## Thomas' Calculus 13th Edition

Published by Pearson

# Chapter 5: Integrals - Practice Exercises - Page 308: 68

#### Answer

a) 2 ; b) $\dfrac{2a}{3}$

#### Work Step by Step

a. The average value is defined as:$\dfrac{1}{3} \int_{0}^3 \sqrt {3x} dx$ $\dfrac{\sqrt 3}{3} [\dfrac{x^{\frac{1}{2}+1}}{\frac{1}{2}+1}]_{0}^3=\dfrac{2\sqrt 3}{9} [3^{\frac{3}{2}}-0]=2$ b. The average value is defined as: $\dfrac{1}{a} \int_{0}^a \sqrt {ax} dx$ $\dfrac{\sqrt a}{a} [\dfrac{x^{\frac{1}{2}+1}}{\frac{1}{2}+1}]_{0}^a=\dfrac{2\sqrt a}{3a} [3^{\frac{a}{2}}-0]=\dfrac{2a}{3}$

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