Chapter 14 - Multiple Integrals - 14.1 Double Integrals - Exercises Set 14.1 - Page 1007: 3

2

$\int^4_{2}\int^1_{0} x^2y dx dy$ Evaluating Inner Integral : $\int^1_{0} x^2y dx$ $[\frac{x^3y}{3}]^1_{0}$ $[\frac{1^3y}{3}-\frac{0^3y}{3}] = \frac{y}{3}$ Evaluating Outer Integral $\int^4_{2} \frac{y}{3} dy$ $[\frac{y^2}{6}]^4_{2}$ $\frac{4^2}{6}-\frac{2^2}{6}$ = $\frac{16}{6}-\frac{4}{6}= \frac{12}{6}$ = 2