## Calculus 8th Edition

$\frac{1}{10}$
We use substitution to allow us to integrate the term $(1-x)^9$ as we cannot reasonably integrate the current term otherwise. Using the substitution $u = 1-x$ and $du = -dx$ we can rewrite the statement as $-\int_{-1}^{0}u^9du$, which evaluates to $-\frac{u^{10}}{10}\Big|_{-1}^0$ We now substitute $1-x$ into $u$ and get $-\frac{(1-x)^{10}}{10}\Big|_{0}^1$. Substituting $1$ and $0$ into $x$ results in $-\frac{(1-1)^{10}}{10}$ - $(-\frac{(1)^{10}}{10}) = 0$ - ($-\frac{1}{10})$ = $\frac{1}{10}$