5.5 The Substitution Rule/61
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Revision as of 04:00, 22 September 2022 by Ricardom59381@students.laalliance.org (talk | contribs)
= = = = = = = = = \\[2ex]
![{\displaystyle \int _{0}^{13}{\frac {dx}{\sqrt[{3}]{(1+2x)^{2}}}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/366746df0224b60295f774b7c9869a3c355329ff)
![{\displaystyle {\frac {1}{2}}\int _{0}^{13}{\frac {1}{\sqrt[{3}]{t^{2}}}}dt}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/1d54296bace91b1eb68b7a819d224cf901cf5b7d)
![{\displaystyle {\frac {1}{2}}3{\sqrt[{3}]{t}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/16a3b59a69048f671074fd0201320c2ffb6b137e)
![{\displaystyle {\frac {1}{2}}3{\sqrt[{3}]{1+2x}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/20415fd42cdb3705a79026c6b206e0d8eae227e1)
![{\displaystyle {\frac {3}{2}}{\sqrt[{3}]{1+2x}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/af311e7d8c3cf04c63a34118cdc43ae40922fd8a)
![{\displaystyle {\frac {3}{2}}{\sqrt[{3}]{1+2x}}{\bigg |}_{0}^{13}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/a499de07f4b55f145653fc30d4d5ebc73aaa144d)
![{\displaystyle {\frac {3}{2}}{\sqrt[{3}]{1+2x*13}}-{\frac {3}{2}}{\sqrt[{3}]{1+2*0}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/6b11b0f2aaf612f19578317fe0894d48b4ce3756)
![{\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: Failed to parse (syntax error): {\displaystyle \27 = 1+2(13)}
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}\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]&={\frac {1}{2}}\sin {(u)}{\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/9f2d3be53b6c782ed6e625f9cba9b37677f15c3e)