF}{4.1943in}{0.0623in}{0pt}{}{}{maroon.wmf}{\special% {language "Scientific Word";type "GRAPHIC";display "PICT";valid_file "F";width 4.1943in;height 0.0623in;depth 0pt;original-width 290.5625pt;original-height 1pt;cropleft "0";croptop "0.6633";cropright "1";cropbottom "0.6621";filename 'graphics/maroon.wmf';file-properties "XNPEU";}} \begin{example} A rocket is fired straight up from a mountain top 5,000 ft high with an initial velocity of 1,100 ft/sec. \ What is the height of the rocket after 25 seconds? after 50 seconds? What is the maximum hight of the rocket? \lbrack \emph{Hint\/}:\ Recall the relations $v^{\prime }(t)=a(t)$ and $% s^{\prime }(t)=v(t)$. So, to find $s(t)$, two integrations must be carried out.\rbrack\ \end{example} \emph{Solution. }\ From physics we know that the acceleration of gravity is $% -32$ feet/second/second. This is the case because \begin{enumerate} \item \begin{enumerate} \item We are taking the positive direction in this problem to be up. \item The force of gravity (and hence acceleration) is down.\FRAME{dtbpF}{% 2.559in}{2.0003in}{0pt}{}{}{rocket.wmf}{\special{language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio TRUE;display "PICT";valid_file "F";width 2.559in;height 2.0003in;depth 0pt;original-width 251.5pt;original-height 210.6875pt;cropleft "0";croptop "1";cropright "1";cropbottom "0";filename 'graphics/rocket.wmf';file-properties "XNPEU";}} \end{enumerate} \end{enumerate} \vspace{1pt}To find the velocity we find the antiderivative of accereration, i.e. $v\left( t\right) =\int a\left( t\right) \,dt+C$. Thus $v\left( t\right) =-32t+C.$ \ Since we know $v\left( 0\right) =1100$ \ (Why?), it is easy to see that $C=1100$, and thus \[ v\left( t\right) =-32t+1100 \] \ To find the position we find the antiderivative of velocity, i.e. $s\left( t\right) =\int v\left( t\right) \,dt+C$. Thus \[ s\left( t\right) =\int \left( -32t+1100\,\right) dt+C=-16t^{2}+1100t+C \] $\func{Si}$nce $s\left( 0\right) =5000,$ we determine from $s\left( 0\right) =-16\left( 0\right) ^{2}+1100\left( 0\right) +C=5000$ that $C=5000.$ \ Hence, the position is \[ s\left( t\right) =-16t^{2}+1100t+5000 \] Now to answer the first question compute $s\left( 25\right) $% \[ s\left( 25\right) =22\,,500\allowbreak \ \;\text{feet} \] and \[ s\left( 50\right) =\allowbreak 20\,,000\;\text{feet} \] The maximum height of the rocket occurs when the velocity is zero, that is, when it stops going up and begins to fall. \ So, solve $v\left( t\right) =-32t+1100=0$ for $t.$ \ Thus, \[ t=\dfrac{1100}{32}=34.\,\allowbreak 375\text{\ seconds} \]% The height will be $s\left( \dfrac{1100}{32}\right) =\allowbreak 23906.\,\allowbreak 25$\ feet. \begin{center} \vspace{1pt}\vspace{0in} \end{center} \subsection{Antiderivatives --- Exercises} In Problems 1--24 evaluate the indefinite integral. \begin{enumerate} \item $\int 3x^{4}\ dx\;\;$\dotfill \CustomNote[\hyperref{\TCIIcon{BITMAPSETAnswer}{0.1609in}{0.1487in}{0in}}{}{% }{}]{Margin Hint}{$\int 3x^{4}\ dx=\allowbreak \dfrac{3}{5}x^{5}+C$} \item $\int (2x+2)dx\;\;$\dotfill \CustomNote[\hyperref{\TCIIcon{BITMAPSETAnswer}{0.1609in}{0.1487in}{0in}}{}{% }{}]{Margin Hint}{$\int (2x+2)dx=\allowbreak x^{2}+2x+C$} \item $\int 2e^{3x}\ dx\;\;$\dotfill \CustomNote[\hyperref{\TCIIcon{BITMAPSETAnswer}{0.1609in}{0.1487in}{0in}}{}{% }{}]{Margin Hint}{$\int 2e^{3x}\ dx\;\;=\allowbreak \dfrac{2}{3}e^{3x}+C$} \item $\int 5e^{-u}\ du\;\;$\dotfill \CustomNote[\hyperref{\TCIIcon{BITMAPSETAnswer}{0.1609in}{0.1487in}{0in}}{}{% }{}]{Margin Hint}{$\int 5e^{-u}\ du=\allowbreak -5e^{-u}+C$} \item $\int t^{100}\ dt\;\;$\dotfill \CustomNote[\hyperref{\TCIIcon{BITMAPSETAnswer}{0.1609in}{0.1487in}{0in}}{}{% }{}]{Margin Hint}{$\int t^{100}\ dt=\allowbreak \dfrac{1}{101}t^{101}$ $+C$} \item $\int x^{-6}\ dx\;\;$\dotfill \CustomNote[\hyperref{\TCIIcon{BITMAPSETAnswer}{0.1609in}{0.1487in}{0in}}{}{% }{}]{Margin Hint}{$\int x^{-6}\ dx=\allowbreak -\dfrac{1}{5x^{5}}+C$} \item $\int \sqrt{x}\ dx\;\;$\dotfill \CustomNote[\hyperref{\TCIIcon{BITMAPSETAnswer}{0.1609in}{0.1487in}{0in}}{}{% }{}]{Margin Hint}{$\int \sqrt{x}\ dx=\allowbreak \frac{2}{3}\left( \sqrt{x}% \right) ^{3}+C=\allowbreak \frac{2}{3}\left( x\right) ^{\dfrac{3}{2}}+C$} \item $\int x^{3/2}\ dx\;\;$\dotfill \CustomNote[\hyperref{\TCIIcon{BITMAPSETAnswer}{0.1609in}{0.1487in}{0in}}{}{% }{}]{Margin Hint}{$\int x^{3/2}\ dx=\allowbreak \frac{2}{5}\left( \sqrt{x}% \right) ^{5}+C=\dfrac{2}{5}x^{5/2}+C$} \item $\int t^{1/3}\ dt\;\;$\dotfill \CustomNote[\hyperref{\TCIIcon{BITMAPSETAnswer}{0.1609in}{0.1487in}{0in}}{}{% }{}]{Margin Hint}{$\int t^{1/3}\ dt=\allowbreak \dfrac{3}{4}\left( \sqrt[3]{t% }\right) ^{4}+C=\dfrac{3}{4}t^{4/3}+C$} \item $\int (t+(1/t))dt\;\;$\dotfill \CustomNote[\hyperref{\TCIIcon{BITMAPSETAnswer}{0.1609in}{0.1487in}{0in}}{}{% }{}]{Margin Hint}{$\int (t+(1/t))dt=\allowbreak \dfrac{1}{2}t^{2}+\ln \left| t\right| +C$} \item $\int x^{17}\ dx\;\;$\dotfill \CustomNote[\hyperref{\TCIIcon{BITMAPSETAnswer}{0.1609in}{0.1487in}{0in}}{}{% }{}]{Margin Hint}{$\int x^{17}\ dx=\allowbreak \dfrac{1}{18}x^{18}+C$} \item $\int x^{23}\ dx\;\;$\dotfill \CustomNote[\hyperref{\TCIIcon{BITMAPSETAnswer}{0.1609in}{0.1487in}{0in}}{}{% }{}]{Margin Hint}{$\int x^{23}\ dx=\allowbreak \dfrac{1}{24}x^{24}$ $+C$} \item $\int (\sqrt{x}+\sqrt[4]{x})dx\;\;$\dotfill \CustomNote[\hyperref{\TCIIcon{BITMAPSETAnswer}{0.1609in}{0.1487in}{0in}}{}{% }{}]{Margin Hint}{$\int (\sqrt{x}+\sqrt[4]{x})dx=\allowbreak \dfrac{2}{3}% \left( \sqrt{x}\right) ^{3}+\dfrac{4}{5}\left( \sqrt[4]{x}\right) ^{5}+C$ $% =\allowbreak \dfrac{2}{3}x^{3/2}+\dfrac{4}{5}x^{5/2}+C$ :} \item $\int (t^{1/3}+2t^{1/4})dt\;\;$\dotfill \CustomNote[\hyperref{\TCIIcon{BITMAPSETAnswer}{0.1609in}{0.1487in}{0in}}{}{% }{}]{Margin Hint}{$\int (t^{1/3}+2t^{1/4})dt=\allowbreak \frac{3}{4}\left( \sqrt[3]{t}\right) ^{4}+\frac{8}{5}\left( \sqrt[4]{t}\right) ^{5}+C=\dfrac{3% }{4}t^{4/3}+\dfrac{8}{5}t^{5/4}+C$} \item $\int 2u^{-1}\ du\;\;$\dotfill \CustomNote[\hyperref{\TCIIcon{BITMAPSETAnswer}{0.1609in}{0.1487in}{0in}}{}{% }{}]{Margin Hint}{$\int 2u^{-1}\ du=\allowbreak 2\ln \left| u\right| $ $+C$} \item $\int \dfrac{3}{x}dx\;\;$\dotfill \CustomNote[\hyperref{\TCIIcon{BITMAPSETAnswer}{0.1609in}{0.1487in}{0in}}{}{% }{}]{Margin Hint}{$\int \dfrac{3}{x}dx=\allowbreak 3\ln \left| x\right| +C$} \item $\int \sin 4x\,dx\;\;$\dotfill \CustomNote[\hyperref{\TCIIcon{BITMAPSETAnswer}{0.1609in}{0.1487in}{0in}}{}{% }{}]{Margin Hint}{$\int \sin 4x\,dx=\allowbreak -\frac{1}{4}\cos 4x+C$} \item $\int 3\cos \dfrac{1}{2}x\,dx\;\;$\dotfill \CustomNote[\hyperref{\TCIIcon{BITMAPSETAnswer}{0.1609in}{0.1487in}{0in}}{}{% }{}]{Margin Hint}{$\int 3\cos \dfrac{1}{2}x\,dx=\allowbreak 6\sin \frac{1}{2}% x+C$} \item $\int \left( 4\cos 3+x^{3}\right) \,dx\;\;$\dotfill \CustomNote[\hyperref{\TCIIcon{BITMAPSETAnswer}{0.1609in}{0.1487in}{0in}}{}{% }{}]{Margin Hint}{$\int \left( 4\cos 3+x^{3}\right) \,dx=\allowbreak 4\left( \cos 3\right) x+\frac{1}{4}x^{4}+C$} \item $\int \left( 5\sin 6x-\dfrac{1}{x}\right) \,dx\;\;$\dotfill \CustomNote[\hyperref{\TCIIcon{BITMAPSETAnswer}{0.1609in}{0.1487in}{0in}}{}{% }{}]{Margin Hint}{$\int \left( 5\sin 6x-\dfrac{1}{x}\right) \,dx=\allowbreak -\frac{5}{6}\cos 6x-\ln x+C$} \item $\int (e^{x}+e^{-x})dx\;\;$\dotfill \CustomNote[\hyperref{\TCIIcon{BITMAPSETAnswer}{0.1609in}{0.1487in}{0in}}{}{% }{}]{Margin Hint}{$\int (e^{x}+e^{-x})dx=\allowbreak e^{x}-e^{-x}+C$} \item $\int (e^{-x}+2x)dx\;\;$\dotfill \CustomNote[\hyperref{\TCIIcon{BITMAPSETAnswer}{0.1609in}{0.1487in}{0in}}{}{% }{}]{Margin Hint}{$\int (e^{-x}+2x)dx=\allowbreak -e^{-x}+x^{2}+C$} \item $\int (3/t^{3})dt\;\;$\dotfill \CustomNote[\hyperref{\TCIIcon{BITMAPSETAnswer}{0.1609in}{0.1487in}{0in}}{}{% }{}]{Margin Hint}{$\int (3/t^{3})dt\;=\allowbreak -\dfrac{3}{2t^{2}}+C$} \item $\int (1/2x)dx$ \ \ \dotfill \CustomNote[\hyperref{\TCIIcon{BITMAPSETAnswer}{0.1609in}{0.1487in}{0in}}{}{% }{}]{Margin Hint}{$\int (1/2x)dx=\allowbreak \frac{1}{2}\ln \left| x\right| +C$}\newline \ \ \noindent In Problems 25--32 find the antiderivative $F(x)$ of the given function $f(x)$ having the specified value. \newline (Remember, $F\left( x\right) =\int f\left( x\right) \,dx$) \item $f(x)=x^{3},\quad F(1)=0\;\;$\dotfill \CustomNote[\hyperref{\TCIIcon{BITMAPSETAnswer}{0.1609in}{0.1487in}{0in}}{}{% }{}]{Margin Hint}{} \item $f(x)=x+1/x,\quad F(1)=2\;\;$\dotfill \CustomNote[\hyperref{\TCIIcon{BITMAPSETAnswer}{0.1609in}{0.1487in}{0in}}{}{% }{}]{Margin Hint}{$f(x)=x+1/x,\quad F(1)=2\;$% \par $F\left( x\right) =\int f\left( x\right) \,dx=\int \left( x+1/x\right) \,dx=\allowbreak \frac{1}{2}x^{2}+\ln \left| x\right| +C$% \par $F\left( 1\right) =\dfrac{1}{2}1^{2}+\ln \left| 1\right| +C=\dfrac{1}{2}+C=2$% \par So, $C=\dfrac{3}{2}$ and $F\left( x\right) =\allowbreak \frac{1}{2}x^{2}+\ln \left| x\right| +\dfrac{3}{2}$} \item $f(x)=\sqrt{x},\quad F(0)=4\;\;$\dotfill \CustomNote[\hyperref{\TCIIcon{BITMAPSETAnswer}{0.1609in}{0.1487in}{0in}}{}{% }{}]{Margin Hint}{$f(x)=\sqrt{x},\quad F(0)=4$% \par \vspace{1pt}$F\left( x\right) =\allowbreak \frac{2}{3}\left( \sqrt{x}\right) ^{3}+4$} \item $f(x)=x+e^{x},\quad F(0)=0\;\;$\dotfill \CustomNote[\hyperref{\TCIIcon{BITMAPSETAnswer}{0.1609in}{0.1487in}{0in}}{}{% }{}]{Margin Hint}{$f(x)=x+e^{x},\quad F(0)=0$% \par $F\left( x\right) =\allowbreak \frac{1}{2}x^{2}+e^{x}-1$} \item $f(x)=e^{2x},\quad F(1)=2\;\;$\dotfill \CustomNote[\hyperref{\TCIIcon{BITMAPSETAnswer}{0.1609in}{0.1487in}{0in}}{}{% }{}]{Margin Hint}{$f(x)=e^{2x},\quad F(1)=2\;$% \par $F\left( x\right) =\dfrac{1}{2}e^{2x}+(2-\dfrac{1}{2}e^{2})$% \par {}} \item $f(x)=x^{2}+1+x,\quad F(0)=3\;\;$\dotfill \CustomNote[\hyperref{\TCIIcon{BITMAPSETAnswer}{0.1609in}{0.1487in}{0in}}{}{% }{}]{Margin Hint}{$f(x)=x^{2}+1+x,\quad F(0)=3\;$% \par $F\left( x\right) x=\allowbreak \frac{1}{3}x^{3}+x+\frac{1}{2}x^{2}+3$} \item $f\left( x\right) =1+\sin x,\;\;F\left( \pi /6\right) =4\;\;$\dotfill \CustomNote[\hyperref{\TCIIcon{BITMAPSETAnswer}{0.1609in}{0.1487in}{0in}}{}{% }{}]{Margin Hint}{$f\left( x\right) =1+\sin x,\;\;F\left( \pi /6\right) =4$% \par $F\left( x\right) =x-\cos x+(4-$ $\frac{1}{6}\pi +\frac{1}{2}\sqrt{3})$% \par $\;$} \item $f\left( x\right) =x-\cos 4x,\;\;F\left( \pi /8\right) =0\;\;$\dotfill \CustomNote[\hyperref{\TCIIcon{BITMAPSETAnswer}{0.1609in}{0.1487in}{0in}}{}{% }{}]{Margin Hint}{$f\left( x\right) =x-\cos 4x,\;\;F\left( \pi /8\right) =0$% \par $F\left( x\right) =\dfrac{1}{2}x^{2}-\dfrac{1}{4}\sin 4x-\left( \allowbreak \frac{1}{128}\pi ^{2}-\frac{1}{4}\right) $% \par {}} In Problems 33-45, determine the quantity asked for. \item The marginal cost $C_{M}(x)$ of producing $x$ tons of sheet steel is given by $C_{M}(x)=165+0.01x^{1/2}$ dollars. Production start-up costs are \$1500. What is the cost of producing $x$ tons of steel?\dotfill \CustomNote[\hyperref{\TCIIcon{BITMAPSETAnswer}{0.1609in}{0.1487in}{0in}}{}{% }{}]{Margin Hint}{$C_{M}(x)=165+0.01x^{1/2}$ $,\;C(0)=1500$% \par $C\left( x\right) \ =\allowbreak 165x+\dfrac{,02}{3}x^{3/2}+1500$% \par $\ $} \item Suppose the marginal cost of producing a type of bookcase is $% 200+0.01x^{3/2}$, with start-up costs of \$500. What is the cost function?\dotfill \CustomNote[\hyperref{\TCIIcon{BITMAPSETAnswer}{0.1609in}{0.1487in}{0in}}{}{% }{}]{Margin Hint}{$C_{M\left( x\right) }=200+0.01x^{3/2}$ ,$\;C(0)=500$% \par $C\left( x\right) =\allowbreak 200x+.00\,\allowbreak 4\left( \sqrt{x}\right) ^{5}+500$} \item The production engineer for an automobile parts supplier has determined that the marginal cost for the production of a new roller bearing assembly is given by $C_{M}(x)=100+50e^{x}$. If the start-up costs are known to be \$450, determine the cost function.\ \dotfill \CustomNote[\hyperref{\TCIIcon{BITMAPSETAnswer}{0.1609in}{0.1487in}{0in}}{}{% }{}]{Margin Hint}{$C_{M}(x)=100+50e^{x},\;C\left( 0\right) =450$% \par $C\left( x\right) =\allowbreak \ 100x-50e^{x}+450$} \item The marginal cost of producing $x$ units of a new facial cream is known to be $1+0.01e^{.01x}$, and start-up costs are \$10,000. What is the cost functions?\ \dotfill \CustomNote[\hyperref{\TCIIcon{BITMAPSETAnswer}{0.1609in}{0.1487in}{0in}}{}{% }{}]{Margin Hint}{$C_{M}\left( x\right) =1+0.01e^{.01x},\;\;C\left( 0\right) =10000$% \par $C\left( x\right) =\allowbreak x+\exp \left( .0\,\allowbreak 1x\right) +9999$% \par {}} \item The mass of a solid tumor grown under laboratory conditions was found to be increasing at the rate given by $R(t)=0.1+0.02t^{-1/2}$ grams per day. (Assume that time $t$ is measured in days.) If the mass of the tumor was known to be $12$ grams one Monday, what was the mass of the tumor the following Monday?\ \dotfill \CustomNote[\hyperref{\TCIIcon{BITMAPSETAnswer}{0.1609in}{0.1487in}{0in}}{}{% }{}]{Margin Hint}{$R(t)=0.1+0.02t^{-1/2}$% \par The mass of the tumor is $m\left( t\right) =\int R\left( t\right) dt=\allowbreak .\,\allowbreak 1t+.0\,\allowbreak 4\allowbreak \sqrt{t}+C.$% \par We are given that $m\left( 0\right) =12$ grams.\ So,$\;C=12.$ and \[ m\left( t\right) \ =\allowbreak .\,\allowbreak 1t+.0\,\allowbreak 4\allowbreak \sqrt{t}+12. \]% \par The following Monday is seven days later. \ We have \[ m\left( 7\right) \allowbreak =12.\,\allowbreak 7+.0\,\allowbreak 4\sqrt{7}% =\allowbreak 12.\,\allowbreak 806\text{ grams} \]% \par {}} \item The mass $m(t),t$ in days, of a solid tumor decreases after chemotherapy. If the rate at which it shrinks is given by $m^{\prime }(t)=-0.5t-0.1t^{2}$ grams per day and if the mass is 230 grams at the onset of treatment, what is the mass of the tumor after 5 days? After $T$ days?\ \dotfill \CustomNote[\hyperref{\TCIIcon{BITMAPSETAnswer}{0.1609in}{0.1487in}{0in}}{}{% }{}]{Margin Hint}{$m^{\prime }(t)=-0.5t-0.1t^{2}\ ,\;m\left( 0\right) =230$% \par $m(t)=\int \left( -0.5t-0.1t^{2}\right) dt+C=-\dfrac{1}{4}t^{2}-\dfrac{0.1}{3% }t^{3}+230$} \item Suppose the price $P(t)$ of a new Rolls Royce Silver Shadow has been increasing in recent years according to the equation $P^{\prime }(t)=15,000e^{.15t}+300t^{2}$, where $t$ is measured in years, and $t=0$ corresponds to 1980, $t=1$ to 1981, and so on. If the 1980 price was \$100,000, what will the price be in the year 2000?\ \dotfill \CustomNote[\hyperref{\TCIIcon{BITMAPSETAnswer}{0.1609in}{0.1487in}{0in}}{}{% }{}]{Margin Hint}{$P^{\prime }(t)=15,000e^{.15t}+300t^{2}$ ,\ $\ \;\;P\left( 0\right) =100000$% \par $P\left( t\right) =100000e^{.15t}\ +100\allowbreak t^{3}$% \par The constant $C$ \ is zero in this problem.} \item The diameter $D(t)$ of the very rare and long-lived Expoe Nut tree grows at a rate given by $D^{\prime }(t)=e^{-.1t}$. Suppose the diameter of the tree was 5 centimeters in 1900 $(t=0)$. What was the diameter in the years: $\left( a.\right) $~1950, $\left( b.\right) $~2000, and $\left( c.\right) $~3000? If the tree lives forever, will its diameter become infinitely large? If not, what is its limiting diameter?\ \dotfill \CustomNote[\hyperref{\TCIIcon{BITMAPSETAnswer}{0.1609in}{0.1487in}{0in}}{}{% }{}]{Margin Hint}{$D^{\prime }(t)=e^{-.1t}$ , \ $D\left( 0\right) =5$% \par $D\left( t\right) =-10e^{-0.1t}+15$% \par $D\left( 50\right) \ =\allowbreak 14.\,\allowbreak 933\;\ D\left( 100\right) =\allowbreak $ $14.\,\allowbreak 999\,546$% \par $D\left( 1100\right) =\allowbreak 15.0\ $(to 40 decimal places) \par If the tree lives forever, the diameter will never exceed, but will approach $15$ feet.$\ $} \item The purchase price of a new car was \$10,000. Its value $V(t)$ decreases at the rate given by $V^{\prime }(t)=-2000e^{-.2t}$, $t$ in years. When will the car be worth 10\% of its original value?\ \dotfill \CustomNote[\hyperref{\TCIIcon{BITMAPSETAnswer}{0.1609in}{0.1487in}{0in}}{}{% }{}]{Margin Hint}{$V^{\prime }(t)=-2000e^{-.2t},$ \ $V\left( 0\right) =$ $% 10000$% \par $V\left( t\right) =\allowbreak 10000e^{-.2t}$% \par Solve $V\left( t\right) =0.1\left( 10000\right) =1000$ for $t.$ \ This will be the time when $V\left( t\right) $ is $10\%$ of the original value. \par \begin{eqnarray*} 10000e^{-.2t}=1000 \\ e^{-.2t}=0.1 \\ -.2t=\ln \left( 0.1\right) =-2.\,\allowbreak 3023 \\ t=\dfrac{\ln \left( 0.1\right) }{-.2}=11.\,\allowbreak 513\text{ \ years} \end{eqnarray*}% } \item The marginal gain on an investment is given by the formula $% G_{M}(P)=6P^{1/4}$. Assuming that the gain $G$ when $P=0$ is \$256, find the gain for any investment $P$.\thinspace \dotfill \CustomNote[\hyperref{\TCIIcon{BITMAPSETAnswer}{0.1609in}{0.1487in}{0in}}{}{% }{}]{Margin Hint}{$G_{M}(P)=6P^{1/4}$, \ $G\left( 0\right) =$ $256$% \par $G\left( P\right) =\allowbreak \dfrac{24}{5}P^{5/4}+256$} \item The acceleration of a space shuttle is given by $a(t)=10t$, $t\geq 0$. If its velocity at time $t$ is 0 and it is launched from a height of 700 feet above sea level, what is its height after 20 seconds? \lbrack \emph{% Hint\/}:\ Recall the relations $v^{\prime }(t)=a(t)$ and $s^{\prime }(t)=v(t) $. So, to find $s(t)$, two integrations must be carried out.\rbrack\ \dotfill \CustomNote[\hyperref{\TCIIcon{BITMAPSETAnswer}{0.1609in}{0.1487in}{0in}}{}{% }{}]{Margin Hint}{$a(t)=10t,$ \ $v\left( 0\right) =0,\;s\left( 0\right) =700$% \par $v\left( t\right) =5t^{2}$% \par $s\left( t\right) =\dfrac{5}{3}t^{3}+700$% \par {}} \item The depth $d(t)$ of a drilling bit is increasing at the rate of $% 1/t^{1/4}$ feet per minute, for $t\geq 1$. If the depth of the bit is $10$ feet when $t=1$, how long will it take for the drill to reach a depth of 15,000 feet?\ \dotfill \CustomNote[\hyperref{\TCIIcon{BITMAPSETAnswer}{0.1609in}{0.1487in}{0in}}{}{% }{}]{Margin Hint}{$d^{\prime }\left( t\right) =1/t^{1/4}=$ $\int 1/t^{1/4}dt$% \par $d\left( t\right) =\allowbreak \dfrac{4}{3}t^{3/4}+C$% \par $d\left( 1\right) =10=\dfrac{4}{3}+C.$ \ So, $C=\dfrac{26}{3}$% \par Thus $d\left( t\right) =\allowbreak \dfrac{4}{3}t^{3/4}+\dfrac{26}{3}.$ \ Now solve $d\left( t\right) =15000.$ \ We have \begin{eqnarray*} \allowbreak \dfrac{4}{3}t^{3/4}+\dfrac{26}{3}=15000 \\ \dfrac{4}{3}t^{3/4}=15000-\dfrac{26}{3}=\frac{44\,974}{3} \\ t^{3/4}=\frac{44\,974}{3}\cdot \dfrac{3}{4}=\frac{22\,487}{2} \\ t=\left( \frac{22\,487}{2}\right) ^{4/3} \end{eqnarray*} or about minutes. \par {} \par {}Or $\dfrac{251885}{24\ast 60}=\allowbreak 174.\,\allowbreak 920\,138\,9$ days} \item The velocity in feet per second of a bullet is given by $% v(t)=3500-100t $ after $t$ secs, $0\leq t\leq 35$. How far has the bullet traveled after 3 seconds?\ \dotfill \CustomNote[\hyperref{\TCIIcon{BITMAPSETAnswer}{0.1609in}{0.1487in}{0in}}{}{% }{}]{Margin Hint}{$v(t)=3500-100t$% \par $s\left( t\right) =3500t-50t^{2}+C$% \par $s\left( 3\right) -s\left( 0\right) =\allowbreak 10,\,050$ feet \par There is no need to know $C$ for this problem. $\ Why?