5.5 The Substitution Rule/61: Difference between revisions

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New upper limit: <math>\pi = 1+2(13) = 27</math><br>
New upper limit: <math>\27 = 1+2(13)</math><br>
New lower limit: <math>0 = 1+2(0) = 1</math>
New lower limit: <math>1 = 1+2(0)</math>
   
   


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\begin{align}
\begin{align}


\int_{0}^{13}\frac{dx}{\sqrt[3]{(1+2x)^2}}\,dx &= \int_{0}^{\sqrt{\pi}} (xdx)\cos{(x^2)} \\[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_{0}^{\pi} \left(\frac{1}{2}du\right)\cos{(u)} = \frac{1}{2}\int_{0}^{\pi} \cos{(u)}du \\[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{(u)}\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]

Revision as of 04:00, 22 September 2022

= = = = = = = = = \\[2ex]



New upper limit: Failed to parse (syntax error): {\displaystyle \27 = 1+2(13)}
New lower limit: