5.3 The Fundamental Theorem of Calculus/28: Difference between revisions

Revision as of 20:48, 23 August 2022

$\int _{0}^{1}\left(3+x{\sqrt {x}}\right)dx=\int _{0}^{1}\left(3+x^{1}{x}^{\frac {1}{2}}\right)dx=\int _{0}^{1}\left(3+x^{1+{\frac {1}{2}}}\right)dx$

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 \int_{0}^{1}\left(3+x\sqrt{x}\right)dx = \int_{0}^{1}\left(3+x^{1}{x}^{\frac{1}{2}}\right)dx
+= \int_{0}^{1}\left(3+x^{1+\frac{1}{2}}\right)dx
 </math>