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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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| New upper limit: <math>\27 = 1+2(13)</math><br> | | New upper limit: <math>27 = 1+2(13)</math><br> |
| New lower limit: <math>1 = 1+2(0)</math> | | New lower limit: <math>1 = 1+2(0)</math> |
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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} \int_{1}^{27}\frac{1}{\sqrt[3]{(1+2x)^2}}\left(\frac{1}{2}du\right) = \frac{1}{2}\int_{0}^{\pi} \cos{(u)}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 |
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| \end{align} | | \end{align} |
| </math> | | </math> |
![{\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}}{\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{aligned}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/a7e3038f4c272e5690d6bfe67139c94691aa395f)