Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 8 - Integration Techniques, L'Hopital's Rule, and Improper Integrals - 8.1 Exercises: 79

Answer

The $a>0$ that gives the area of $2/3$ is $$a=\frac{1}{2}.$$

Work Step by Step

The area bounded by these functions is given by the integral of the absolute value of their difference between the points of intersection: $$A=\int_0^{1/a}|f(x)-g(x)|dx.$$ When $x<1/a$ i.e. when $a>\frac{1}{x}$ then $g(x)=ax^2$
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