Calculus (3rd Edition)

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

Chapter 5 - The Integral - 5.3 The Indefinite Integral - Exercises: 6

Answer

$\int4x^7- 3cos(x)dx = \frac{1}{2}x^8 -3sin(x) + C$

Work Step by Step

Rules to remember: $\int sin(x)dx = -cos(x)$, $\int cos(x)dx = sin(x)$ $sin(x)' = cos(x)$, $cos(x)' = -sin(x)$ Anti-differentiation: $\int4x^7- 3cos(x)dx = 4\times\frac{1}{8}x^8 -3sin(x) + C= \frac{1}{2}x^8 -3sin(x) + C$ Checking with differentiation: $(\frac{1}{2}x^8 -3sin(x) + C)'= \frac{1}{2}\times8x^7 -3cos(x)= 4x^7 -3cos(x)$
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