5.4 Indefinite Integrals and the Net Change Theorem/43 - Revision history

Dvaezazizi@laalliance.org: Protected "5.4 Indefinite Integrals and the Net Change Theorem/43" ([Edit=Allow only administrators] (indefinite) [Move=Allow only administrators] (indefinite))

2022-09-21T19:42:02Z

Protected "5.4 Indefinite Integrals and the Net Change Theorem/43" ([Edit=Allow only administrators] (indefinite) [Move=Allow only administrators] (indefinite))

← Older revision	Revision as of 19:42, 21 September 2022
(No difference)

2022-09-21T17:48:26Z

← Older revision		Revision as of 17:48, 21 September 2022
Line 1:		Line 1:
	<math>		<math>
	\begin{align}		\begin{align}
			\int_{-1}^{2}(x-2\|x\|)dx &= \int_{-1}^{0}(x-2(-x))dx + \int_{0}^{2}(x-2(x))dx = \int_{-1}^{0}3x\,dx - \int_{0}^{2}x\,dx \\[2ex]


	\~~int~~\~~limits_~~{-1}^{2}~~(x-2\|x\|~~)~~dx &=~~ \~~int\limits_~~{-1}^{0}(x-2~~(-x)~~)~~dx +~~ \~~int\limits_~~{0}^{2}~~(x-2(x))dx~~ \\[2ex]		&= \left(\frac{3x^2}{2} \right)\bigg\|_{-1}^{0} - \left(\frac{x^2}{2} \right)\bigg\|_{0}^{2} \\[2ex]

	&= \left(\frac{1}{2} {x^2} ~~+ x^~~2 \right~~)\bigg\|_{~~-~~1}^{0} +~~ \left(\frac{~~1}{~~2} {x^2} - ~~x^2~~ \~~right)\bigg\|_~~{0}^{2} \\[2ex]		&= \left[\frac{3(0)^2}{2}-\frac{3(-1)^2}{2}\right]-\left[\frac{(2)^2}{2} - \frac{(0)^2}{2}\right] \\[2ex]

	&= 0- \left~~(\frac{1}{2} (~~-~~1)^2 + (-1)^2 \right) + \left(~~\frac{1}{2} ~~(2)^2 - (2)^2~~ \right~~) - 0 \\[2ex~~]		&= \left[-\frac{3}{2}\right]-\left[\frac{4}{2}\right] \\[2ex]

	~~&= \left(-\frac{1}{2}~~ - ~~1\right) +~~ \left~~(2 - 4\right) \\~~[~~2ex]~~

	~~&=-~~\frac{1}{2}~~-1-2 = -~~\~~frac{1}{2}-3 \\[2ex~~]

	~~&= -3.5~~ \\[2ex]

			&= -\frac{7}{2} \\[2ex]
	\end{align}		\end{align}
	</math>		</math>

2022-08-31T18:40:19Z

← Older revision		Revision as of 18:40, 31 August 2022
Line 3:		Line 3:


	\int\limits_{-1}^{2}(x-2\|x\|)dx = \int\limits_{-1}^{0}(x-2(-x))dx + \int\limits_{0}^{2}(x-2(x))dx \\[2ex]		\int\limits_{-1}^{2}(x-2\|x\|)dx &= \int\limits_{-1}^{0}(x-2(-x))dx + \int\limits_{0}^{2}(x-2(x))dx \\[2ex]

	&= \left(\frac{1}{2} {x^2} + x^2 \right)\bigg\|_{-1}^{0} + \left(\frac{1}{2} {x^2} - x^2 \right)\bigg\|_{0}^{2} \\[2ex]		&= \left(\frac{1}{2} {x^2} + x^2 \right)\bigg\|_{-1}^{0} + \left(\frac{1}{2} {x^2} - x^2 \right)\bigg\|_{0}^{2} \\[2ex]

2022-08-30T19:12:37Z

2022-08-30T19:12:19Z

← Older revision		Revision as of 19:12, 30 August 2022
Line 9:		Line 9:
	&= 0- \left(\frac{1}{2} (-1)^2 + (-1)^2 \right) + \left(\frac{1}{2} (2)^2 - (2)^2 \right) - 0 \\[2ex]		&= 0- \left(\frac{1}{2} (-1)^2 + (-1)^2 \right) + \left(\frac{1}{2} (2)^2 - (2)^2 \right) - 0 \\[2ex]

	&= \left(-\frac{1}{2} - 1\right) + \left(~~\frac{1}{~~2~~} 4)~~ - 4\right) \\[2ex]		&= \left(-\frac{1}{2} - 1\right) + \left(2 - 4\right) \\[2ex]

