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| <math>\int_{0}^{13}\frac{dx}{\sqrt[3]{(1+2x)^2}}</math>
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| = <math>\int_{0}^{13}\frac{1}{2^3\sqrt{t^2}}dt</math>
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| = <math>\frac{1}{2}\int_{0}^{13}\frac{1}{\sqrt[3]{t^2}}dt</math>
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| = <math>\frac{1}{2}\int_{0}^{13}\frac{1}{t^\frac{2}{3}}dt</math>
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| = <math>\frac{1}{2}3\sqrt[3]{t}</math>
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| = <math>\frac{1}{2}3\sqrt[3]{1+2x}</math>
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| = <math>\frac{3}{2}\sqrt[3]{1+2x}</math>
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| = <math>\frac{3}{2}\sqrt[3]{1+2x}\bigg|_{0}^{13}</math>
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| = <math>\frac{3}{2}\sqrt[3]{1+2x* 13}-\frac{3}{2}\sqrt[3]{1+2*0}</math>
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| = <math>\ 3 </math> \\[2ex]
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| <math> | | <math> |
| \int_{0}^{13}\frac{1}{\sqrt[3]{(1+2x)^2}}\,dx | | \int_{0}^{13}\frac{1}{\sqrt[3]{(1+2x)^2}}\,dx |
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| \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}{u}^{1/3}\bigg|_{0}^{\pi} \\[2ex] |
| &= \frac{1}{2}\sin{(\pi)} - \frac{1}{2}\sin{(0)} \\[2ex] | | &= \frac{1}{2}\sin{(\pi)} - \frac{1}{2}\sin{(0)} \\[2ex] |
| &= 0 | | &= 0 |
Revision as of 04:13, 22 September 2022
![{\displaystyle \int _{0}^{13}{\frac {1}{\sqrt[{3}]{(1+2x)^{2}}}}\,dx}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/e9f38b04c5de29c05e4c1854ebafe81175b77978)
![{\displaystyle {\begin{aligned}u&=1+2x\\[2ex]du&=2dx\\[2ex]{\frac {1}{2}}du&=dx\\[2ex]\end{aligned}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/a737fc278965117d70e906c0f6291b759f13ac66)
New upper limit: 
New lower limit: 
![{\displaystyle {\begin{aligned}\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}}{u}^{1/3}{\bigg |}_{0}^{\pi }\\[2ex]&={\frac {1}{2}}\sin {(\pi )}-{\frac {1}{2}}\sin {(0)}\\[2ex]&=0\end{aligned}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/d73406fc8154f9ec1f4c135f26ef1c4ca95c2256)