Answer
$\frac{1}{e+1}+e-1$
Work Step by Step
$\int_0^1(e^x+x^e)dx$
$=(e^x+\frac{x^{e+1}}{e+1})|_0^1$
$=(e^1+\frac{1^{e+1}}{e+1})-(e^0+\frac{0^{e+1}}{e+1})$
$=(e+\frac{1}{e+1})-(1+0)$
$=e+\frac{1}{e+1}-1$
$=\frac{1}{e+1}+e-1$
Calculus 8th Edition
by Stewart, James
Published by
Cengage
ISBN 10:
1285740629
ISBN 13:
978-1-28574-062-1
Chapter 6 - Inverse Functions - 6.2 Exponential Functions and Their Derivatives - 6.2 Exercises - Page 420: 83
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