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5.6 Solution to Example 5.5

In the first case, the inner integral is

\int_{0}^{1}xy^{2}\,dy=\,{\left[\frac{xy^{3}}{3}\right]_{0}^{1}=}\,{\frac{x}{3% }\left(1^{3}-0^{3}\right)=}\,{\frac{x}{3}.}

Thus the outer integral is ${\int_{1}^{2}\frac{x}{3}\,dx=}\,{\left[\frac{x^{2}}{6}\right]_{1}^{2}=}\,{% \frac{2^{2}-1^{2}}{6}=}\,{\frac{1}{2}.}$

In the second case, the inner integral is

\int_{1}^{2}xy^{2}\,dx=\,{\left[\frac{x^{2}y^{2}}{2}\right]_{1}^{2}=}\,{\frac{% 3y^{2}}{2}.}

So the outer integral is ${\int_{0}^{1}\frac{3y^{2}}{2}\,dy=}\,{\left[\frac{y^{3}}{2}\right]_{0}^{1}=}\,% {\frac{1}{2}.}$