## Calculus: Early Transcendentals 8th Edition

$\frac{e^2}{e+1}$
$\int^1_0 (x^e + e^x)dx$ 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 $1^{e+1} = 1$): $= [\frac{1}{e+1}- 0] + [e - 1]$ $= \frac{1}{e+1} + e - 1$ $= \frac{1}{e+1} + \frac{e^2 + e}{e+1} - \frac{e+1}{e+1}$ $= \frac{e^2}{e+1}$