5.5 The Substitution Rule/61 - Revision history

Ricardom59381@students.laalliance.org at 04:20, 22 September 2022

2022-09-22T04:20:27Z

← Older revision		Revision as of 04:20, 22 September 2022
Line 25:		Line 25:
	&= \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}\frac{{u}^{1/3}}{\frac{1}{3}}\bigg\|_{1}^{27} = \frac{3}{2}{u}^{1/3}\bigg\|_{1}^{27}\\[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]		&= \frac{9}{2}-\frac{3}{2}\\[2ex]
	~~&= \frac{3}{2}{(27)}^{1/3} - \frac{3}{2}{(1)}^{1/3} \\[2ex]~~
	&= 3		&= 3

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

Ricardom59381@students.laalliance.org at 04:20, 22 September 2022

2022-09-22T04:20:06Z

← Older revision		Revision as of 04:20, 22 September 2022
Line 25:		Line 25:
	&= \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}\frac{{u}^{1/3}}{\frac{1}{3}}\bigg\|_{1}^{27} = \frac{3}{2}{u}^{1/3}\bigg\|_{1}^{27}\\[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{9}{2}-\frac{3}{2}\\[2ex]
	&= \frac{3}{2}{(27)}^{1/3} - \frac{3}{2}{(1)}^{1/3} \\[2ex]		&= \frac{3}{2}{(27)}^{1/3} - \frac{3}{2}{(1)}^{1/3} \\[2ex]
	&= 3		&= 3

Ricardom59381@students.laalliance.org at 04:18, 22 September 2022

2022-09-22T04:18:35Z

← Older revision		Revision as of 04:18, 22 September 2022
Line 25:		Line 25:
	&= \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}\frac{{u}^{1/3}}{\frac{1}{3}}\bigg\|_{1}^{27} = \frac{3}{2}{u}^{1/3}\bigg\|_{1}^{27}\\[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		&= 3

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

Ricardom59381@students.laalliance.org at 04:16, 22 September 2022

2022-09-22T04:16:38Z

← Older revision		Revision as of 04:16, 22 September 2022
Line 24:		Line 24:
	\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}\frac{{u}^{1/3}}{3}\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{1}{2}\sin{(\pi)} - \frac{1}{2}\sin{(0)} \\[2ex]
	&= 0		&= 0

Ricardom59381@students.laalliance.org at 04:14, 22 September 2022

2022-09-22T04:14:33Z

← Older revision		Revision as of 04:14, 22 September 2022
Line 24:		Line 24:
	\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}{u}^{1/3}\bigg\|_{0}^{\pi} \\[2ex]		&= \frac{1}{2}\frac{{u}^{1/3}}{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

Ricardom59381@students.laalliance.org at 04:13, 22 September 2022

2022-09-22T04:13:42Z

← Older revision		Revision as of 04:13, 22 September 2022
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 35:		Line 24:
	\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

Ricardom59381@students.laalliance.org at 04:04, 22 September 2022

2022-09-22T04:04:22Z

← Older revision		Revision as of 04:04, 22 September 2022
Line 34:		Line 34:

	\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}\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]

Ricardom59381@students.laalliance.org at 04:03, 22 September 2022

2022-09-22T04:03:22Z

← Older revision		Revision as of 04:03, 22 September 2022
Line 34:		Line 34:

	\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]{(~~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}\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]

Ricardom59381@students.laalliance.org at 04:01, 22 September 2022

2022-09-22T04:01:27Z

← Older revision		Revision as of 04:01, 22 September 2022
Line 26:		Line 26:


	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>

Line 34:		Line 34:

	\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]{(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]

Ricardom59381@students.laalliance.org at 04:00, 22 September 2022

2022-09-22T04:00:44Z

← Older revision		Revision as of 04:00, 22 September 2022
Line 26:		Line 26:


	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>


Line 33:		Line 33:
	\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]