Answer
$ -\frac{1}{6}(e^{-17}-e ) . $
Work Step by Step
We have $$ \int_0^3 e^{1-6t} dt= -\frac{1}{6}\int_0^3-6e^{1-6t} dt = -\frac{1}{6}\left[e^{1-6t}\right]_0^3= -\frac{1}{6}(e^{-17}-e ) . $$
Calculus (3rd Edition)
by Rogawski, Jon; Adams, Colin
Published by
W. H. Freeman
ISBN 10:
1464125260
ISBN 13:
978-1-46412-526-3
Chapter 7 - Exponential Functions - 7.1 Derivative of f(x)=bx and the Number e - Exercises - Page 328: 73
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