5.5 The Substitution Rule/51

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

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

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