\hfill \thepage} %} \newtheorem{acknowledgement}[theorem]{Acknowledgement} \newtheorem{algorithm}[theorem]{Algorithm} \newtheorem{case}[theorem]{Case} \newtheorem{claim}[theorem]{Claim} \newtheorem{conclusion}[theorem]{Conclusion} \newtheorem{condition}[theorem]{Condition} \newtheorem{criterion}[theorem]{Criterion} \newtheorem{notation}[theorem]{Notation} \newtheorem{problem}[theorem]{Problem} \newtheorem{solution}[theorem]{Solution} \newtheorem{summary}[theorem]{Summary} \newenvironment{proof}[1][Proof]{\textbf{#1.} }{\ \rule{0.5em}{0.5em}} \input{tcilatex} \begin{document} \section{Ma 115 Review of Basic Differentiation and Integration} \vspace{1pt} \subsection{Differentiation} \section{Formulas for Derivatives} \subsection{The Derivative Rules --- Summary} \vspace{0in} \begin{quotation} \textbf{Constant Multiplier Rule} \begin{equation*} \dfrac{d}{dx}\left( cf\,\left( x\right) \right) =cf\,^{\prime }\left( x\right) \end{equation*} \textbf{Power Rule} \begin{equation*} \frac{d}{dx}(x^{r})=rx^{r-1},\qquad r\neq 0 \end{equation*} \textbf{Sum Rule} \begin{equation*} \frac{d}{dx}\lbrack f(x)\pm g(x)\rbrack =f\,^{\prime }(x)\pm g\,^{\prime }(x) \end{equation*} \textbf{Product Rule} \begin{equation*} \frac{d}{dx}\lbrack f(x)\cdot g(x)\rbrack =f(x)\cdot g\,^{\prime }(x)+g(x)\cdot f\,^{\prime }(x) \end{equation*} \textbf{Reciprocal Rule} \begin{equation*} \frac{d}{dx}\left[ \frac{1}{g(x)}\right] =\frac{-g\,^{\prime }(x)}{g^{2}(x)} \end{equation*} \textbf{Quotient Rule} \begin{equation*} \frac{d}{dx}\left[ \frac{f(x)}{g(x)}\right] =\frac{g(x)f\,^{\prime }(x)-f(x)g\,^{\prime }(x)}{g^{2}(x)} \end{equation*} \textbf{Extended Power Rule} \begin{equation*} \frac{d}{dx}\lbrack f\,^{r}(x)\rbrack =rf\,^{r-1}(x)\cdot f\,^{\prime }(x),\qquad r\neq 0 \end{equation*} \end{quotation} \vspace{0in} \begin{quotation} \textbf{Chain Rule}~~~If $g(x)$ and $f(x)$ are two differentiable functions, the derivative of the composition of $f$ \ with $g$ \ is given by \begin{equation*} \frac{d}{dx}[f(g(x))]=f^{\prime }(g(x))g^{\prime }(x) \end{equation*}% for all $x$ \ in the domain of $f(g(x))$. \end{quotation} By applying the Chain Rule, we also have the following differentiation formulas for the trigonometric functions. \begin{eqnarray*} &&\frac{d}{dt}\sin f(t)=[\cos f(t)]f\,\,^{\prime }(t) \\ &&\frac{d}{dt}\cos f(t)=-[\sin f(t)]f\,\,^{\prime }(t) \\ &&\frac{d}{dt}\tan f(t)=[\sec ^{2}f(t)]f\,\,^{\prime }(t) \\ &&\frac{d}{dt}\cot f(t)=-[\csc ^{2}f(t)]f\,\,^{\prime }(t) \\ &&\frac{d}{dt}\sec f(t)=[\sec f(t)\tan f(t)]f\,\,^{\prime }(t) \\ &&\frac{d}{dt}\csc f(t)=-[\csc f(t)\cot f(t)]f\,\,^{\prime }(t) \end{eqnarray*}% \QTR{red}{Memorize these!} \paragraph{Inverse Trig Functions} The following table summarizes what we have learned about the \emph{inverse trigonometric} functions. \begin{center} \begin{tabular}{cccccc} Function & Domain & & Range & & Derivative \\ \hline $\sin ^{-1}x$ & $\left[ -1,1\right] $ & & $\left[ -\dfrac{\pi }{2},\dfrac{% \pi }{2}\right] $ & & $\dfrac{1}{\sqrt{1-x^{2}}}$ for $\left| x\right| <1$ \\ \hline $\cos ^{-1}x$ & $\left[ -1,1\right] $ & & $\left[ 0,\pi \right] $ & & $% \dfrac{-1}{\sqrt{1-x^{2}}}$ for $\left| x\right| <1$ \\ \hline $\tan ^{-1}x$ & $\left( -\infty ,\infty \right) $ & & $\left( -\dfrac{\pi }{% 2},\dfrac{\pi }{2}\right) $ & & $\dfrac{1}{1+x^{2}}$ for all $x$ \\ \hline $\cot ^{-1}x$ & $\left( -\infty ,\infty \right) $ & & $\left( 0,\pi \right) $ & & $\dfrac{-1}{1+x^{2}}$ for all $x$ \\ \hline $\sec ^{-1}x$ & $\left| x\right| \geq 1$ & & $\lbrack 0,\dfrac{\pi }{2}% )\cup (\dfrac{\pi }{2},\pi \rbrack $ & & $\dfrac{1}{\left| x\right| \sqrt{% x^{2}-1}}$ for $\left| x\right| >1$ \\ \hline $\csc ^{-1}x$ & $\left| x\right| \geq 1$ & & $\lbrack \dfrac{-\pi }{2}% ,0)\cup (0,\dfrac{\pi }{2}\rbrack $ & & $\dfrac{-1}{\left| x\right| \sqrt{% x^{2}-1}}$ for $\left| x\right| >1$ \\ \hline \end{tabular} \end{center} \vspace{1pt} \emph{Remember} that the \emph{ranges} of these inverse functions equal the \emph{restricted domains} of functions for which they are the inverse. \paragraph{Log Functions} \begin{quotation} If $f(x)>0$ and is differentiable, then \begin{equation*} \frac{d}{dx}[\log _{a}f(x)]=\frac{f^{\prime }(x)}{f(x)\ln a} \end{equation*} \end{quotation} \noindent \begin{quotation} If $a>0$ and $f(x)$ is differentiable, then \begin{equation*} \frac{d}{dx}[a^{f(x)}]=(\ln a)a^{f(x)}f^{\prime }(x) \end{equation*} \end{quotation} \paragraph{Fundamental Theorems of Calculus} \begin{theorem} (\emph{The First Fundamental Theorem of Calculus} ) Let $f\,\left( x\right) $ be continuous on the interval $\left[ a,\,b\right] $. \ Then for any $x$ in $% \left( a,\,b\right) ,$ the derivative of the integral $\int_{a}^{x}f\,\left( t\right) \,dt$ exists, and \begin{equation*} \dfrac{d}{dx}\int_{a}^{x}f\,\left( t\right) \,dt=f\left( x\right) \end{equation*} \medskip [Thus the function $\int_{a}^{x}f\,\left( t\right) \,dt$ is an antiderivative of $f\,\left( x\right) .$] \end{theorem} \begin{theorem} (\emph{The Second Fundamental Theorem of Calculus} ) \ Let $f\,\left( x\right) $ be continuous on the interval $\left[ a,\,b\right] $. \ Let $% F\left( x\right) $ be an aintiderivative of $f(x)$\ Then \begin{equation*} \int_{a}^{b}f\,\left( t\right) \,dt=F\left( b\right) -F\left( a\right) \end{equation*} \end{theorem} \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 "0.6969";cropright "1";cropbottom "0.6964";filename '/document/graphics/maroon.wmf';file-properties "XNPEU";}} \QTR{blue}{Notation}. We also write this as \begin{equation*} \int_{a}^{b}f\,\left( t\right) \,dt=F\left( x\right) {\Large |}% _{a}^{b}=F\left( b\right) -F\left( a\right) \end{equation*} \subsubsection{Examples:} \subsubsection{The Derivative Rules --- How to Use Them} \vspace{0in}If you are given an expression to differentiate, say $% f(x)=\left( 3x^{4}-2x\right) \left( x^{3}+5x^{2}\right) ^{5}+\dfrac{4x^{3}}{% 2x^{5}-3x+4}$, first write down the derivative symbol $\dfrac{d}{dx}$ in front of the expression and add an equal sign: \begin{eqnarray*} \dfrac{df}{dx} &=&\dfrac{d}{dx}\left[ \left( 3x^{4}-2x\right) \left( x^{3}+5x^{2}\right) ^{5}+\dfrac{4x^{3}}{2x^{5}-3x+4}\right] \\ &=& \end{eqnarray*} \vspace{0in}Then you must determine if the expression is basically a sum, a difference, a product, a quotient or a power. \ To do this, think about evaluating the expression at a number. \ What is the very last operation you would perform? \ A sum, a difference, a product, a quotient or a power? \ That is the basic type of the expression. \vspace{0in} \begin{itemize} \item If the expression is basically a sum or difference, use the Sum Rule and differentiate each term. \ If the expressions are complicated, do not attempt to differentiate, just write a derivative symbol in front of each term. \ For example, \begin{eqnarray*} \dfrac{df}{dx} &=&\dfrac{d}{dx}\left[ \left( 3x^{4}-2x\right) \left( x^{3}+5x^{2}\right) ^{5}+\dfrac{4x^{3}}{2x^{5}-3x+4}\right] \\ &=&\dfrac{d}{dx}\left[ \left( 3x^{4}-2x\right) \left( x^{3}+5x^{2}\right) ^{5}\right] +\dfrac{d}{dx}\left[ \dfrac{4x^{3}}{2x^{5}-3x+4}\right] \end{eqnarray*} \item If the expression is basically a product, and one factor is a constant, use the Constant Multiplier Rule by writing down the constant and differentiating the other factor. \ For example, \begin{equation*} \dfrac{d}{dx}\left[ \dfrac{4x^{3}}{2x^{5}-3x+4}\right] =4\dfrac{d}{dx}\left[ \dfrac{x^{3}}{2x^{5}-3x+4}\right] \end{equation*} If neither factor is a constant, use the Product Rule by writing down each factor and multiplying by the derivative of the other factor. \ Again, if the expressions are complicated, do not attempt to differentiate, just write a derivative symbol in front of each factor. \ For example, \begin{equation*} \dfrac{d}{dx}\left[ \left( 3x^{4}-2x\right) \left( x^{3}+5x^{2}\right) ^{5}% \right] =\left( 3x^{4}-2x\right) \dfrac{d}{dx}\left[ \left( x^{3}+5x^{2}\right) ^{5}\right] +\left( x^{3}+5x^{2}\right) ^{5}\dfrac{d}{dx}% \left( 3x^{4}-2x\right) \end{equation*} \item If the expression is basically a quotient, and the denominator is a constant, you really have a product of a constant fraction and a function, so use the use the Constant Multiplier Rule. \ If the numerator is a constant, use the Constant Multiplier and Reciprocal rules by writing down minus the constant divided by the square of the denominator and multiplied by the derivative of