Answer
$0.1$
Work Step by Step
$\int^1_{0}(1-x)^9dx$
$=[-1\times\frac{1}{10}(1-x)^{10}] ^1_{0}$
$=[-\frac{1}{10}(1-x)^{10}] ^1_{0}$.
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}) $
$=0.1$.