\ $}\FRAME{dtbpF}{% 4.1943in}{0.0623in}{0pt}{}{}{maroon.wmf}{\special{language "Scientific Word";type "GRAPHIC";display "PICT";valid_file "F";width 4.1943in;height 0.0623in;depth 0pt;original-width 290.5625pt;original-height 1pt;cropleft "0";croptop "0.5012";cropright "1";cropbottom "0.5";filename 'graphics/maroon.wmf';file-properties "XNPEU";}} \end{enumerate} \begin{center} \vspace{1pt} \end{center} \section{Introduction to the Integral} \vspace{1pt} \subsection{\QTR{yellowbig}{What is} \QTR{yellowbig}{area?}} \vspace{1pt} \vspace{1pt}The goal of this discussion is to clarify and develop ideas about \QTR{red}{area}. \ We consider the region under the graph of the function $f\left( x\right) $ between $x=a$ and $x=b.$ \ See below. \[ \FRAME{itbpF}{1.5748in}{1.1606in}{0in}{}{}{function.wmf}{\special{language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio TRUE;display "PICT";valid_file "F";width 1.5748in;height 1.1606in;depth 0in;original-width 107.875pt;original-height 79pt;cropleft "0";croptop "1";cropright "1";cropbottom "0";filename 'graphics/function.wmf';file-properties "XNPEU";}} \] Two big questions are these: \begin{description} \item \begin{description} \item \begin{itemize} \item What is the \QTR{red}{area} of this region? \item How can we find this \QTR{red}{area}? \end{itemize} \end{description} \end{description} When you think about it for a moment, the only figures we really know the area of are triangles and rectangles. \ (Count circles, too, if you wish.) \ So, the question: ``What \QTR{red}{is} the area?'' is not a simple as it sounds. More than what is the number of square units, we really need to know just what area means? Historically, dating back more than two thousand years, area has been one of the most difficult measures to make. \ With calculus, though, it's easy. \ Let's begin. \vspace{1pt}\FRAME{dtbpF}{4.1459in}{0.0735in}{0pt}{}{}{maroon2.wmf}{\special% {language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio TRUE;display "PICT";valid_file "F";width 4.1459in;height 0.0735in;depth 0pt;original-width 287.1875pt;original-height 3.1875pt;cropleft "0";croptop "0.8313";cropright "1";cropbottom "0.1687";filename 'graphics/maroon2.wmf';file-properties "XNPEU";}} Regarding the second question, if you were asked to find the area, you would probably draw a grid over the region and count the squares, getting an approximation to the area the accuracy of which depends on the grid size and the irregularity of the function. \ For the region above, examine the three overlayed grids below. \begin{center} \FRAME{itbpF}{1.5748in}{1.1606in}{0.4099in}{}{}{function_1.wmf}{\special% {language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio TRUE;display "PICT";valid_file "F";width 1.5748in;height 1.1606in;depth 0.4099in;original-width 107.875pt;original-height 79pt;cropleft "0";croptop "1";cropright "1";cropbottom "0";filename 'graphics/function_1.wmf';file-properties "XNPEU";}}$\rightarrow \FRAME{itbpF% }{1.5748in}{1.1606in}{0.3961in}{}{}{function_2.wmf}{\special{language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio TRUE;display "PICT";valid_file "F";width 1.5748in;height 1.1606in;depth 0.3961in;original-width 107.875pt;original-height 79pt;cropleft "0";croptop "1";cropright "1";cropbottom "0";filename 'graphics/function_2.wmf';file-properties "XNPEU";}}\rightarrow $\FRAME{itbpF% }{1.5748in}{1.1606in}{0.4238in}{}{}{function_3.wmf}{\special{language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio TRUE;display "PICT";valid_file "F";width 1.5748in;height 1.1606in;depth 0.4238in;original-width 107.875pt;original-height 79pt;cropleft "0";croptop "1";cropright "1";cropbottom "0";filename 'graphics/function_3.wmf';file-properties "XNPEU";}} \end{center} We approximate the area by the sum of the rectangles the are completely inside the region. Of course, this is a \QTR{red}{lower bound} for the area. \ Notice that the ``finer'' the grid the better the approximation to the area. \begin{remark} We could just as well have taken the covering set of rectangles. \ In this way we get a lower bound for the area and then an upper bound for the area. \ See below. \end{remark} \begin{center} \vspace{1pt}\FRAME{itbpF}{1.5748in}{1.1606in}{0.4722in}{}{}{function_4.wmf}{% \special{language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio TRUE;display "PICT";valid_file "F";width 1.5748in;height 1.1606in;depth 0.4722in;original-width 107.875pt;original-height 79pt;cropleft "0";croptop "1";cropright "1";cropbottom "0";filename 'graphics/function_4.wmf';file-properties "XNPEU";}}$\rightarrow \FRAME{itbpF% }{1.5748in}{1.1606in}{0.4653in}{}{}{function_5.wmf}{\special{language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio TRUE;display "PICT";valid_file "F";width 1.5748in;height 1.1606in;depth 0.4653in;original-width 107.875pt;original-height 79pt;cropleft "0";croptop "1";cropright "1";cropbottom "0";filename 'graphics/function_5.wmf';file-properties "XNPEU";}}\rightarrow \FRAME{itbpF% }{1.5748in}{1.1606in}{0.4583in}{}{}{function_6.wmf}{\special{language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio TRUE;display "PICT";valid_file "F";width 1.5748in;height 1.1606in;depth 0.4583in;original-width 107.875pt;original-height 79pt;cropleft "0";croptop "1";cropright "1";cropbottom "0";filename 'graphics/function_6.wmf';file-properties "XNPEU";}}$ \end{center} \QTR{bluebig}{Finding areas by summing} the areas of the rectangles of a covering grid is what we shall do to not only find the areas but in fact to define the area itself. \vspace{1pt} Our first task will be to devise special notation for the sums we will need. \FRAME{dtbpF}{4.1459in}{0.0735in}{0pt}{}{}{maroon2.wmf}{\special{language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio TRUE;display "PICT";valid_file "F";width 4.1459in;height 0.0735in;depth 0pt;original-width 287.1875pt;original-height 3.1875pt;cropleft "0";croptop "0.8313";cropright "1";cropbottom "0.1687";filename 'graphics/maroon2.wmf';file-properties "XNPEU";}} \vspace{1pt}We have just discussed the importance of summing the areas of rectangles to finding the area under the graph of a function. \ In this section we introduce some notation to help us express the sums we need.\medskip\ \ Two different notations are needed: the ``\QTR{red}{dots}'' notation and the ``\QTR{red}{sigma}'' notation. \subsection{\QTR{yellowbig}{Sums and the ``dots'' notation}} \vspace{1pt} \QTR{blue}{Problem}. \ Suppose it is desired to sum the first one hundred integers. \ How can this be expressed? \vspace{1pt} \emph{Solution 1.} Write the whole sum. \[ 1+2+3+4+5+6+\text{ and so on up to }100. \] This is way too long. \ Even the sum of the first ten integers is just about too long to write out fully. \[ 1+2+3+4+5+6+7+8+9+10 \] \vspace{1pt}\emph{Solution 2.} Another way to express it is using the ``dots'' notation. \ We write \[ 1+2+3+\cdots +100 \] Here the ``$+\cdots +"$represents all the integers in between. $\ $You, the reader, are to see the arithmetic sequence $1+2+3$ and \QTR{red}{deduce} that what comes next is $4+5+6$ and so on up to $100.$ \ Thus, you are required to \QTR{red}{see the pattern}. \ (This isn't always easy.) \ The second sum above is written as \[ 1+2+3+\cdots +10 \] \vspace{1pt} With the\QTR{blue}{\ ``dots'' notation} for summation, you must \begin{itemize} \item \begin{itemize} \item \begin{itemize} \item See/write the first few terms. \item Identify the pattern. \item Recognize/write the last term. \end{itemize} \end{itemize} \end{itemize} Let's try a few examples.\medskip \begin{example} What does $2+4+6+\cdots +20$ mean? \end{example} \emph{Solution.} \ This one is easy. \ The pattern is the even numbers. \ Sum them from $2$ to $20.\medskip \medskip $ \begin{example} Express the sum of the odd integers up to $101.$ \end{example} \emph{Solution.} \ We write $1+3+5+$ and so on up to $101.$ \ Using the dots notation this is \[ 1+3+5+\cdots +101 \] \vspace{1pt} \FRAME{dtbpF}{4.1943in}{0.0692in}{0pt}{}{}{maroon.wmf}{\special{language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio TRUE;display "PICT";valid_file "F";width 4.1943in;height 0.0692in;depth 0pt;original-width 290.5625pt;original-height 1pt;cropleft "0";croptop "0.7968";cropright "1";cropbottom "0.5965";filename 'graphics/maroon.wmf';file-properties "XNPEU";}} \vspace{1pt} \subsection{\QTR{yellowbig}{Sigma Notation}} \vspace{1pt} \vspace{1pt}In many cases where the arithmetic pattern is not evident, or even non-existant, but where the terms are know by some formula, the ``Sigma'' notation is effective. \ To use it we need to know four terms: \begin{itemize} \item \begin{itemize} \item The formula to sum over. \item The starting term. \item The ending term. \item A symbol for the variable term. \end{itemize} \end{itemize} Let's look at this notation for the sum of the first $10$ integers.\FRAME{% dtbpF}{3.6037in}{1.5532in}{0pt}{}{}{sigma.wmf}{\special{language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio TRUE;display "PICT";valid_file "F";width 3.6037in;height 1.5532in;depth 0pt;original-width 249.3125pt;original-height 106.375pt;cropleft "0";croptop "1";cropright "1";cropbottom "0";filename 'graphics/sigma.wmf';file-properties "XNPEU";}} In this graphic, the $\Sigma $ (Greek letter ``sigma'') is the symbol that tells us to sum. \ The ``formula'' here is $f\left( x\right) =x$. \ The summation variable (sometimes called the \emph{index}) is symbolized by $i$, \ we start at $i=1,$ and we end at $i=10.