5.3 The Fundamental Theorem of Calculus/29: Difference between revisions

Revision as of 02:27, 7 September 2022

$\int _{1}^{9}{\frac {x-1}{\sqrt {x}}}dx=\int _{1}^{9}{\frac {x}{\sqrt {x}}}-{\frac {1}{\sqrt {x}}}=\int _{1}^{9}$

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 <math>
-\int_{1}^{9}\frac{x-1}{\sqrt{x}} dx =
+\int_{1}^{9}\frac{x-1}{\sqrt{x}} dx = \int_{1}^{9}\frac{x}{\sqrt{x}}-\frac{1}{\sqrt{x}} = \int_{1}^{9}
 </math>