5.5 The Substitution Rule/61: Difference between revisions
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&= \int_{1}^{27}\frac{1}{\sqrt[3]{u^2}}\left(\frac{1}{2}du\right) = \frac{1}{2}\int_{1}^{27} {u}^{-2/3}du \\[2ex] | &= \int_{1}^{27}\frac{1}{\sqrt[3]{u^2}}\left(\frac{1}{2}du\right) = \frac{1}{2}\int_{1}^{27} {u}^{-2/3}du \\[2ex] | ||
&= \frac{1}{2}\frac{{u}^{1/3}}{\frac{1}{3}}\bigg|_{1}^{27} = \frac{3}{2}{u}^{1/3}\bigg|_{1}^{27}\\[2ex] | &= \frac{1}{2}\frac{{u}^{1/3}}{\frac{1}{3}}\bigg|_{1}^{27} = \frac{3}{2}{u}^{1/3}\bigg|_{1}^{27}\\[2ex] | ||
&= \frac{ | &= \frac{3}{2}{(27)}^{1/3} - \frac{3}{2}{(1)}^{1/3} \\[2ex] | ||
&= | &= 3 | ||
\end{align} | \end{align} | ||
</math> | </math> | ||
Revision as of 04:18, 22 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_{0}^{13}\frac{1}{\sqrt[3]{(1+2x)^2}}\,dx }
Failed to parse (Conversion error. Server ("https://en.wikipedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\begin{aligned}u&=1+2x\\[2ex]du&=2dx\\[2ex]{\frac {1}{2}}du&=dx\\[2ex]\end{aligned}}}
New upper limit: 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 27 = 1+2(13)}
New lower limit: 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 1 = 1+2(0)}
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} \int_{0}^{13}\frac{1}{\sqrt[3]{(1+2x)^2}}\,dx &= \int_{0}^{13}\frac{1}{\sqrt[3]{(1+2x)^2}}\,(dx) \\[2ex] &= \int_{1}^{27}\frac{1}{\sqrt[3]{u^2}}\left(\frac{1}{2}du\right) = \frac{1}{2}\int_{1}^{27} {u}^{-2/3}du \\[2ex] &= \frac{1}{2}\frac{{u}^{1/3}}{\frac{1}{3}}\bigg|_{1}^{27} = \frac{3}{2}{u}^{1/3}\bigg|_{1}^{27}\\[2ex] &= \frac{3}{2}{(27)}^{1/3} - \frac{3}{2}{(1)}^{1/3} \\[2ex] &= 3 \end{align} }