5.5 The Substitution Rule/51: Difference between revisions

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Tag: Manual revert
Line 4: Line 4:
& \int_{0}^{2} ({x-1})^{25} dx \\[2ex]
& \int_{0}^{2} ({x-1})^{25} dx \\[2ex]


&= \int {t^{25}} dt \\[2ex]
& \int {t^{25}} dt \\[2ex]


&= \cfrac{t^{25}} {26} \\[2ex]
& \cfrac{t^{25}} {26} \\[2ex]


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


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


&= 0
&= 0

Revision as of 17:37, 31 August 2022

Failed to parse (Conversion error. Server ("https://en.wikipedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\begin{aligned}&\int _{0}^{2}({x-1})^{25}dx\\[2ex]&\int {t^{25}}dt\\[2ex]&{\cfrac {t^{25}}{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}}}