5.5 The Substitution Rule/51: Difference between revisions

Revision as of 17:22, 31 August 2022

${\begin{aligned}&\int _{0}^{2}({x-1})^{25}dx\\[2ex]&\int {t^{25}}dt\\[2ex]&{\cfrac {t^{25}}{26}}\end{aligned}}$

@@ Line 4: / Line 4: @@
 & \int_{0}^{2} ({x-1})^{25} dx \\[2ex]
-& \int {t^{25}} dt
+& \int {t^{25}} dt \\[2ex]
+& \cfrac{t^{25}} {26}