5.5 The Substitution Rule/65: Difference between revisions

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


</math>
</math>
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\begin{align}
\begin{align}


u &= x-1
u &= x-1 \\ u+1 &= x \\[2ex]
  \\[2ex]
du &= 1 dx \\[2ex]
du &= 2 dx \\[2ex]
du &= dx
\frac{1}{2} du &= dx
\end{align}
</math>
 
 
<math>
\begin{align}
\int_{1}^{2} (x \sqrt{x-1}\,)\;dx &= \int_{0}^{1} ((u+1) \sqrt{u})\;du = \int_{0}^{1} (u^ \frac{3}{2} + u^ \frac{1}{2})\;du \\[2ex]
&= (\frac{2}{5} u^\frac{5}{2} + \frac{2}{3} u^\frac{3}{2})\bigg| _{0}^{1}  =\frac{2}{5} + \frac{2}{3} \\[2ex]
&= \frac{16}{15}\\[2ex]
\end{align}
\end{align}
</math>
</math>

Latest revision as of 23:11, 13 September 2022

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle \int_{1}^{2} x \sqrt{x-1} dx }


Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle \begin{align} u &= x-1 \\ u+1 &= x \\[2ex] du &= 1 dx \\[2ex] du &= dx \end{align} }


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