Answer
See explanation.
Work Step by Step
(a) $F^{\prime}(x)=-3+3 x^{2}$
(b) $F(x)=\int_{1}^{x}\left(-3+3 t^{2}\right) d t=\left(-3 t+t^{3}\right)_{1}^{x}=-1+3+x^{3}-3 x=2-3 x+x^{3}$
Differentiating then gives $F^{\prime}(x)=-3+3 x^{2}$
Calculus, 10th Edition (Anton)
by Anton, Howard
Published by
Wiley
ISBN 10:
0-47064-772-8
ISBN 13:
978-0-47064-772-1
Chapter 4 - Integration - 4.6 The Fundamental Theorem Of Calculus - Exercises Set 4.6 - Page 321: 47
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