Answer
$\int_0^42^{s}ds=\frac{15}{ln2}$
Work Step by Step
Evaluate the integral $\int_0^42^{s}ds$.
$\int_0^42^{s}ds=|\frac{2^{s}}{ln2}|_0^4$
$=\frac{2^{4}-2^{0}}{ln2}$
$=\frac{16-1}{ln2}$
Hence, $\int_0^42^{s}ds=\frac{15}{ln2}$
Calculus 8th Edition
by Stewart, James
Published by
Cengage
ISBN 10:
1285740629
ISBN 13:
978-1-28574-062-1
Chapter 6 - Inverse Functions - 6.4 Derivatives of Logarithmic Functions - 6.4 Exercises - Page 438: 81
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