University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 4 - Practice Exercises - Page 279: 118


$\dfrac{2^{\pi+r}}{\ln 2} +C$

Work Step by Step

Calculate the anti-derivative. Since, we know $\int x^{n} dx=\dfrac{x^{n+1}}{n+1}+c$ and $ \int a^x dx=\dfrac{1}{\ln a} a^x+c$ where $c$ is a constant of proportionality. Then $\int 2^{\pi+r} dr=(\dfrac{1}{\ln 2})(2^{\pi+r})+c$ Thus, $= \dfrac{2^{\pi+r}}{\ln 2} +c$
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