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\textbf{Multiple Integration}

\textit{\textbf{Iteration of Double Integrals}}
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\textbf{Question}

The functions $f(x,t)$ and $f_1(x,t)$ are continuous on the rectangle
$a \le x \le b$, $c \le t \le d$.
\begin{eqnarray*}
g(x) & = & \int_c^d f(x,t)\,dt\\
G(x) & = & \int_c^d f_1(x,t) \,dt
\end{eqnarray*}

Show that $g'(x)=G(x)$ if $a < x < b$.

[Hint: evaluate $\int_a^x G(u) \,du$ by reversing the order of
iteration. Then differentiate the result.]


\textbf{Answer}

\begin{eqnarray*}
\int_a^x G(u) \,du & = & \int_a^x \,du \int_c^d f_1(u,t) dt\\
& = & \int_c^d \,dt \int_a^x f_1(u,t) \,du\\
& = & \int_c^d (f(x,t) - f(a,t)) \,dt \\
& = & g(x)-C
\end{eqnarray*}
With $C = \ds \int_c^d f(a,t) \,dt$ being independent of $x$. If we
apply the fundamental theorem of Calculus then it can be seen that
$$g'(x)=\frac{d}{dx} \int_a^x G(u) \,du = G(x)$$

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