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%TCIDATA{<META NAME="Title" CONTENT="Substitution">}
%TCIDATA{Created=Mon Aug 19 14:52:24 1996}
%TCIDATA{LastRevised=Thu Feb 13 17:12:43 1997}
%TCIDATA{Language=American English}
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\begin{document}


\section{Substitution 
\index{Substitution@Substitution}\label{Substitution}}

To substitute a number or new expression for a variable, enclose an
expression in square brackets, enter an assignment for the variable in a
subscript, and choose \textsf{Evaluate}.

\begin{quote}
$\blacktriangleright $ \textsf{Evaluate}
\end{quote}

\begin{quotation}
$\left[ x^{2}+2x-3\right] _{x=a}=\allowbreak a^{2}+2a-3%
\vspace{6pt}$

$\left[ x^{2}+2x-3\right] _{x=5}=\allowbreak 32\vspace{6pt}$

$\left[ \frac{x^{2}-3x}{5}\right] _{x=y-z}=\allowbreak \frac{1}{5}\left(
y-z\right) ^{2}-\frac{3}{5}y+\frac{3}{5}z$
\end{quotation}

\strut

The expression in the subscript is an assignment for the variable on the
left of the equals sign. Notice that, in particular, $x=y+z$, $y=x-z$ and $%
z=x-y$ are not equivalent assignments.

\begin{quote}
$\blacktriangleright $ \textsf{Evaluate}
\end{quote}

\begin{quotation}
$\left[ x+y\right] _{x=y+z}\vspace{6pt}=\allowbreak 2y+z$

$\left[ x+y\right] _{y=x-z}\vspace{6pt}=\allowbreak 2x-z$

$\left[ x+y\right] _{z=x-y}=\allowbreak x+y$
\end{quotation}

\strut

To substitute two numbers or new expressions for a variable and find the
difference, enclose an expression in square brackets 
\[
\left[ x\right] _{x=a}^{x=b} 
\]
enter the numbers or new expressions in a subscript and superscript, and
choose \textsf{Evaluate}.

\begin{quote}
$\blacktriangleright $ \textsf{Evaluate}
\end{quote}

\begin{quotation}
$\left[ x\right] _{x=a}^{x=b}=\allowbreak b-a\vspace{4pt}$

$\left[ x^{2}+2x-3\right] _{x=a}^{x=b}=\allowbreak b^{2}+2b-a^{2}-2a\vspace{%
4pt}$

$\left[ x^{2}+2x-3\right] _{x=3}^{x=5}=\allowbreak 20$
\end{quotation}

\begin{quote}
$\blacktriangleright $ \textsf{Evaluate, Factor}
\end{quote}

\begin{quotation}
$\left[ \frac{x^{2}-3x}{5}\right] _{x=y-z}^{x=y+z}=\allowbreak \frac{1}{5}%
\left( y+z\right) ^{2}-\frac{6}{5}z-\frac{1}{5}\left( y-z\right)
^{2}=\allowbreak \frac{2}{5}z\left( 2y-3\right) $
\end{quotation}

\strut

From the expanding brackets panel, you can choose a nonprinting dashed
vertical line for the left bracket; and a square bracket or vertical line
for the right bracket. You can also enter from the keyboard a right vertical
line only, with subscript, or with both subscript and superscript.

\begin{quote}
$\blacktriangleright $ \textsf{Evaluate}
\end{quote}

\begin{quotation}
$\left[ x^{2}-3\right] _{x=a}^{x=b}=\allowbreak b^{2}-a^{2}\vspace{6pt}$

$\left. x^{2}-3\right] _{x=a}^{x=b}=b^{2}-a^{2}\vspace{6pt}$

$\left. x^{2}-3\right| _{x=a}^{x=b}=\allowbreak b^{2}-a^{2}\vspace{6pt}$

$x+3|_{x=y+z}=\allowbreak y+z+3\vspace{6pt}$

$x+3|_{x=y+z}^{x=u-v}=\allowbreak u-v-y-z$
\end{quotation}

\strut

\subsection{\ Related topics}

\begin{itemize}
\item  \hyperref{Fundamental Theorem of Calculus}{}{}{DM7-9.tex#Fundamental
theorem}

\item  \hyperref{Exercises and Solutions}{}{}{DM3-6.tex#Exercises}
\end{itemize}

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