5.5 The Substitution Rule/51

${\begin{aligned}&\int _{0}^{2}({x-1})^{25}dx\\[2ex]\end{aligned}}$

${\begin{aligned}u&=x-1\\[2ex]du&=dx\\[2ex]\end{aligned}}$

& \int {u^{25}} du &= \frac{1}{26}{u^{26}} \\[2ex]

&= \cfrac{(x-1)^{26}} {26}\bigg|_{0}^{2} \\[2ex]

& \cfrac{(2-1)^{26}} {26} - \cfrac {(0-1)^{26}} {26} \\[2ex]

&= 0

\end{align} </math>