5.5 The Substitution Rule/61: Difference between revisions

From Mr. V Wiki Math
Jump to navigation Jump to search
No edit summary
No edit summary
 
(6 intermediate revisions by the same user not shown)
Line 1: Line 1:
<math>\int_{0}^{13}\frac{dx}{\sqrt[3]{(1+2x)^2}}</math>
=  <math>\int_{0}^{13}\frac{1}{2^3\sqrt{t^2}}dt</math>
=  <math>\frac{1}{2}\int_{0}^{13}\frac{1}{\sqrt[3]{t^2}}dt</math>
=  <math>\frac{1}{2}\int_{0}^{13}\frac{1}{t^\frac{2}{3}}dt</math>
=  <math>\frac{1}{2}3\sqrt[3]{t}</math>
=  <math>\frac{1}{2}3\sqrt[3]{1+2x}</math>
=  <math>\frac{3}{2}\sqrt[3]{1+2x}</math>
=  <math>\frac{3}{2}\sqrt[3]{1+2x}\bigg|_{0}^{13}</math>
=  <math>\frac{3}{2}\sqrt[3]{1+2x* 13}-\frac{3}{2}\sqrt[3]{1+2*0}</math>
=  <math>\ 3 </math> \\[2ex]
<math>
<math>
\int_{0}^{13}\frac{1}{\sqrt[3]{(1+2x)^2}}\,dx
\int_{0}^{13}\frac{1}{\sqrt[3]{(1+2x)^2}}\,dx
Line 34: Line 23:


\int_{0}^{13}\frac{1}{\sqrt[3]{(1+2x)^2}}\,dx &= \int_{0}^{13}\frac{1}{\sqrt[3]{(1+2x)^2}}\,(dx) \\[2ex]
\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]
&= \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}\sin{(u)}\bigg|_{0}^{\pi} \\[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}\sin{(\pi)} - \frac{1}{2}\sin{(0)} \\[2ex]
&= \frac{3}{2}{(27)}^{1/3} - \frac{3}{2}{(1)}^{1/3} \\[2ex]
&= 0
&= \frac{9}{2}-\frac{3}{2}\\[2ex]
&= 3


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

Latest revision as of 04:20, 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 (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 &= 1+2x \\[2ex] du &= 2dx \\[2ex] \frac{1}{2}du &= dx \\[2ex] \end{align} }


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] &= \frac{9}{2}-\frac{3}{2}\\[2ex] &= 3 \end{align} }

Retrieved from "https://wiki.dvaezazizi.com/index.php?title=5.5_The_Substitution_Rule/61&oldid=4679"