Calculus (3rd Edition)

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

Chapter 5 - The Integral - 5.5 The Fundamental Theorem of Calculus, Part II - Exercises - Page 262: 8


$f(x)= 4x^3-4x^2-16.$

Work Step by Step

We have $$ f(x)=\int_2^x12t^2-8t \ dt=\frac{12}{3}t^3-\frac{8}{2}t^2|_2^x=4x^3-4x^2-16. $$
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