5.3 The Fundamental Theorem of Calculus/25 - Revision history

Dvaezazizi@laalliance.org: Protected "5.3 The Fundamental Theorem of Calculus/25" ([Edit=Allow only administrators] (indefinite) [Move=Allow only administrators] (indefinite))

2022-09-06T21:01:11Z

Protected "5.3 The Fundamental Theorem of Calculus/25" ([Edit=Allow only administrators] (indefinite) [Move=Allow only administrators] (indefinite))

← Older revision

Revision as of 21:01, 6 September 2022

2022-09-06T21:00:50Z

← Older revision		Revision as of 21:00, 6 September 2022
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	<math>		<math>
	\begin{align}		\begin{align}

	\int_{1}^{2}\left(\frac{3}{t^4}\right)dt &= \int_{1}^{2}\left(3t^{-4}\right)\,dt \\[2ex]		\int_{1}^{2}\left(\frac{3}{t^4}\right)dt &= \int_{1}^{2}\left(3t^{-4}\right)\,dt \\[2ex]
	&= \frac{3t^{-4}}{-3}\bigg\|_{1}^{2} = -t^{-3} \bigg\|_{1}^{2} \\[2ex]
			&= \frac{3t^{-4+1}}{-4+1}\bigg\|_{1}^{2} = -t^{-3} \bigg\|_{1}^{2} \\[2ex]

	&= \left[-(2)^{-3}\right]-\left[-(1)^{-3}\right] \\[2ex]		&= \left[-(2)^{-3}\right]-\left[-(1)^{-3}\right] \\[2ex]
	&= 1-\frac{1}{8} = \frac{7}{8}
			&= \left[-\frac{1}{8}\right]+[1] = \frac{7}{8}

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

2022-08-25T20:09:44Z

← Older revision		Revision as of 20:09, 25 August 2022
Line 7:		Line 7:
	\end{align}		\end{align}
	</math>		</math>


	~~\int_{1}^{2}\left(\frac{3}{t^4}\right)dt = \int_{1}^{2}\left(\{3t^-4}\right) \\~~

	~~&= \frac{3t^-4}{-3}\bigg\|_{1}^{2} = -t^-3 \bigg\|_{1}^{2} \\~~

	~~&= \left[-(2)^-3\right]-\left[-(1)^-3\right] \\[2ex]~~

	~~&= 1-\frac{1}{8} = \frac{7}{8}~~

2022-08-25T20:09:35Z

← Older revision		Revision as of 20:09, 25 August 2022
Line 2:		Line 2:
	\begin{align}		\begin{align}
	\int_{1}^{2}\left(\frac{3}{t^4}\right)dt &= \int_{1}^{2}\left(3t^{-4}\right)\,dt \\[2ex]		\int_{1}^{2}\left(\frac{3}{t^4}\right)dt &= \int_{1}^{2}\left(3t^{-4}\right)\,dt \\[2ex]
	&= \frac{3t^{-4}}{-3}\bigg\|_{1}^{2} = -t^{-3} \bigg\|_{1}^{2} \\		&= \frac{3t^{-4}}{-3}\bigg\|_{1}^{2} = -t^{-3} \bigg\|_{1}^{2} \\[2ex]
	&= \left[-(2)^{-3}\right]-\left[-(1)^{-3}\right] \\[2ex]		&= \left[-(2)^{-3}\right]-\left[-(1)^{-3}\right] \\[2ex]
	&= 1-\frac{1}{8} = \frac{7}{8}		&= 1-\frac{1}{8} = \frac{7}{8}

2022-08-25T20:09:13Z

2022-08-25T20:09:03Z

2022-08-25T20:08:14Z

2022-08-25T20:07:51Z

← Older revision		Revision as of 20:07, 25 August 2022
Line 3:		Line 3:
	\int_{1}^{2}\left(\frac{3}{t^4}\right)dt &= \int_{1}^{2}\left(3t^{-4}\right)\,dt \\[2ex]		\int_{1}^{2}\left(\frac{3}{t^4}\right)dt &= \int_{1}^{2}\left(3t^{-4}\right)\,dt \\[2ex]
	&= \frac{3t^{-4}}{-3}\bigg\|_{1}^{2} = -t^{-3} \bigg\|_{1}^{2} \\		&= \frac{3t^{-4}}{-3}\bigg\|_{1}^{2} = -t^{-3} \bigg\|_{1}^{2} \\
	&= \left[-(2)^-3\right]-\left[-(1)^-3\right] \\[2ex]		&= \left[-(2)^{-3}\right]-\left[-(1)^{-3}\right] \\[2ex]
	\end{align}		\end{align}
	</math>		</math>

2022-08-25T20:07:30Z

← Older revision		Revision as of 20:07, 25 August 2022
Line 3:		Line 3:
	\int_{1}^{2}\left(\frac{3}{t^4}\right)dt &= \int_{1}^{2}\left(3t^{-4}\right)\,dt \\[2ex]		\int_{1}^{2}\left(\frac{3}{t^4}\right)dt &= \int_{1}^{2}\left(3t^{-4}\right)\,dt \\[2ex]
	&= \frac{3t^{-4}}{-3}\bigg\|_{1}^{2} = -t^{-3} \bigg\|_{1}^{2} \\		&= \frac{3t^{-4}}{-3}\bigg\|_{1}^{2} = -t^{-3} \bigg\|_{1}^{2} \\
			&= \left[-(2)^-3\right]-\left[-(1)^-3\right] \\[2ex]
	\end{align}		\end{align}
	</math>		</math>

2022-08-25T20:07:06Z

← Older revision		Revision as of 20:07, 25 August 2022
Line 2:		Line 2:
	\begin{align}		\begin{align}
	\int_{1}^{2}\left(\frac{3}{t^4}\right)dt &= \int_{1}^{2}\left(3t^{-4}\right)\,dt \\[2ex]		\int_{1}^{2}\left(\frac{3}{t^4}\right)dt &= \int_{1}^{2}\left(3t^{-4}\right)\,dt \\[2ex]
	&= \frac{3t^{-4}}{-3}\bigg\|_{1}^{2} = -t^-3 \bigg\|_{1}^{2} \\		&= \frac{3t^{-4}}{-3}\bigg\|_{1}^{2} = -t^{-3} \bigg\|_{1}^{2} \\
	\end{align}		\end{align}
	</math>		</math>