## University Calculus: Early Transcendentals (3rd Edition)

a) 2 and b) $\dfrac{2a}{3}$
a) Average value can be calculated as: $\dfrac{1}{3} \int_{0}^3 \sqrt {3x} dx$ $=\dfrac{\sqrt 3}{3} [\dfrac{x^{3/2}}{3/2}]_{0}^3$ or, $\dfrac{2\sqrt 3}{9} [3^{3/2}-0]=2$ b) Average value can be calculated as: $\dfrac{1}{a} \int_{0}^a \sqrt {ax} dx$ $=\dfrac{\sqrt a}{a} [\dfrac{x^{3/2}}{3/2}]_{0}^a$ or, $\dfrac{2\sqrt a}{3a} [3^{a/2}-0]=\dfrac{2a}{3}$ Hence, a) 2 and b) $\dfrac{2a}{3}$