University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 5 - Section 5.4 - The Fundamental Theorem of Calculus - Exercises: 48

Answer

$2x^2sin(x^6)+\int_{2}^{x^2} {sin (t^3)} dt$

Work Step by Step

Let's take the derivative of both sides, use the product rule and let's apply the fundamental theorem of Calculus (substitute the limits, and multiply by their derivatives)! $\frac{dy}{dx}=x(sin(x^6)\cdot 2x)+1\cdot\int_{2}^{x^2} {sin (t^3)}dt=2x^2sin(x^6)+\int_{2}^{x^2} {sin (t^3)} dt$
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