\medskip $ \begin{tabular}{c} \FRAME{itbpF}{0.7636in}{0.5172in}{-0.0138in}{}{}{wow.wmf}{\special{language "Scientific Word";type "GRAPHIC";display "PICT";valid_file "F";width 0.7636in;height 0.5172in;depth -0.0138in;original-width 40.6875pt;original-height 30.5pt;cropleft "0";croptop "1";cropright "1";cropbottom "0";filename 'graphics/wow.wmf';file-properties "XNPEU";}}% \end{tabular} \begin{tabular}{|l|} \hline You can see that this notation is very powerful because \\ it expresses \QTR{red}{exactly} what is to be summed and over what range. \\ There is no pattern to figure out! \\ \hline \end{tabular} \medskip \begin{description} \item \vspace{1pt}Here how we express the sum of function evaluations of $% f\left( x\right) $ at the integers $1,\,2,\,...,\,n.$ \end{description} \[ \sum_{i\,=1^{{}}}^{n}f\left( i\right) =f\left( 1\right) +f\left( 2\right) \,+\,f\left( 3\right) +\cdots +\,f\left( n\right) \] \begin{description} \item Here's how we express the sum of the sequence numbers $% a_{1},\,a_{2},\,a_{3},\ldots ,\,a_{n}.$% \[ \sum_{i\,=1^{{}}}^{n}a_{i}=a_{1}+\,a_{2}+\,a_{3}+\cdots +\,a_{n} \] \FRAME{dtbpF}{4.1952in}{0.0623in}{0pt}{}{}{maroon.wmf}{\special{language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio TRUE;display "PICT";valid_file "F";width 4.1952in;height 0.0623in;depth 0pt;original-width 290.5625pt;original-height 1pt;cropleft "0";croptop "0.4129";cropright "0.9981";cropbottom "0.9125";filename 'graphics/maroon.wmf';file-properties "XNPEU";}} \end{description} \begin{example} \vspace{1pt}Using the summation notation write the sum of the even integers from $2$ to $50$. \end{example} \emph{Solution.} \ Here the function is $f\left( x\right) =2x.$ \ \CustomNote[{\fbox{Why}}]{Margin Hint}{% Because we can get just the even integers this way with this ``doubling'' function.}\ \ \ Make the summation variable $i.$ \ The sum should start from $i=1$ and end at $i=25.$ (This will give the value $f\left( 25\right) =2\left( 25\right) =50.$) \ We have \[ \sum_{i=1}^{25}2i \] \begin{exercise} Using the summation notation write the sum of the integer multiples of $5$ from integers from $0$ to $50$\dotfill \CustomNote[\hyperref{\TCIIcon{BITMAPSETAnswer}{0.1609in}{0.1487in}{0in}}{}{% }{}]{Margin Hint}{% Here the function is $f\left( x\right) =5x.$ \ \ Make the summation variable $i.$ \ The sum should start from $i=0$ and end at $i=10.$ (This will give the value $50.$) \ We have \[ \sum_{i=1}^{10}5i \]% } \end{exercise} \begin{center} \vspace{1pt}\FRAME{dtbpF}{4.1459in}{0.0735in}{0pt}{}{}{maroon2.wmf}{\special% {language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio TRUE;display "PICT";valid_file "F";width 4.1459in;height 0.0735in;depth 0pt;original-width 287.1875pt;original-height 3.1875pt;cropleft "0";croptop "0.6626";cropright "1";cropbottom "0.3374";filename 'graphics/maroon2.wmf';file-properties "XNPEU";}} \end{center} \vspace{1pt} \subsection{\QTR{yellowbig}{Properties of the Sigma Notation}} \vspace{1pt} \vspace{1pt}The sigma notation is powerful indeed as it has several useful properites. \ Since it is really just a ``glorified'' sum, it obeys all of the rules of sums. \begin{center} \vspace{1pt} \begin{tabular}{|cc|} \hline & \QTR{blue}{Properties of the Sigma Notation} \\ $(i)$ & \multicolumn{1}{l|}{$\dsum\limits_{i=1}^{n}ca_{i}=c\dsum% \limits_{i=1}^{n}a_{i}\;,$ where $c$ is any constant} \\ $(ii)$ & \multicolumn{1}{l|}{$\dsum\limits_{i=1}^{n}\left( a_{i}+b_{i}\right) =\dsum\limits_{i=1}^{n}a_{i}+\dsum\limits_{i=1}^{n}b_{i}$} \\ $(iii)$ & \multicolumn{1}{l|}{$\dsum\limits_{i=1}^{n}\left( a_{i}-b_{i}\right) =\dsum\limits_{i=1}^{n}a_{i}-\dsum\limits_{i=1}^{n}b_{i}$} \\ \hline \end{tabular} \medskip \end{center} The power of these properties will be evident once we have a few formulas to apply them to. \ Here are four\QTR{purple}{\ very useful formulas}. \begin{center} \vspace{1pt} \begin{tabular}{|llllll|} \hline & & & \QTR{blue}{Useful Summation Formulas} & & \\ \hline\hline & $(i)$ & & $\dsum\limits_{i=1}^{n}1=n$ & & Sum of ones \\ & $(ii)$ & & $\dsum\limits_{i=1}^{n}i=\dfrac{n\left( n+1\right) }{2}$ & & Sum of integers \\ & $(iii)$ & & $\dsum\limits_{i=1}^{n}i^{2}=\dfrac{n\left( n+1\right) \left( 2n+1\right) }{6}$ & & Sum of squares \\ & $(iv)$ & & $\dsum\limits_{i=1}^{n}i^{3}=\left[ \dfrac{n\left( n+1\right) }{2}\right] ^{2}$ & & Sum of cubes \\ \hline \end{tabular} \end{center} \begin{description} \item \begin{itemize} \item \vspace{1pt}Here how we express the sum of function evaluations of $% f\left( x\right) $ at the integers $1,\,2,\,...,\,n.