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Examples 57
In the previous Examples 3 and 4, the way we specified $D$ and $R$
suggested how to write the integral as a iterated integral.
Sometimes conditions are best interpreted graphically before
deciding on whether to evaluate as Type I or Type II. 
Solution 5: $D$ is enclosed by the straight line $y = x$ and the parabola $y = x^2$ as shown here. To determine the limits of integration we first need to find the points of intersection of $y = x$ and $y = x^2$. These occur when $x^2 = x$. This means $x(x1)=0$, so $x = 0,\, 1$. Treating $D$ as a Type I region, we fix $x$ between $x=0$ and $x=1$, and integrate with respect to $y$ along the black vertical line, getting the iterated integral $\displaystyle I \ = \ \int_0^1\left(\int_{x^2}^{x}\, (3x + 4y)\, dy\right)\,dx$ DO: Evaluate $I$. 

Example 6: The region $D$ that
is shown here is Type I but is not Type II: Fixing $x$
and integrating first with respect to $y$ along the vertical
black line makes good sense because then we have the same
curve $f_1(x)$ along the bottom and $f_2(x)$ along the top:
$$ D \ = \ \Bigl\{\,(x,\,y) : f_1(x) \le y \le f_2(x),\ \ a
\le x \le b\,\Bigl\}$$ for suitable choices of $a,\, b$ and
functions $f_1(x),\, f_2(x)$ giving us: $$ \iint_D\,
f(x,\,y)\, dA = \int_a^b \left(\int_{f_1(x)}^{f_2(x)}\,
f(x,\,y)\, dy\right) dx\,.$$But if we had chosen to fix $y$,
then the integral with respect to $x$ would sometimes split
into two parts, as is shown with the red horizontal
lines. This would make evaluate of this integral more
complicated  for one thing, there would be more than one
integral. 

Example 7: Similarly, the region $D$ shown here is Type II but not Type I. Fixing $y$ and integrating first with respect to $x$ along the horizontal black line makes good sense because then $$ D \ = \ \Bigl\{\,(x,\,y) : g_1(y) \le x \le g_2(y),\ \ c \le y \le d\,\Bigl\}$$ for suitable choices of $c,\, d$ and functions $g_1(y),\, g_2(y)$. In this case $$ \iint_D\, f(x,\,y)\, dxdy = \int_c^d \left(\int_{g_1(y)}^{g_2(y)}\, f(x,\,y)\, dx\right) dy\,.$$But if we had chosen to fix $x$, then the integral with respect to $y$ would sometimes splits into two parts as shown by the red vertical lines. Again, this would make the integral(s) more complicated. 