Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 6 - Applications of the Integral - Chapter Review Exercises - Page 319: 7



Work Step by Step

The area $(A)$ can be calculated as: $A=\int_{0}^{1} [4-(4-x^2)] \ dx +\int_1^{4/3} [4-3x] \ dx \\= \int_{0}^{1} (-x^2) \ dx +\int_1^{4/3} [4-3x] \ dx \\=\dfrac{1}{3} (x^3)_0^1 +[4x-\dfrac{3x^2}{2}]_1^{4/3} \\=\dfrac{1}{3}+\dfrac{8}{3}-\dfrac{5}{2} \\=\dfrac{1}{2}$
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