Answer
$32(e^{4}-1)\approx 1715.14$
Work Step by Step
Apply Formula (5)
$\displaystyle \iint_{R}g(x)h(x)dA=\int_{a}^{b}g(x)dx\int_{c}^{d}h(y)dy, $ where $R=[a,b]\times[c,d]$.
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$\displaystyle \int_{0}^{2}\int_{0}^{4}y^{3}e^{2x}dy dx=\displaystyle \int_{0}^{2}e^{2x}dx\int_{0}^{4}y^{3}dy$
$=[\displaystyle \frac{1}{2}e^{2x}]_{0}^{2}\ \cdot\ [\frac{y^{4}}{4}]_{0}^{4}$
$=\displaystyle \frac{1}{8}(e^{4}-e^{0})(256-0)$
$=\displaystyle \frac{256}{8}(e^{4}-1)$
$=32(e^{4}-1)\approx 1715.14$
