5.4 Indefinite Integrals and the Net Change Theorem/30 - Revision history

Dvaezazizi@laalliance.org: Protected "5.4 Indefinite Integrals and the Net Change Theorem/30" ([Edit=Allow only administrators] (indefinite) [Move=Allow only administrators] (indefinite))

2022-08-26T18:25:39Z

Protected "5.4 Indefinite Integrals and the Net Change Theorem/30" ([Edit=Allow only administrators] (indefinite) [Move=Allow only administrators] (indefinite))

← Older revision	Revision as of 18:25, 26 August 2022
(No difference)

2022-08-26T18:16:20Z

← Older revision		Revision as of 18:16, 26 August 2022
Line 9:		Line 9:
	&= \left(-\frac{1}{(2)}+(2)^5\right) - \left(-\frac{1}{(1)}+(1)^5\right) \\[2ex]		&= \left(-\frac{1}{(2)}+(2)^5\right) - \left(-\frac{1}{(1)}+(1)^5\right) \\[2ex]

	&= \left(-\frac{1}{2}+32\right)		&= \left(-\frac{1}{2}+32\right) = \left(-\frac{1}{2}+\frac{64}{2}\right) = \frac{63}{2}
	\end{align}		\end{align}
	</math>		</math>

2022-08-26T18:15:46Z

← Older revision		Revision as of 18:15, 26 August 2022
Line 7:		Line 7:
	&= \left(\frac{y^{-2+1}}{-2+1}+\frac{5y^{4+1}}{4+1}\right)\bigg\|_{1}^{2} = \left(\frac{y^{-1}}{-1}+y^5\right)\bigg\|_{1}^{2} = \left(-\frac{1}{y}+y^5\right)\bigg\|_{1}^{2} \\[2ex]		&= \left(\frac{y^{-2+1}}{-2+1}+\frac{5y^{4+1}}{4+1}\right)\bigg\|_{1}^{2} = \left(\frac{y^{-1}}{-1}+y^5\right)\bigg\|_{1}^{2} = \left(-\frac{1}{y}+y^5\right)\bigg\|_{1}^{2} \\[2ex]

	&= \left(-\frac{1}{(2)}+(2)^5\right) - \left(-\frac{1}{(1)}+(1)^5\right)		&= \left(-\frac{1}{(2)}+(2)^5\right) - \left(-\frac{1}{(1)}+(1)^5\right) \\[2ex]

	&= \left(-\frac{1}{2}+32\right)		&= \left(-\frac{1}{2}+32\right)
	\end{align}		\end{align}
	</math>		</math>

2022-08-26T18:15:38Z

← Older revision		Revision as of 18:15, 26 August 2022
Line 7:		Line 7:
	&= \left(\frac{y^{-2+1}}{-2+1}+\frac{5y^{4+1}}{4+1}\right)\bigg\|_{1}^{2} = \left(\frac{y^{-1}}{-1}+y^5\right)\bigg\|_{1}^{2} = \left(-\frac{1}{y}+y^5\right)\bigg\|_{1}^{2} \\[2ex]		&= \left(\frac{y^{-2+1}}{-2+1}+\frac{5y^{4+1}}{4+1}\right)\bigg\|_{1}^{2} = \left(\frac{y^{-1}}{-1}+y^5\right)\bigg\|_{1}^{2} = \left(-\frac{1}{y}+y^5\right)\bigg\|_{1}^{2} \\[2ex]

	&= \left(-\frac{1}{(2)}+(2)^5\right) - \left(-\frac{1}{(1)}+(1)^5\right) = \left(-\frac{1}{2}+32\right)		&= \left(-\frac{1}{(2)}+(2)^5\right) - \left(-\frac{1}{(1)}+(1)^5\right)

			&= \left(-\frac{1}{2}+32\right)
	\end{align}		\end{align}
	</math>		</math>

2022-08-26T18:15:26Z

← Older revision		Revision as of 18:15, 26 August 2022
Line 7:		Line 7:
	&= \left(\frac{y^{-2+1}}{-2+1}+\frac{5y^{4+1}}{4+1}\right)\bigg\|_{1}^{2} = \left(\frac{y^{-1}}{-1}+y^5\right)\bigg\|_{1}^{2} = \left(-\frac{1}{y}+y^5\right)\bigg\|_{1}^{2} \\[2ex]		&= \left(\frac{y^{-2+1}}{-2+1}+\frac{5y^{4+1}}{4+1}\right)\bigg\|_{1}^{2} = \left(\frac{y^{-1}}{-1}+y^5\right)\bigg\|_{1}^{2} = \left(-\frac{1}{y}+y^5\right)\bigg\|_{1}^{2} \\[2ex]

	&= \left(-\frac{1}{(2)}+(2)^5\right) - \left(-\frac{1}{(1)}+(1)^5\right)		&= \left(-\frac{1}{(2)}+(2)^5\right) - \left(-\frac{1}{(1)}+(1)^5\right) = \left(-\frac{1}{2}+32\right)
	\end{align}		\end{align}
	</math>		</math>

