∫ 0 2 ( x − 1 ) 25 d x {\displaystyle {\begin{aligned}&\int _{0}^{2}({x-1})^{25}dx\\[2ex]\end{aligned}}}
u = x − 1 d u = d x {\displaystyle {\begin{aligned}u&=x-1\\[2ex]du&=dx\\[2ex]\end{aligned}}}
∫ 0 2 u 25 d u = 1 26 u 26 = ( x − 1 ) 26 26 | 0 2 = ( 2 − 1 ) 26 26 − ( 0 − 1 ) 26 26 = 1 26 26 − − 1 26 26 = 0 {\displaystyle {\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}}}
![{\displaystyle {\begin{aligned}&\int _{0}^{2}({x-1})^{25}dx\\[2ex]\end{aligned}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/e2e72a726a76074d092e09f6837a6700bbe2b2bb)
![{\displaystyle {\begin{aligned}u&=x-1\\[2ex]du&=dx\\[2ex]\end{aligned}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/12e939d2357ce4f786ecee1e47b1b540d0bd3c6f)
![{\displaystyle {\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}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/f289a23e04228cf12857a6f85664c56cca8c67d1)