$ \end{itemize} \end{description} \[ \sum_{i\,=1^{{}}}^{n}f\left( i\right) =f\left( 1\right) +f\left( 2\right) \,+\,f\left( 3\right) +\cdots +\,f\left( n\right) \] \begin{description} \item \begin{itemize} \item Here's how we express the sum of the sequence numbers $% a_{1},\,a_{2},\,a_{3},\ldots ,\,a_{n}.$% \[ \sum_{i\,=1^{{}}}^{n}a_{i}=a_{1}+\,a_{2}+\,a_{3}+\cdots +\,a_{n} \] \end{itemize} \end{description} \FRAME{dtbpF}{4.1459in}{0.0735in}{0pt}{}{}{maroon2.wmf}{\special{language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio TRUE;display "PICT";valid_file "F";width 4.1459in;height 0.0735in;depth 0pt;original-width 287.1875pt;original-height 3.1875pt;cropleft "0";croptop "0.8313";cropright "1";cropbottom "0.1687";filename 'graphics/maroon2.wmf';file-properties "XNPEU";}} \begin{example} Find $\dsum\limits_{i=1}^{100}i.$ \end{example} \emph{Solution.} $\ $Using the sum of integers \CustomNote[{\fbox{\underline{formula}}}]{Margin Hint}{$\dsum% \limits_{i=1}^{n}i=\dfrac{n\left( n+1\right) }{2}$} we have $% \dsum\limits_{i=1}^{100}i=\dfrac{100\left( 100+1\right) }{2}=\allowbreak 5050.$ \begin{example} Find $\dsum\limits_{i=1}^{100}\left( 2i^{2}+4i\right) .$ \end{example} \emph{Solution.} $\ $First break the sum into two sums using the summation property $\left( ii\right) $ above. \ Then use the multiplier rule $\left( i\right) $. \ We have \begin{eqnarray*} \dsum\limits_{i=1}^{100}\left( 2i^{2}+4i\right) &=&\dsum\limits_{i=1}^{100}2i^{2}+\dsum\limits_{i=1}^{100}4i \\ &=&2\dsum\limits_{i=1}^{100}i^{2}+4\dsum\limits_{i=1}^{100}i \end{eqnarray*}% On the first term use the sum of squares $\dsum\limits_{i=1}^{n}i^{2}=\dfrac{% n\left( n+1\right) \left( 2n+1\right) }{6}$ $\ $and on the second use the sum of integers $\dsum\limits_{i=1}^{n}i=\dfrac{n\left( n+1\right) }{2}$ to get \begin{eqnarray*} \dsum\limits_{i=1}^{100}\left( 2i^{2}+4i\right) &=&2\dsum\limits_{i=1}^{100}i^{2}+4\dsum\limits_{i=1}^{100}i \\ &=&2\dfrac{100\left( 100+1\right) \left( 2\left( 100\right) +1\right) }{6}+4% \dfrac{100\left( 100+1\right) }{2} \\ &=&20\,200+676\,700=\allowbreak 696\,900 \end{eqnarray*} \begin{example} Find $\dsum\limits_{i=1}^{15}\left( i+1\right) ^{2}$ \end{example} \emph{Solution.} $\ $First expand the expression $\left( i+1\right) ^{2}=i^{2}+2i+1.$ \ Insert this into the sum and apply the properties to get \begin{eqnarray*} \dsum\limits_{i=1}^{15}\left( i+1\right) ^{2} &=&\dsum\limits_{i=1}^{15}\left( i^{2}+2i+1\right) \\ &=&\dsum\limits_{i=1}^{15}i^{2}+2\dsum\limits_{i=1}^{15}i+\dsum% \limits_{i=1}^{15}1 \end{eqnarray*} Now apply the formulas to get \begin{eqnarray*} \dsum\limits_{i=1}^{15}\left( i+1\right) ^{2} &=&\dfrac{15\left( 15+1\right) \left( 2\left( 15\right) +1\right) }{6}+2\dfrac{15\left( 15+1\right) }{2}+15 \\ &=&1240+240+15=1495 \end{eqnarray*} \begin{exercise} What is $\dsum\limits_{i=1}^{15}\left( 2i+8\right) \,$?\dotfill \CustomNote[\hyperref{\TCIIcon{BITMAPSETAnswer}{0.1609in}{0.1487in}{0in}}{}{% }{}]{Margin Hint}{% \begin{eqnarray*} \dsum\limits_{i=1}^{15}\left( 2i+8\right) =\dsum\limits_{i=1}^{15}2i+\dsum\limits_{i=1}^{15}8 \\ =2\dsum\limits_{i=1}^{15}i+8\dsum\limits_{i=1}^{15}1 \\ =2\dfrac{15(15+1)}{2}+8\left( 15\right) \\ =360 \end{eqnarray*}% \par {}} \end{exercise} \begin{center} \vspace{1pt}\FRAME{dtbpF}{4.1459in}{0.0761in}{0pt}{}{}{maroon2.wmf}{\special% {language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio TRUE;display "PICT";valid_file "F";width 4.1459in;height 0.0761in;depth 0pt;original-width 287.1875pt;original-height 3.1875pt;cropleft "0";croptop "0.5686";cropright "1";cropbottom "0.4686";filename 'graphics/maroon2.wmf';file-properties "XNPEU";}} \end{center} \vspace{1pt} \subsection{\QTR{yellowbig}{Areas and sums}} \vspace{1pt} \vspace{1pt}{\Large Recall} our goal: Given the region under the graph of the function $f\left( x\right) $ between $x=a$ and $x=b.$ Find its area.% \FRAME{dtbpF}{2.8963in}{2.0453in}{0pt}{}{}{sum_integral.wmf}{\special% {language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio TRUE;display "PICT";valid_file "F";width 2.8963in;height 2.0453in;depth 0pt;original-width 200.375pt;original-height 185.5625pt;cropleft "0";croptop "1";cropright "1";cropbottom "0";filename 'graphics/sum_integral.wmf';file-properties "XNPEU";}} \ \[ \] We begin in earnest to express areas as sums of regular areas. \ The ``graph paper'' is not the simplistic kind used in the \hyperref{chapter introduction% }{}{}{./sintintro0101.tex#graphpaper} but rather is customized for each function. \ \ Here's how we do it. \begin{description} \item[Step 1.] Divide the interval $\left[ a,\,b\right] $ into $n$ equal pieces. \ This is called a \QTR{red}{subdivision} of $\left[ a,b\right] .$ \ Our notation will be \[ a=x_{0}