5.5 The Substitution Rule/51

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

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

Failed to parse (unknown function "\begin{align}"): {\displaystyle \begin{align} \int_{0}^{2} {u^{25}} du = \frac{1}{26}{u^{26}} &= \cfrac{(x-1)^{26}} {26}\bigg|_{0}^{2} = \cfrac{(2-1)^{26}} {26} - \cfrac {(0-1)^{26}} {26} = \frac{1}^{26} {26} - \frac{-1}^{26} {26} = 0 \\[2ex] \end{align} }