SOLUTION (a) Regarding \(x\) as a constant, we obtain \[
\int_1^2 x^2y dy = \left[ x^2 \frac{y^2}{2} \right]_{y=1}^{y=2} =
x^2\left(\frac{2^2}{2}\right) - x^2\left(\frac{1^2}{2}\right) =
\frac{3}{2}x^2
\] Thus the function \(A\) in
the preceding discussion is given by \(A(x) =
\frac{3}{2}x^2\) in this example. We now integrate this function
of \(x\) from 0 to 3: \[
\int_0^3 \int_1^2 x^2y dy dx = \int_0^3 \left[ \int_1^2 x^2y dy \right]
dx = \int_0^3 \frac{3}{2}x^2 dx = \frac{3}{2} \left[ \frac{x^3}{3}
\right]_0^3 = \frac{3}{2} \frac{27}{3} = \frac{27}{2}
\] (b) Here we first integrate with respect to \(x\): \[
\int_1^2 \int_0^3 x^2y dx dy = \int_1^2 \left[ \int_0^3 x^2y dx \right]
dy = \int_1^2 \left[ \frac{x^3}{3}y \right]_{x=0}^{x=3} dy = \int_1^2 9y
dy = 9 \left[ \frac{y^2}{2} \right]_1^2 = \frac{27}{2}
\]
Notice that in Example 4 we obtained the same answer whether we
integrated with respect to \(y\) or
\(x\) first. In general, it turns out
(see Theorem 10) that the two iterated integrals in Equations (\(\ref{eq:8}\)) and (\(\ref{eq:9}\)) are always equal; that is,
the order of integration does not matter.