the denominator. If neither the numerator nor the denominator is a constant, use the Quotient Rule be writing down the numerator times the derivative of the denominator minus the denominator times the derivative of the numerator. \ For example, \begin{equation*} \dfrac{d}{dx}\left[ \dfrac{x^{3}}{2x^{5}-3x+4}\right] =\dfrac{\left( 2x^{5}-3x+4\right) 3x^{2}-x^{3}\dfrac{d}{dx}\left( 2x^{5}-3x+4\right) }{% \left( 2x^{5}-3x+4\right) ^{2}} \end{equation*} If the expression is a quotient, always think about converting it into a product by putting a negative power on the denominator. \ For example, \begin{eqnarray*} \dfrac{d}{dx}\left[ \dfrac{x^{3}}{2x^{5}-3x+4}\right] &=&\dfrac{d}{dx}\left[ x^{3}\left( 2x^{5}-3x+4\right) ^{-1}\right] \\ &=&x^{3}\dfrac{d}{dx}\left[ \left( 2x^{5}-3x+4\right) ^{-1}\right] +\left( 2x^{5}-3x+4\right) ^{-1}3x^{2} \end{eqnarray*} \item If the expression is basically a power, use the Power Rule or the Extended Power Rule.\ Again, if the expressions are complicated, just write a derivative symbol in front of each quantity. \ For example, \begin{equation*} \dfrac{d}{dx}\left[ \left( x^{3}+5x^{2}\right) ^{5}\right] =5\left( x^{3}+5x^{2}\right) ^{4}\dfrac{d}{dx}\left( x^{3}+5x^{2}\right) \end{equation*} \end{itemize} Repeat this process on each subexpression, perfoming one step on each subexpression on each line. \ The complete process for the above example would look like: \begin{eqnarray*} \dfrac{df}{dx} &=&\dfrac{d}{dx}\left[ \left( 3x^{4}-2x\right) \left( x^{3}+5x^{2}\right) ^{5}+\dfrac{4x^{3}}{2x^{5}-3x+4}\right] \\ &=&\dfrac{d}{dx}\left[ \left( 3x^{4}-2x\right) \left( x^{3}+5x^{2}\right) ^{5}\right] +\dfrac{d}{dx}\left[ \dfrac{4x^{3}}{2x^{5}-3x+4}\right] \\ &=&\left( 3x^{4}-2x\right) \dfrac{d}{dx}\left[ \left( x^{3}+5x^{2}\right) ^{5}\right] +\left( x^{3}+5x^{2}\right) ^{5}\dfrac{d}{dx}\left( 3x^{4}-2x\right) \\ &&+\dfrac{\left( 2x^{5}-3x+4\right) \dfrac{d}{dx}\left( 4x^{3}\right) -4x^{3}% \dfrac{d}{dx}\left( 2x^{5}-3x+4\right) }{\left( 2x^{5}-3x+4\right) ^{2}} \\ &=&\left( 3x^{4}-2x\right) 5\left( x^{3}+5x^{2}\right) ^{4}\left( 3x^{2}+10x\right) +\left( x^{3}+5x^{2}\right) ^{5}\left( 12x^{3}-2\right) \\ &&+\dfrac{\left( 2x^{5}-3x+4\right) \left( 12x^{2}\right) -4x^{3}\left( 10x^{4}-3\right) }{\left( 2x^{5}-3x+4\right) ^{2}} \end{eqnarray*} In each case below find the derivative: \vspace{1pt} 1. $y=\cos ^{3}\left( x^{4}+\sin 6x\right) $ Use the chain rule: \vspace{1pt}$y\prime =3\cos ^{2}\left( x^{4}+\sin 6x\right) \left( -\sin \left( x^{4}+\sin 6x\right) \right) \left( 4x^{3}+6\cos 6x\right) $ 2. $y=10^{\sec 3x}$ $\qquad \qquad \qquad \qquad \Longrightarrow \ln y=(\sec 3x)(\ln 10).$ $\ $% Now differentiate implicitly: $\qquad \qquad \qquad \qquad \dfrac{1}{y}\dfrac{dy}{dx}=(3\ln 10)(\sec 3x\tan 3x)\Longrightarrow \dfrac{dy}{dx}=(3\ln 10)(\sec 3x\tan 3x)(10^{\sec 3x})$ or $\qquad \qquad \qquad \qquad $If $y=a^{u},\qquad \dfrac{dy}{dx}=(a^{u}\ln a)% \dfrac{du}{dx}\Longrightarrow \dfrac{dy}{dx}=(3\ln 10)(\sec 3x\tan 3x)(10^{\sec 3x})$ \vspace{1pt} 3. $\ y=\dint_{x}^{5}\dfrac{e^{2t}}{3t^{2}+5t+4}dt$ $y=-\dint_{5}^{x}\dfrac{e^{2t}}{3t^{2}+5t+4}dt\rightarrow \dfrac{dy}{dx}% =-\left( \dfrac{e^{2x}}{3x^{2}+5x+4}\right) $ by the Fundamental Theorem of Calculus. \vspace{1pt} 4. $y=\dint_{3}^{\sin x}\dfrac{e^{t}}{2t^{2}+5}dt$ Let $u=\sin x.$ Then $y\left( u\right) =\dint_{3}^{u}\dfrac{e^{t}}{2t^{2}+5}% dt,$ and \begin{equation*} \dfrac{dy}{dx}=\dfrac{dy}{du}\dfrac{du}{dx} \end{equation*} By the Fundamental Theorem of Calculus \begin{equation*} \dfrac{dy}{du}=\dfrac{e^{u}}{2u^{2}+5}=\dfrac{e^{\sin x}}{2\sin ^{2}x+5} \end{equation*} Also \begin{equation*} \dfrac{du}{dx}=\cos x \end{equation*} Thus \begin{equation*} \dfrac{dy}{dx}=\cos x\left( \dfrac{e^{\sin x}}{2\sin ^{2}x+5}\right) \end{equation*} 5. $y=\arcsin x^{2}$ \begin{equation*} \dfrac{dy}{dx}=\allowbreak 2\dfrac{x}{\sqrt{\left( 1-x^{4}\right) }} \end{equation*} 6. $y=\dfrac{x}{x^{2}+\sec ^{2}(x)}$ \begin{equation*} \vspace{1pt}\dfrac{dy}{dx}=\dfrac{1(x^{2}+\sec ^{2}x)-(2x+2\sec x\sec x\tan x)x}{(x^{2}+\sec ^{2}x)^{2}} \end{equation*} \vspace{1pt} 7. $y=x^{\tan x}$ Take the natural log of both sides to get \begin{equation*} \ln y=\tan x\left( \ln