5.5 The Substitution Rule/65: Difference between revisions

From Mr. V Wiki Math
Jump to navigation Jump to search
(Created page with "<math> \int_{1}^{2} x \sqrt{x-1} dx = \int_{0}^{1} u+1 \sqrt{u}du = \int_{0}^{1}(u + 1)(\sqrt{u}) = \int_{0}^{1} u^ \frac{2}{3} + \sqrt{u}du = \frac{2}{5} + \frac{2}{3} = \frac{16}{15} </math>")
 
No edit summary
Line 1: Line 1:
<math>
<math>


\int_{1}^{2} x \sqrt{x-1} dx = \int_{0}^{1} u+1 \sqrt{u}du = \int_{0}^{1}(u + 1)(\sqrt{u}) = \int_{0}^{1} u^ \frac{2}{3} + \sqrt{u}du = \frac{2}{5} + \frac{2}{3} = \frac{16}{15}
\int_{1}^{2} x \sqrt{x-1} dx = \int_{0}^{1} u+1 \sqrt{u}du = \int_{0}^{1}(u + 1)(\sqrt{u}) = \int_{0}^{1} u^ \frac{3}{2} + \sqrt{u}du = \frac{2}{5} U^\frac{5}{2} + \frac{2}{3} U^\frac{3}{2}| _{1}^{2}  =\frac{2}{5} + \frac{2}{3} = \frac{16}{15}  
 
</math>
</math>

Revision as of 09:26, 7 September 2022

Failed to parse (Conversion error. Server ("https://en.wikipedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \int _{1}^{2}x{\sqrt {x-1}}dx=\int _{0}^{1}u+1{\sqrt {u}}du=\int _{0}^{1}(u+1)({\sqrt {u}})=\int _{0}^{1}u^{\frac {3}{2}}+{\sqrt {u}}du={\frac {2}{5}}U^{\frac {5}{2}}+{\frac {2}{3}}U^{\frac {3}{2}}|_{1}^{2}={\frac {2}{5}}+{\frac {2}{3}}={\frac {16}{15}}}