Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 5 - Discovery Project - Area Functions - Page 391: 1

Answer

(a) A = 10 (b) $A(x) = x^2+x-2$ (c) $A'(x) = 2x+1$ The derivative of $A(x)$ has the same form as $~~y = 2t+1~~$ in part a.

Work Step by Step

(a) The area consists of a triangle located above a rectangle. The total area is the sum of each part: $A = \frac{1}{2}(2)(4)+(3)(2) = 4+6 = 10$ (b) The area consists of a triangle located above a rectangle. The total area is the sum of each part: $A(x) = \frac{1}{2}(x-1)(2x+1-3)+(3)(x-1)$ $A(x) = \frac{1}{2}(x-1)(2x-2)+(3)(x-1)$ $A(x) = (x-1)(x-1)+(3)(x-1)$ $A(x) = (x^2-2x+1)+(3x-3)$ $A(x) = x^2+x-2$ (c) $A(x) = x^2+x-2$ $A'(x) = 2x+1$ The derivative of $A(x)$ has the same form as $~~y = 2t+1~~$ in part a.
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