5.5 The Substitution Rule/51

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

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

${\begin{aligned}\int _{0}^{2}{u^{25}}du={\frac {1}{26}}{u^{26}}&={\cfrac {(x-1)^{26}}{26}}{\bigg |}_{0}^{2}={\cfrac {(2-1)^{26}}{26}}-{\cfrac {(0-1)^{26}}{26}}={\frac {1^{26}}{26}}-{\frac {-1^{26}}{26}}=0\\[2ex]\end{aligned}}$