x\right) \end{equation*} Now differentiate both sides implicitly. \begin{equation*} \dfrac{1}{y}y^{\prime }=\sec ^{2}x\left( \ln x\right) +\tan x\left( \dfrac{1% }{x}\right) \end{equation*} Solving for $y^{\prime }$ we have \begin{eqnarray*} y^{\prime } &=&y\left( \sec ^{2}x\left( \ln x\right) +\tan x\left( \dfrac{1}{% x}\right) \right) \\ &=&x^{\tan x}\left( \sec ^{2}x\left( \ln x\right) +\tan x\left( \dfrac{1}{x}% \right) \right) \end{eqnarray*} 8. $y=\cos ^{3}\left( x^{4}+\sin 6x\right) $ Use the chain rule: \vspace{1pt}% \begin{equation*} y\prime =3\cos ^{2}\left( x^{4}+\sin 6x\right) \left( -\sin \left( x^{4}+\sin 6x\right) \right) \left( 4x^{3}+6\cos 6x\right) \end{equation*} \vspace{1pt} 9. Find the derivative of$\ \ y=\sqrt[4]{x^{3}+\left( \dfrac{x+1}{x-1}% \right) ^{6}}.$ Do \textsl{not} simplify your result. \vspace{1pt}Think of this as \ $\left( x^{3}+\left( \dfrac{x+1}{x-1}\right) ^{6}\right) ^{\frac{1}{4}}$ and use the chain rule where \ $u=x^{3}+\left( \dfrac{x+1}{x-1}\right) ^{6}$ \vspace{1pt}then $\ \ y=\left( x^{3}+\left( \dfrac{x+1}{x-1}\right) ^{6}\right) ^{\frac{1}{4}}=u^{\frac{1}{4}}$ \ and \ $y^{\prime }=\frac{1}{4}% u^{-\frac{3}{4}}\cdot u^{\prime }.$ But \ \ $u^{\prime }$ \ can be found by using the sum rule and another application of the power and chain rule and use of the quotient rule: \vspace{1pt}$u^{\prime }=3x^{2}+6\left( \dfrac{x+1}{x-1}\right) ^{5}\cdot \left( \dfrac{1\cdot \left( x-1\right) -\left( x+1\right) \cdot 1}{\left( x-1\right) ^{2}}\right) $ Thus \begin{equation*} y^{\prime }=\frac{1}{4}\left( x^{3}+\left( \dfrac{x+1}{x-1}\right) ^{6}\right) ^{-\frac{3}{4}}\left[ 3x^{2}+6\left( \dfrac{x+1}{x-1}\right) ^{5}\cdot \left( \dfrac{1\cdot \left( x-1\right) -\left( x+1\right) \cdot 1}{% \left( x-1\right) ^{2}}\right) \right] \end{equation*} \vspace{1pt} 10. $y=\dfrac{e}{\sqrt{x^{3}+2x}}$ \vspace{1pt} Think \ \begin{equation*} \frac{d}{dx}\left( e\left( x^{3}+2x\right) ^{-\frac{1}{2}}\right) =e\left[ \frac{-1}{2}\left( x^{3}+2x\right) ^{-\frac{3}{2}}\left( 3x^{2}+2\right) % \right] \end{equation*} (Note: \ $e$ is a constant not a variable and not a function in the same way as $\pi $ is a constant: \ \ $e\approx \allowbreak 2.\,\allowbreak 718\,281\,828.$\ ) \vspace{1pt} 11. $y=\dfrac{\sin ^{2}\left( t^{3}+t\right) }{t^{3}+t}$ \vspace{1pt} This is a quotient rule with a chain rule with a power rule with a sum rule ... \vspace{1pt} \begin{equation*} \frac{d}{dx}\left( \dfrac{\sin ^{2}\left( t^{3}+t\right) }{t^{3}+t}\right) =% \frac{2\sin \left( t^{3}+t\right) \cdot \cos \left( t^{3}+t\right) \cdot \left( 3t^{2}+1\right) \left( t^{3}+t\right) -\sin ^{2}\left( t^{3}+t\right) \left( 3t^{2}+1\right) }{\left( t^{3}+t\right) ^{2}} \end{equation*} 12. $y=\left( \tan \left( x^{2}+2\right) \right) \cos \left( \dfrac{e}{\sqrt{% x^{3}+2x}}\right) $ \vspace{1pt}Product rule, chain rule ... \vspace{1pt} \begin{eqnarray*} \frac{d}{dx}\left[ \left( \tan \left( x^{2}+2\right) \right) \cos \left( \dfrac{e}{\sqrt{x^{3}+2x}}\right) \right] &=&\sec ^{2}\left( x^{2}+2\right) \cdot 2x\left( \cos \left( \dfrac{e}{\sqrt{x^{3}+2x}}\right) \right) \\ &&+\left( \tan \left( x^{2}+2\right) \right) \left[ -\sin \left( \dfrac{e}{% \sqrt{x^{3}+2x}}\right) \cdot \left\{ \frac{-e}{2}\left( x^{3}+2x\right) ^{-% \frac{3}{2}}\left( 3x^{2}+2\right) \right\} \right] \end{eqnarray*} \vspace{1pt} 13. Find the equation of the line tangent to $y=\sqrt{x^{2}+1}\left( \dfrac{1% }{2x-1}\right) ^{2}\left( x+4\right) ^{3}$ at $(0,64).$ \vspace{1pt} \vspace{1pt}First apply the above rule to find \ \begin{eqnarray*} y^{\prime } &=&\left[ \frac{1}{2}\left( x^{2}+1\right) ^{-\frac{1}{2}}\cdot 2x\right] \left( \dfrac{1}{2x-1}\right) ^{2}\left( x+4\right) ^{3}+\sqrt{% x^{2}+1}\left[ 2\left( \dfrac{1}{2x-1}\right) \cdot \left( -1\right) \left( 2x-1\right) ^{-2}\cdot 2\right] \left( x+4\right) ^{3} \\ &&\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }+\sqrt{x^{2}+1% }\left( \dfrac{1}{2x-1}\right) ^{2}\left[ 3\left( x+4\right) ^{2}\cdot 1% \right] \end{eqnarray*} When \ \ $x=0$ ,\ \ \ \ $y^{\prime }=\left[ \frac{1}{2}\left( 1\right) \cdot 2\left( 0\right) \right] \left( -1\right) \left( 4\right) ^{3}+\left( 1\right) \left[ 