\documentclass[a4paper,12pt]{article}
\usepackage{epsfig}
\newcommand{\un}{\underline}
\newcommand{\ds}{\displaystyle}
\begin{document}
\parindent=0pt

\begin{center}
\textbf{Multiple Integration}

\textit{\textbf{Iteration of Double Integrals}}
\end{center}

\textbf{Question}

Suppose $F'(x)=f(x)$ and $G'(x)=g(x)$ on the interval $a \le x \le b$.

$T$ is the triangle defined by the vertices $(a,a)$, $(b,a)$ and
$(b,b)$.

Iterate $\int\!\!\!\int_T f(x)g(x) \,dA$ in both directions to show
that
\begin{eqnarray*}
\int_a^b f(x)G(x) \,dx & = & F(b)G(b) - F(a)G(a)\\
& & - \int_a^b g(y)F(y) \,dy
\end{eqnarray*}


\textbf{Answer}

$F'(x)=f(x)$ and $G'(x)=g(x)$ on $a \le x \le b$
\begin{eqnarray*}
\Rightarrow \int\!\!\!\int_T f(x)g(x) \,dA & = & \int_a^b f(x) \,dx
\int_a^x G'(y) \,dy\\
& = & \int_a^b f(x) (G(x)-G(a)) \,dx\\
& = & \int_a^b f(x)G(x) \,dx - G(a)F(b)+G(a)F(a)\\
\int\!\!\!\int_T f(x)g(x) \,dA & = & \int_a^b g(y) \,dt \int_y^b F'(x)
\,dx\\
& = & \int_a^b g(y)(F(b)-F(y)) \,dy\\
& = & F(b)G(b) - F(b)G(a) - \int_a^b F(y)g(y) \,dx
\end{eqnarray*}
\begin{eqnarray*}
\Rightarrow \Rightarrow \int_a^b f(x)G(x) \,dx & = & F(b)G(b) -
F(a)G(a)\\
& & - \int_a^b g(y)F(y) \,dy
\end{eqnarray*} 

\end{document}











