5.5 The Substitution Rule/2 - Revision history

Dvaezazizi@laalliance.org: Protected "5.5 The Substitution Rule/2" ([Edit=Allow only administrators] (indefinite) [Move=Allow only administrators] (indefinite))

2022-08-26T19:00:49Z

Protected "5.5 The Substitution Rule/2" ([Edit=Allow only administrators] (indefinite) [Move=Allow only administrators] (indefinite))

← Older revision	Revision as of 19:00, 26 August 2022
(No difference)

2022-08-26T19:00:03Z

← Older revision		Revision as of 19:00, 26 August 2022
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	<math>		<math>
	\begin{align}		\begin{align}
	\int x^3(2+x^4)^5dx &= \int (x^~~3\cdot dx~~)(2+x^4) = \int \left(\frac{1}{4}du\right)(u) = \frac{1}{4}\int u\,du \\[2ex]		\int x^3(2+x^4)^5dx &= \int (x^3dx)(2+x^4) = \int \left(\frac{1}{4}du\right)(u) = \frac{1}{4}\int u\,du \\[2ex]

	&= \frac{1}{4} \left[\frac{u^{1+1}}{1+1}\right] + C = \frac{u^2}{8} + C \\[2ex]		&= \frac{1}{4} \left[\frac{u^{1+1}}{1+1}\right] + C = \frac{u^2}{8} + C \\[2ex]

2022-08-26T18:59:48Z

← Older revision		Revision as of 18:59, 26 August 2022
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	<math>		<math>
	\begin{align}		\begin{align}
	\int x^3(2+x^4)^5dx &= \int x^~~3dx~~(2+x^4) = \int \left(\frac{1}{4}du\right)(u) = \frac{1}{4}\int u\,du \\[2ex]		\int x^3(2+x^4)^5dx &= \int (x^3\cdot dx)(2+x^4) = \int \left(\frac{1}{4}du\right)(u) = \frac{1}{4}\int u\,du \\[2ex]

	&= \frac{1}{4} \left[\frac{u^{1+1}}{1+1}\right] + C = \frac{u^2}{8} + C \\[2ex]		&= \frac{1}{4} \left[\frac{u^{1+1}}{1+1}\right] + C = \frac{u^2}{8} + C \\[2ex]

2022-08-26T18:59:08Z

← Older revision		Revision as of 18:59, 26 August 2022
Line 19:		Line 19:
	\int x^3(2+x^4)^5dx &= \int x^3dx(2+x^4) = \int \left(\frac{1}{4}du\right)(u) = \frac{1}{4}\int u\,du \\[2ex]		\int x^3(2+x^4)^5dx &= \int x^3dx(2+x^4) = \int \left(\frac{1}{4}du\right)(u) = \frac{1}{4}\int u\,du \\[2ex]

	&= \frac{1}{4} \left[\frac{u^{1+1}}{1+1}\right] + C = \frac{u^2}{8} + C = \frac{(2+x^4)^2}{8} + C		&= \frac{1}{4} \left[\frac{u^{1+1}}{1+1}\right] + C = \frac{u^2}{8} + C \\[2ex]

			&= \frac{(2+x^4)^2}{8} + C

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

2022-08-26T18:58:31Z

← Older revision		Revision as of 18:58, 26 August 2022
Line 19:		Line 19:
	\int x^3(2+x^4)^5dx &= \int x^3dx(2+x^4) = \int \left(\frac{1}{4}du\right)(u) = \frac{1}{4}\int u\,du \\[2ex]		\int x^3(2+x^4)^5dx &= \int x^3dx(2+x^4) = \int \left(\frac{1}{4}du\right)(u) = \frac{1}{4}\int u\,du \\[2ex]

	&= \frac{1}{4} \left(\frac{u^{1+1}}{1+1}\right) + C = \frac{u^2}{8} + C = \frac{(2+x^4)^2}{8} + C		&= \frac{1}{4} \left[\frac{u^{1+1}}{1+1}\right] + C = \frac{u^2}{8} + C = \frac{(2+x^4)^2}{8} + C

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

2022-08-26T18:58:22Z

← Older revision		Revision as of 18:58, 26 August 2022
Line 19:		Line 19:
	\int x^3(2+x^4)^5dx &= \int x^3dx(2+x^4) = \int \left(\frac{1}{4}du\right)(u) = \frac{1}{4}\int u\,du \\[2ex]		\int x^3(2+x^4)^5dx &= \int x^3dx(2+x^4) = \int \left(\frac{1}{4}du\right)(u) = \frac{1}{4}\int u\,du \\[2ex]

	&= \frac{1}{4}\~~cdot~~ \frac{u^{1+1}}{1+1} + C = \frac{u^2}{8} + C = \frac{(2+x^4)^2}{8} + C		&= \frac{1}{4} \left(\frac{u^{1+1}}{1+1}\right) + C = \frac{u^2}{8} + C = \frac{(2+x^4)^2}{8} + C

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

2022-08-26T18:57:43Z

← Older revision		Revision as of 18:57, 26 August 2022
Line 19:		Line 19:
	\int x^3(2+x^4)^5dx &= \int x^3dx(2+x^4) = \int \left(\frac{1}{4}du\right)(u) = \frac{1}{4}\int u\,du \\[2ex]		\int x^3(2+x^4)^5dx &= \int x^3dx(2+x^4) = \int \left(\frac{1}{4}du\right)(u) = \frac{1}{4}\int u\,du \\[2ex]

	&= \frac{1}{4}\cdot \frac{u^{1+1}}{1+1} = \frac{u^2}{8}		&= \frac{1}{4}\cdot \frac{u^{1+1}}{1+1} + C = \frac{u^2}{8} + C = \frac{(2+x^4)^2}{8} + C

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

2022-08-26T18:57:07Z

← Older revision		Revision as of 18:57, 26 August 2022
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	<math>		<math>
	\int x^3(2+x^4)^5dx = \int x^3dx(2+x^4) = \int \left(\frac{1}{4}du\right)(u) = \frac{1}{4}\int u\,du		\begin{align}
			\int x^3(2+x^4)^5dx &= \int x^3dx(2+x^4) = \int \left(\frac{1}{4}du\right)(u) = \frac{1}{4}\int u\,du \\[2ex]

			&= \frac{1}{4}\cdot \frac{u^{1+1}}{1+1} = \frac{u^2}{8}

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

2022-08-26T18:55:57Z

← Older revision		Revision as of 18:55, 26 August 2022
Line 16:		Line 16:

	<math>		<math>
	\int x^3(2+x^4)^5dx = \int x^3dx(2+x^4) = \int \left(\frac{1}{4}du\right)(u) = \frac{1}{4}\int ~~udu~~		\int x^3(2+x^4)^5dx = \int x^3dx(2+x^4) = \int \left(\frac{1}{4}du\right)(u) = \frac{1}{4}\int u\,du
	</math>		</math>

2022-08-26T18:55:49Z

← Older revision		Revision as of 18:55, 26 August 2022
Line 16:		Line 16:

	<math>		<math>
	\int x^3(2+x^4)^5dx = \int x^3dx(2+x^4) = \int \left(\frac{1}{4}du\right)(u)		\int x^3(2+x^4)^5dx = \int x^3dx(2+x^4) = \int \left(\frac{1}{4}du\right)(u) = \frac{1}{4}\int udu
	</math>		</math>