5.5 The Substitution Rule/51: Difference between revisions
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& \int_{0}^{2} ({x-1})^{25} dx \\[2ex] | & \int_{0}^{2} ({x-1})^{25} dx \\[2ex] | ||
\end{align} | |||
</math> | |||
& \cfrac{(x-1)^{26}} {26}\bigg|_{0}^{2} \\[2ex] | <math> | ||
\begin{align} | |||
u &= x-1 \\[2ex] | |||
du &= dx \\[2ex] | |||
\end{align} | |||
</math> | |||
& \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] | & \cfrac{(2-1)^{26}} {26} - \cfrac {(0-1)^{26}} {26} \\[2ex] | ||
&= 0 | &= 0 | ||
\end{align} | \end{align} | ||
</math> | </math> | ||
Revision as of 17:21, 7 September 2022
![{\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)
& \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{align} </math>