2\left( -1\right) \cdot \left( -1\right) \left( 1\right) \cdot 2\right] \left( 4\right) ^{3}+\left( 1\right) \left( 1\right) \left[ 3\left( 4\right) ^{2}\right] =\allowbreak 304$ \ is the slope of the tangent line. Thus at the point \ $\left( 0,64\right) $ \ the equation of the tangent line is \ \begin{equation*} \ y-64=304\left( x-0\right) \end{equation*} \qquad \qquad \qquad \qquad \qquad \vspace{1pt} \subsection{Integration} \begin{center} \begin{tabular}{l} \QTR{blue}{Table of Common Integrals}% \end{tabular} \vspace{1pt} \begin{tabular}{llll} \FRAME{itbpF}{0.2205in}{0.2205in}{0.0605in}{}{}{schoice1.bmp}{\special% {language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio TRUE;display "PICT";valid_file "F";width 0.2205in;height 0.2205in;depth 0.0605in;original-width 21.0625pt;original-height 21.0625pt;cropleft "0";croptop "1";cropright "1";cropbottom "0";filename '/document/graphics/Schoice1.bmp';file-properties "XNPEU";}} & $\int u^{n}\,du=\dfrac{u^{n+1}}{n+1}+C,\;n\neq -1$ & \FRAME{itbpF}{0.2205in}{% 0.2205in}{0.0605in}{}{}{schoice1.bmp}{\special{language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio TRUE;display "PICT";valid_file "F";width 0.2205in;height 0.2205in;depth 0.0605in;original-width 21.0625pt;original-height 21.0625pt;cropleft "0";croptop "1";cropright "1";cropbottom "0";filename '/document/graphics/Schoice1.bmp';file-properties "XNPEU";}} & $\int u^{-1}\,du=\ln \left| u\right| +C$ \\ \FRAME{itbpF}{0.2205in}{0.2205in}{0.0605in}{}{}{schoice1.bmp}{\special% {language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio TRUE;display "PICT";valid_file "F";width 0.2205in;height 0.2205in;depth 0.0605in;original-width 21.0625pt;original-height 21.0625pt;cropleft "0";croptop "1";cropright "1";cropbottom "0";filename '/document/graphics/Schoice1.bmp';file-properties "XNPEU";}} & $\int e^{u}\,du=e^{u}+C$ & \FRAME{itbpF}{0.2205in}{0.2205in}{0.0605in}{}{}{% schoice1.bmp}{\special{language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio TRUE;display "PICT";valid_file "F";width 0.2205in;height 0.2205in;depth 0.0605in;original-width 21.0625pt;original-height 21.0625pt;cropleft "0";croptop "1";cropright "1";cropbottom "0";filename '/document/graphics/Schoice1.bmp';file-properties "XNPEU";}} & $\int a^{u}\,du=\dfrac{a^{u}}{\ln a}+C$% \end{tabular} \medskip \medskip \QTR{blue}{Table of Common Trigonometric Integrals} \begin{tabular}{llll} \FRAME{itbpF}{0.2205in}{0.2205in}{0.0605in}{}{}{schoice1.bmp}{\special% {language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio TRUE;display "PICT";valid_file "F";width 0.2205in;height 0.2205in;depth 0.0605in;original-width 21.0625pt;original-height 21.0625pt;cropleft "0";croptop "1";cropright "1";cropbottom "0";filename '/document/graphics/Schoice1.bmp';file-properties "XNPEU";}} & $\int \cos u\ du=\sin u+C$ & \FRAME{itbpF}{0.2205in}{0.2205in}{0.0605in}{}{}{% schoice1.bmp}{\special{language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio TRUE;display "PICT";valid_file "F";width 0.2205in;height 0.2205in;depth 0.0605in;original-width 21.0625pt;original-height 21.0625pt;cropleft "0";croptop "1";cropright "1";cropbottom "0";filename '/document/graphics/Schoice1.bmp';file-properties "XNPEU";}} & $\int \dfrac{1}{\sqrt{1-u^{2}}}\,du=\sin ^{-1}u+C$ \\ \FRAME{itbpF}{0.2205in}{0.2205in}{0.0605in}{}{}{schoice1.bmp}{\special% {language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio TRUE;display "PICT";valid_file "F";width 0.2205in;height 0.2205in;depth 0.0605in;original-width 21.0625pt;original-height 21.0625pt;cropleft "0";croptop "1";cropright "1";cropbottom "0";filename '/document/graphics/Schoice1.bmp';file-properties "XNPEU";}} & $\int \sin u\ du=-\cos u+C$ & \FRAME{itbpF}{0.2205in}{0.2205in}{0.0605in}{}{}{% schoice1.bmp}{\special{language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio TRUE;display "PICT";valid_file "F";width 0.2205in;height 0.2205in;depth 0.0605in;original-width 21.0625pt;original-height 21.0625pt;cropleft "0";croptop "1";cropright "1";cropbottom "0";filename '/document/graphics/Schoice1.bmp';file-properties "XNPEU";}} & $\int \dfrac{1}{1+u^{2}}\,du=\tan ^{-1}u+C$ \\ \FRAME{itbpF}{0.2205in}{0.2205in}{0.0605in}{}{}{schoice1.bmp}{\special% {language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio TRUE;display "PICT";valid_file "F";width 0.2205in;height 0.2205in;depth 0.0605in;original-width 21.0625pt;original-height 21.0625pt;cropleft "0";croptop "1";cropright "1";cropbottom "0";filename '/document/graphics/Schoice1.bmp';file-properties "XNPEU";}} & $\int \sec ^{2}u\ du=\tan u+C$ & \FRAME{itbpF}{0.2205in}{0.2205in}{0.0605in}{}{}{% schoice1.bmp}{\special{language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio TRUE;display "PICT";valid_file "F";width 0.2205in;height 0.2205in;depth 0.0605in;original-width 21.0625pt;original-height 21.0625pt;cropleft "0";croptop "1";cropright "1";cropbottom "0";filename '/document/graphics/Schoice1.bmp';file-properties "XNPEU";}} & $\int \dfrac{1}{u\sqrt{1-u^{2}}}du=\sec ^{-1}u+C$ \\ \FRAME{itbpF}{0.2205in}{0.2205in}{0.0605in}{}{}{schoice1.bmp}{\special% {language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio TRUE;display "PICT";valid_file "F";width 0.2205in;height 0.2205in;depth 0.0605in;original-width 21.0625pt;original-height 21.0625pt;cropleft "0";croptop "1";cropright "1";cropbottom "0";filename '/document/graphics/Schoice1.bmp';file-properties "XNPEU";}} & $\int \csc ^{2}u\ du=-\cot u+C$ & \FRAME{itbpF}{0.2205in}{0.2205in}{0.0605in}{}{}{% schoice1.bmp}{\special{language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio TRUE;display "PICT";valid_file "F";width 0.2205in;height 0.2205in;depth 0.0605in;original-width 21.0625pt;original-height 21.0625pt;cropleft "0";croptop "1";cropright "1";cropbottom "0";filename '/document/graphics/Schoice1.bmp';file-properties "XNPEU";}} & $\int \sec u\tan u\ du=\sec u+C$ \\ & & \FRAME{itbpF}{0.2205in}{0.2205in}{0.0605in}{}{}{schoice1.bmp}{\special% {language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio TRUE;display "PICT";valid_file "F";width 0.2205in;height 0.2205in;depth 0.0605in;original-width 21.0625pt;original-height 21.0625pt;cropleft "0";croptop "1";cropright "1";cropbottom "0";filename '/document/graphics/Schoice1.bmp';file-properties "XNPEU";}} & $\int \csc u\cot u\ du=-\csc u+C$ \\ & & & \end{tabular} \end{center} \noindent Recall also the Constant Multiplier and Sum Rules, \begin{center} \vspace{1pt} \begin{tabular}{lll} \FRAME{itbpF}{0.2205in}{0.2205in}{0.0605in}{}{}{schoice1.bmp}{\special% {language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio TRUE;display "PICT";valid_file "F";width 0.2205in;height 0.2205in;depth 0.0605in;original-width 21.0625pt;original-height 21.0625pt;cropleft "0";croptop "1";cropright "1";cropbottom "0";filename '/document/graphics/Schoice1.bmp';file-properties "XNPEU";}} & $\int kf\,(u)du=k\int f\,(u)du$ & Constant Multiplier Rule \\ \FRAME{itbpF}{0.2205in}{0.2205in}{0.0605in}{}{}{schoice1.bmp}{\special% {language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio TRUE;display "PICT";valid_file "F";width 0.2205in;height 0.2205in;depth 0.0605in;original-width 21.0625pt;original-height 21.0625pt;cropleft "0";croptop "1";cropright "1";cropbottom "0";filename '/document/graphics/Schoice1.bmp';file-properties "XNPEU";}} & $\int \lbrack f\,(u)\pm g(u)\rbrack du=\int f\,(u)du\pm \int \,g(u)du$ & $\text{% Sum Rule}$% \end{tabular} \end{center} \vspace{1pt}which will be used without comment. \FRAME{dtbpF}{4.2384in}{0.0692in}{0pt}{}{}{maroon.wmf}{\special{language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio TRUE;display "PICT";valid_file "F";width 4.2384in;height 0.0692in;depth 0pt;original-width 290.5625pt;original-height 1pt;cropleft "0";croptop "0.9967";cropright "1";cropbottom "0.3966";filename '/document/graphics/maroon.wmf';file-properties "XNPEU";}} \section{Integration by Substitution} \subsection{The Basic Method} Let $F(u)$ be an antiderivative of the function $f(u)$. This means \begin{equation*} F^{\prime }(u)=f(u)\quad \text{and}\quad \int f(u)du=F(u)+C \end{equation*} Now suppose that $u=g(x)$. Then the Chain Rule for differentiation gives \begin{equation*} \frac{d}{dx}\lbrack F(g(x))\rbrack =F^{\prime }(g(x))g^{\prime }(x)=f\,(g(x))g^{\prime }(x) \end{equation*} Writing this in integral form, we obtain \begin{equation*} \int f\,(g(x))g^{\prime }(x)dx=F(g(x))+C=F(u)+C \end{equation*} \vspace{1pt}\FRAME{dtbpF}{4.2384in}{0.0692in}{0pt}{}{}{maroon.wmf}{\special% {language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio TRUE;display "PICT";valid_file "F";width 4.2384in;height 0.0692in;depth 0pt;original-width 290.5625pt;original-height 1pt;cropleft "0";croptop "0.9967";cropright "1";cropbottom "0.3966";filename '/document/graphics/maroon.wmf';file-properties "XNPEU";}} That is, we have the \emph{method of substitution\/} \begin{quotation} \begin{equation*} \int f\,(g(x))g^{\prime }(x)dx=\int f\,(u)du \end{equation*}% where $u=g(x)$ and $du=g^{\prime }(x)dx$ \end{quotation} \subsection{Integration by Parts} \begin{equation*} \int udv=uv-\int vdu \end{equation*} \vspace{1pt} \subsubsection{Examples:} 1. $\int \cos ^{2}12xdx\qquad $ Note: $\cos ^{2}t=\allowbreak \frac{1}{2}\cos 2t+\frac{1}{2}$ and $\sin ^{2}t=\allowbreak \frac{1}{2}-\frac{1}{2}\cos 2t.$ Thus $\qquad \qquad \qquad \qquad $% \begin{eqnarray*} \int \cos ^{2}12xdx &=&\dfrac{1}{2}\int \left( 1+\cos (2\ast 12x)\right) dx \\ &=&\dfrac{1}{2}\int \left( 1+\cos (24x)\right) dx \\ &=&\dfrac{1}{2}\left( x+\dfrac{\sin 24x}{24}\right) +C \end{eqnarray*} 2. $\int \cos ^{5}xdx$ \vspace{1pt}\qquad \qquad \qquad \qquad \qquad $\int \cos ^{5}xdx=\int (1-\sin ^{2}x)^{2}\cos xdx.$ \ Let $u=\sin x,du=\cos xdx.$ \vspace{1pt}\qquad \qquad \qquad \qquad $\Longrightarrow \int (1-u^{2})^{2}du=\int \left( 1-2u^{2}+u^{4}\right) du=u-\dfrac{2u^{3}}{3}+% \dfrac{u^{5}}{5}+C$ \qquad \qquad \qquad \qquad $\Longrightarrow \int \cos ^{5}xdx=\sin x-\dfrac{% 2\sin ^{3}x}{3}+\dfrac{\sin ^{5}x}{5}+C\qquad $ \vspace{1pt} 3. $\dint_{0}^{1}\dfrac{x^{4}+4x^{3}+2}{\left( x^{5}+5x^{4}+10x+1\right) ^{% \tfrac{4}{7}}}dx$ \noindent Let $u=x^{5}+5x^{4}+10x+1,du=5x^{4}+20x^{3}+10.\Longrightarrow \dfrac{du}{5}=\left( x^{4}+4x^{3}+2\right) dx$ $\qquad \qquad \qquad \Longrightarrow \dint \dfrac{x^{4}+4x^{3}+2}{\left( x^{5}+5x^{4}+10x+1\right) ^{\frac{4}{7}}}dx=\dfrac{1}{5}\int u^{\tfrac{-4}{7}% }du=\dfrac{1}{5}\left( \dfrac{7}{3}u^{\tfrac{3}{7}}\right) +C$ $\dint_{0}^{1}\dfrac{x^{4}+4x^{3}+2}{\left( x^{5}+5x^{4}+10x+1\right) ^{% \tfrac{4}{7}}}dx=\dfrac{7}{15}\left. \left( x^{5}+5x^{4}+10x+1\right) ^{% \tfrac{3}{7}}\right| _{0}^{1}=\dfrac{7}{15}\left( \left( 17\right) ^{\tfrac{3% }{7}}-1\right) $ \vspace{1pt} 4. $\dint \left( \dfrac{x^{2}}{\sqrt{3x-5}}\right) dx$ Use substitution. Let $u=3x-5,$ so $x=\dfrac{u+5}{3}.$ Then $dx=\dfrac{du}{3} $ \vspace{1pt} $\dint \left( \dfrac{x^{2}}{\sqrt{3x-5}}\right) dx=\dint \dfrac{\left( \dfrac{u+5}{3}\right) ^{2}}{u^{\frac{1}{2}}}\dfrac{du}{3}=\dfrac{1}{3}\dint \dfrac{\left( \dfrac{u+5}{3}\right) ^{2}}{u^{\frac{1}{2}}}du=\dfrac{1}{27}% \dint \dfrac{\left( u+5\right) ^{2}}{u^{\frac{1}{2}}}du$ $=\dfrac{1}{27}% \dint \left( u^{\frac{3}{2}}+10\sqrt{u}+\dfrac{25}{\sqrt{u}}\right) du$ $\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad =\allowbreak \frac{2}{135}u^{\frac{5}{2}}+\frac{20}{81}u^{\frac{3}{2}}+\frac{50}{27}\sqrt{% u}=\allowbreak \frac{2}{135}\left( 3x-5\right) ^{\frac{5}{2}}+\frac{20}{81}% \left( 3x-5\right) ^{\frac{3}{2}}+\frac{50}{27}\sqrt{\left( 3x-5\right) }% \allowbreak $ $+C$ \vspace{1pt} 5. $\int (\sec ^{2}3x)5^{\tan 3x}dx$ \vspace{1pt}Let \ $u=\tan 3x,du=3\sec ^{2}3xdx\Longrightarrow \dfrac{1}{3}% du=\sec ^{2}3xdx.$ Thus $\int (\sec ^{2}3x)5^{\tan 3x}dx=\dfrac{1}{3}$\vspace{1pt}$\int 5^{u}du=\left( \dfrac{1}{3}\right) \left( \dfrac{5^{u}}{\ln 5}\right) +C=\left( \dfrac{1}{3}\right) \dfrac{5^{\tan 3x}}{\ln 5}+C$ \vspace{1pt} 6. $\int_{0}^{\tfrac{\pi }{24}}\tfrac{1}{2}(1-\cos 6x)dx=\left[ \tfrac{1}{2}% x-\tfrac{1}{12}\sin 6x\right] _{0}^{\tfrac{\pi }{24}}=\left[ \left( \dfrac{% \pi }{48}-\dfrac{1}{12}\sin \tfrac{\pi }{4}\right) -\left( 0-0\right) \right] $ $=\left[ \dfrac{\pi }{48}-\dfrac{1}{12}\dfrac{\sqrt{2}}{2}\right] =\frac{1}{% 48}\pi -\frac{1}{24}\sqrt{2}$ \vspace{1pt} 7. $\dint \dfrac{x^{5}+x^{2}}{x^{6}+2x^{3}+10}dx$ \ Let $u=x^{6}+2x^{3}+10;$ then $du=(6x^{5}+6x^{2})dx$ \ \ \ $\rightarrow (x^{5}+x^{2})dx=\dfrac{1}{6}du\rightarrow \dint \dfrac{% x^{5}+x^{2}}{x^{6}+2x^{3}+10}dx=\dfrac{1}{6}\int \dfrac{du}{u}=\dfrac{1}{6}% \ln \left| u\right| +C=\dfrac{1}{6}\ln \left| x^{6}+2x^{3}+10\right| +C$ \vspace{1pt} 8. $\dint \dfrac{x^{2}}{x^{6}+4x^{3}+13}dx$ $\dint \dfrac{x^{2}}{x^{6}+4x^{3}+13}dx=\dint \dfrac{x^{2}}{(x^{3}+2)^{2}+9}% dx=\dint \left[ \dfrac{\dfrac{x^{2}}{9}}{\left( \dfrac{x^{3}+2}{3}\right) ^{2}+1}\right] dx=\dfrac{1}{9}\dint \dfrac{x^{2}}{\left( \dfrac{x^{3}+2}{3}% \right) ^{2}+1}dx$ \ \ $\rightarrow $ let $u=\dfrac{1}{3}(x^{3}+2);$ then $du=x^{2}dx% \rightarrow \dfrac{1}{9}\int \dfrac{du}{u^{2}+1}$ $=\dfrac{1}{9}\arctan (u)+C $ Thus $\dint \dfrac{x^{2}}{x^{6}+4x^{3}+13}dx=\dfrac{1}{9}\arctan \left( \dfrac{1}{3}(x^{3}+2)\right) +C$ \vspace{1pt} 9. $\int \tan 3xdx$ \ Let $u=3x;$ so $du=\tfrac{1}{3}dx$ $\rightarrow \tfrac{1}{3}\int \tan udu=\dfrac{1}{3}\dint \dfrac{\sin u}{\cos u}du=-\dfrac{1}{3}\ln \left| \cos u\right| +C=\tfrac{1}{3}\ln \left| \sec u\right| +C$ $\qquad \qquad \qquad \qquad \qquad \qquad \qquad =\tfrac{1}{3}\ln \left| \sec 3x\right| +C$ \vspace{1pt} 10. $\dint \dfrac{dy}{4-y^{2}}=\allowbreak $ \begin{equation} \dfrac{1}{4-y^{2}}=\dfrac{1}{\left( 2-y\right) \left( 2+y\right) }=\dfrac{A}{% 2-y}+\dfrac{B}{2+y} \tag{$\left( \ast \right) $} \end{equation} Multiply both sides by $2-y$ \vspace{1pt} \begin{equation*} \dfrac{1}{2+y}=A+\dfrac{B}{2+y}\left( 2-y\right) \end{equation*} Setting $y=+2$ gives $A=\dfrac{1}{4}.$ Similarly, multiplying both sides of $% \left( \ast \right) $ by $2+y$ leads to \begin{equation*} \dfrac{1}{2-y}=\dfrac{\frac{1}{4}}{2-y}\left( 2+y\right) +B \end{equation*} Setting $y=-2$ tells us that $B=\dfrac{1}{4}.$ Thus \begin{equation*} \dfrac{1}{4-y^{2}}=\allowbreak \dfrac{1}{4\left( 2-y\right) }+\dfrac{1}{% 4\left( 2+y\right) } \end{equation*} so \vspace{1pt}$\dint \dfrac{dy}{4-y^{2}}=\int \dfrac{1}{4\left( 2-y\right) }% +\int \dfrac{1}{4\left( 2+y\right) }=\allowbreak \frac{1}{4}\ln \left| 2-y\right| +\frac{1}{4}\ln \left| y+2\right| +C$ \vspace{1pt} 11. $\dint \dfrac{1+3x^{2}}{\sqrt{x^{3}+x+1}}dx$ \ $\qquad $ Let $u=x^{3}+x+1;$ then $du=(3x^{2}+1)dx$ $\rightarrow \dint \dfrac{1+3x^{2}}{\sqrt{x^{3}+x+1}}dx=\dint \dfrac{du}{% \sqrt{u}}=2u^{\tfrac{1}{2}}+C=2(x^{3}+x+1)^{\tfrac{1}{2}}+C\qquad $ $\vspace{1pt}$ 12. $\dint \dfrac{2}{3e^{5x}}dx$ Solution: \qquad \qquad $\dint \dfrac{2}{3e^{5x}}dx=\dfrac{2}{3}\int e^{-5x}dx=\dfrac{2% }{3}\left( \dfrac{e^{-5x}}{-5}\right) +C=-\frac{2}{15}e^{-5x}+C$ \vspace{1pt} 13. $\int \left( x-1\right) \ln xdx$ Solution: Use integration by parts: \ $u=\ln x\qquad dv=x-1,$ then \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \begin{equation*} du=\dfrac{1}{x}dx\qquad v=\dfrac{x^{2}}{2}-x \end{equation*} $\int udv=uv-vdu$ $\Longrightarrow \int \left( x-1\right) \ln xdx=\left( \dfrac{x^{2}}{2}% -x\right) \ln x-\int \left( \frac{x^{2}}{2}-x\right) \frac{1}{x}dx=\left( \dfrac{x^{2}}{2}-x\right) \ln x-\int \left( \dfrac{x}{2}-1\right) dx$ $=\left( \frac{x^{2}}{2}-x\right) \ln x-\frac{x^{2}}{4}+x+C.$ \vspace{1pt} 14. $\dint \dfrac{5x-2}{x^{2}+4}dx$ \ \ \ \ \ \ \ \ \ \ \vspace{1pt}\qquad Solution: \qquad \qquad \qquad \begin{equation*} \dint \dfrac{5x-2}{x^{2}+4}dx=\int \dfrac{5x}{x^{2}+4}dx-\int \dfrac{2}{% x^{2}+4}dx. \end{equation*} Now:\qquad $\dint \dfrac{5x}{x^{2}+4}dx.\qquad $Let \ $u=x^{2}+4,du=2xdx% \Longrightarrow \dfrac{du}{2}=xdx$ Then\qquad $\int \dfrac{5x}{x^{2}+4}dx=\dfrac{5}{2}\int \dfrac{1}{u}du=% \dfrac{5}{2}\ln u=\dfrac{5}{2}\ln \left( x^{2}+4\right) $ Next:\qquad $\dint \dfrac{2}{x^{2}+4}dx=\int \dfrac{2}{4\left( \left( \frac{x% }{2}\right) ^{2}+1\right) }dx.$ \ \ Now let \ $u=\dfrac{x}{2},du=\frac{1}{2}% dx\Longrightarrow 2du=dx.$ Then\qquad $\dint \dfrac{2}{4\left( \left( \frac{x}{2}\right) ^{2}+1\right) }% dx$ \ becomes \ $\dfrac{2}{4}\dint \dfrac{2}{u^{2}+1}du=\dint \dfrac{1}{% u^{2}+1}du=\arctan (u)=\arctan (\frac{x}{2}).$ \vspace{1pt} Thus\qquad $\dint \dfrac{5x-2}{x^{2}+4}dx$ =$\dfrac{5}{2}\ln \left( x^{2}+4\right) +\arctan (\dfrac{x}{2})+C.$\qquad \vspace{1pt} $\qquad \qquad \qquad \qquad \qquad \qquad \qquad $ \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \end{document} %%%%%%%%%%%%%%%%%%% End /document/rev_bdiff_bint.tex %%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%% Start /document/graphics/maroon.wmf %%%%%%%%%%%%%%%%% 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