2022-08-26T18:14:31Z

← Older revision		Revision as of 18:14, 26 August 2022
Line 7:		Line 7:
	&= \left(\frac{y^{-2+1}}{-2+1}+\frac{5y^{4+1}}{4+1}\right)\bigg\|_{1}^{2} = \left(\frac{y^{-1}}{-1}+y^5\right)\bigg\|_{1}^{2} = \left(-\frac{1}{y}+y^5\right)\bigg\|_{1}^{2} \\[2ex]		&= \left(\frac{y^{-2+1}}{-2+1}+\frac{5y^{4+1}}{4+1}\right)\bigg\|_{1}^{2} = \left(\frac{y^{-1}}{-1}+y^5\right)\bigg\|_{1}^{2} = \left(-\frac{1}{y}+y^5\right)\bigg\|_{1}^{2} \\[2ex]

	&= \left(-\frac{1}{2}+(2)^5\right) - \left(-\frac{1}{1}+(1)^5\right)		&= \left(-\frac{1}{(2)}+(2)^5\right) - \left(-\frac{1}{(1)}+(1)^5\right)
	\end{align}		\end{align}
	</math>		</math>

2022-08-26T18:14:15Z

← Older revision		Revision as of 18:14, 26 August 2022
Line 7:		Line 7:
	&= \left(\frac{y^{-2+1}}{-2+1}+\frac{5y^{4+1}}{4+1}\right)\bigg\|_{1}^{2} = \left(\frac{y^{-1}}{-1}+y^5\right)\bigg\|_{1}^{2} = \left(-\frac{1}{y}+y^5\right)\bigg\|_{1}^{2} \\[2ex]		&= \left(\frac{y^{-2+1}}{-2+1}+\frac{5y^{4+1}}{4+1}\right)\bigg\|_{1}^{2} = \left(\frac{y^{-1}}{-1}+y^5\right)\bigg\|_{1}^{2} = \left(-\frac{1}{y}+y^5\right)\bigg\|_{1}^{2} \\[2ex]

	&= \left(-\frac{1}{2}+(2)^5\right)		&= \left(-\frac{1}{2}+(2)^5\right) - \left(-\frac{1}{1}+(1)^5\right)
	\end{align}		\end{align}
	</math>		</math>

2022-08-26T18:13:52Z

← Older revision		Revision as of 18:13, 26 August 2022
Line 7:		Line 7:
	&= \left(\frac{y^{-2+1}}{-2+1}+\frac{5y^{4+1}}{4+1}\right)\bigg\|_{1}^{2} = \left(\frac{y^{-1}}{-1}+y^5\right)\bigg\|_{1}^{2} = \left(-\frac{1}{y}+y^5\right)\bigg\|_{1}^{2} \\[2ex]		&= \left(\frac{y^{-2+1}}{-2+1}+\frac{5y^{4+1}}{4+1}\right)\bigg\|_{1}^{2} = \left(\frac{y^{-1}}{-1}+y^5\right)\bigg\|_{1}^{2} = \left(-\frac{1}{y}+y^5\right)\bigg\|_{1}^{2} \\[2ex]

			&= \left(-\frac{1}{2}+(2)^5\right)
	\end{align}		\end{align}
	</math>		</math>

2022-08-26T18:13:02Z

← Older revision		Revision as of 18:13, 26 August 2022
Line 5:		Line 5:
	\int_{1}^{2}\frac{y+5y^7}{y^3}dy &= \int_{1}^{2}\left(\frac{y}{y^3}+\frac{5y^7}{y^3}\right)dy = \int_{1}^{2}(y^{-2}+5y^{4})dy\\[2ex]		\int_{1}^{2}\frac{y+5y^7}{y^3}dy &= \int_{1}^{2}\left(\frac{y}{y^3}+\frac{5y^7}{y^3}\right)dy = \int_{1}^{2}(y^{-2}+5y^{4})dy\\[2ex]

	&= \left(\frac{y^{-2+1}}{-2+1}+\frac{5y^{4+1}}{4+1}\right)\bigg\|_{1}^{2} = \left(\frac{y^{-1}}{-1}+y^5\right)\bigg\|_{1}^{2} \\[2ex]		&= \left(\frac{y^{-2+1}}{-2+1}+\frac{5y^{4+1}}{4+1}\right)\bigg\|_{1}^{2} = \left(\frac{y^{-1}}{-1}+y^5\right)\bigg\|_{1}^{2} = \left(-\frac{1}{y}+y^5\right)\bigg\|_{1}^{2} \\[2ex]

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

2022-08-26T18:11:33Z

← Older revision		Revision as of 18:11, 26 August 2022
Line 5:		Line 5:
	\int_{1}^{2}\frac{y+5y^7}{y^3}dy &= \int_{1}^{2}\left(\frac{y}{y^3}+\frac{5y^7}{y^3}\right)dy = \int_{1}^{2}(y^{-2}+5y^{4})dy\\[2ex]		\int_{1}^{2}\frac{y+5y^7}{y^3}dy &= \int_{1}^{2}\left(\frac{y}{y^3}+\frac{5y^7}{y^3}\right)dy = \int_{1}^{2}(y^{-2}+5y^{4})dy\\[2ex]

	&= \left(\frac{y^{-2+1}}{-2+1}+\frac{5y^{4+1}}{4+1}\right)\bigg\|_{1}^{2} = \left(\frac{y^{-1}{-1}+y^5\right)\bigg\|_{1}^{2} \\[2ex]		&= \left(\frac{y^{-2+1}}{-2+1}+\frac{5y^{4+1}}{4+1}\right)\bigg\|_{1}^{2} = \left(\frac{y^{-1}}{-1}+y^5\right)\bigg\|_{1}^{2} \\[2ex]

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