Calculus Concepts: An Informal Approach to the Mathematics of Change 5th Edition

Published by Brooks Cole
ISBN 10: 1-43904-957-2
ISBN 13: 978-1-43904-957-0

Chapter 5 - Accumulating Change: Limits of Sums and the Definite Integral - 5.4 Activities - Page 364: 22

Answer

$$F\left( p \right) = \frac{{25{p^2}}}{2} + 3p - 53$$

Work Step by Step

$$\eqalign{ & f\left( p \right) = 25p + 3;\,\,\,F\left( 2 \right) = 3 \cr & \cr & {\text{Write a formula }}F\left( p \right){\text{ for the antiderivative of }}f\left( p \right) \cr & F\left( p \right) = \int {\left( {25p + 3} \right)} dp \cr & {\text{integrate}} \cr & F\left( p \right) = \frac{{25{p^2}}}{2} + 3p + C \cr & \cr & {\text{Use the condition }}F\left( 2 \right) = 3{\text{ to find }}C \cr & 3 = \frac{{25{{\left( 2 \right)}^2}}}{2} + 3\left( 2 \right) + C \cr & 3 = 50 + 6 + C \cr & 3 - 56 = C \cr & C = - 53 \cr & \cr & {\text{The specific antiderivative of }}f{\text{ is}} \cr & F\left( p \right) = \frac{{25{p^2}}}{2} + 3p - 53 \cr} $$
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