## Calculus: Early Transcendentals 8th Edition

$0.1$
$\int^1_{0}(1-x)^9dx$ $=[-1\times\frac{1}{10}(1-x)^{10}] ^1_{0}$ $=[-\frac{1}{10}(1-x)^{10}] ^1_{0}$. Sub in the bounds and subtract the lower bound (0) from the upper bound (1): $=[-\frac{1}{10}(1-(1))^{10}]- [-\frac{1}{10}(1-(0))^{10}]$ $=0- (-\frac{1}{10})$ $=0.1$.