Answer
$\frac{e^2}{e+1}$
Work Step by Step
$\int^1_0 (x^e + e^x)dx$
The first step is to separate into 2 definite integrals:
$ = \int^1_0 x^e dx + \int^1_0 e^x dx$
Solve both definite integrals:
$ = (\frac{x^{e+1}}{e+1})|^1_0 + (e^x)|^1_0$
Plug in the limits of integration:
$ = [\frac{(1)^{e+1}}{e+1} - \frac{(0)^{e+1}}{e+1}] + [e^1 - e^0]$
Simplify until final answer is reached (remember that $1^{e+1} = 1$):
$= [\frac{1}{e+1}- 0] + [e - 1]$
$= \frac{1}{e+1} + e - 1$
$= \frac{1}{e+1} + \frac{e^2 + e}{e+1} - \frac{e+1}{e+1}$
$= \frac{e^2}{e+1}$