			&=-\frac{1}{2}-1-2 = -\frac{1}{2}-3

			&= -3.5

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

2022-08-30T19:09:49Z

← Older revision		Revision as of 19:09, 30 August 2022
Line 9:		Line 9:
	&= 0- \left(\frac{1}{2} (-1)^2 + (-1)^2 \right) + \left(\frac{1}{2} (2)^2 - (2)^2 \right) - 0 \\[2ex]		&= 0- \left(\frac{1}{2} (-1)^2 + (-1)^2 \right) + \left(\frac{1}{2} (2)^2 - (2)^2 \right) - 0 \\[2ex]

	&= \left(\frac{1}{2} + 1\right) + \left(\frac{1}{2} 4) - 4\right) \\[2ex]		&= \left(-\frac{1}{2} - 1\right) + \left(\frac{1}{2} 4) - 4\right) \\[2ex]
	\end{align}		\end{align}
	</math>		</math>

2022-08-30T19:04:14Z

← Older revision		Revision as of 19:04, 30 August 2022
Line 3:		Line 3:


	\int\limits_{-1}^{2}(x-2\|x\|)dx = \int\limits_{-1}^{0}(x-2(-x))dx + \int\limits_{0}^{2}(x-2(x))dx \\[~~1ex~~]		\int\limits_{-1}^{2}(x-2\|x\|)dx = \int\limits_{-1}^{0}(x-2(-x))dx + \int\limits_{0}^{2}(x-2(x))dx \\[2ex]

	&= \left(\frac{1}{2} {x^2} + x^2 \right)\bigg\|_{-1}^{0} + \left(\frac{1}{2} {x^2} - x^2 \right)\bigg\|_{0}^{2} \\[~~1ex~~]		&= \left(\frac{1}{2} {x^2} + x^2 \right)\bigg\|_{-1}^{0} + \left(\frac{1}{2} {x^2} - x^2 \right)\bigg\|_{0}^{2} \\[2ex]

	&= 0- \left(\frac{1}{2} (-1)^2 + (-1)^2 \right) + \left(\frac{1}{2} (2)^2 - (2)^2 \right) - 0 \\[~~1ex~~]		&= 0- \left(\frac{1}{2} (-1)^2 + (-1)^2 \right) + \left(\frac{1}{2} (2)^2 - (2)^2 \right) - 0 \\[2ex]

	&= \left(\frac{1}{2} + 1\right) + \left(\frac{1}{2} 4) - 4\right) \\[~~1ex~~]		&= \left(\frac{1}{2} + 1\right) + \left(\frac{1}{2} 4) - 4\right) \\[2ex]
	\end{align}		\end{align}
	</math>		</math>

2022-08-30T19:01:06Z

← Older revision		Revision as of 19:01, 30 August 2022
Line 5:		Line 5:
	\int\limits_{-1}^{2}(x-2\|x\|)dx = \int\limits_{-1}^{0}(x-2(-x))dx + \int\limits_{0}^{2}(x-2(x))dx \\[1ex]		\int\limits_{-1}^{2}(x-2\|x\|)dx = \int\limits_{-1}^{0}(x-2(-x))dx + \int\limits_{0}^{2}(x-2(x))dx \\[1ex]

	&= \left(\frac{1}{2} {x^2} + x^2 \right)\bigg\|_{-1}^{0} + \left(\frac{1}{2} {x^2} - x^2 \right)\bigg\|_{0}^{2} \\[~~2ex~~]		&= \left(\frac{1}{2} {x^2} + x^2 \right)\bigg\|_{-1}^{0} + \left(\frac{1}{2} {x^2} - x^2 \right)\bigg\|_{0}^{2} \\[1ex]

	&= 0- \left(\frac{1}{2} (-1)^2 + (-1)^2 \right) + \left(\frac{1}{2} (2)^2 - (2)^2 \right) - 0 \\[~~2ex~~]		&= 0- \left(\frac{1}{2} (-1)^2 + (-1)^2 \right) + \left(\frac{1}{2} (2)^2 - (2)^2 \right) - 0 \\[1ex]

	&= \left(\frac{1}{2} + 1\right) + \left(\frac{1}{2} 4) - 4\right) \\[~~2ex~~]		&= \left(\frac{1}{2} + 1\right) + \left(\frac{1}{2} 4) - 4\right) \\[1ex]
	\end{align}		\end{align}
	</math>		</math>

2022-08-30T19:00:55Z

← Older revision		Revision as of 19:00, 30 August 2022
Line 3:		Line 3:


	\int\limits_{-1}^{2}(x-2\|x\|)dx = \int\limits_{-1}^{0}(x-2(-x))dx + \int\limits_{0}^{2}(x-2(x))dx \\[~~2ex~~]		\int\limits_{-1}^{2}(x-2\|x\|)dx = \int\limits_{-1}^{0}(x-2(-x))dx + \int\limits_{0}^{2}(x-2(x))dx \\[1ex]

	&= \left(\frac{1}{2} {x^2} + x^2 \right)\bigg\|_{-1}^{0} + \left(\frac{1}{2} {x^2} - x^2 \right)\bigg\|_{0}^{2} \\[2ex]		&= \left(\frac{1}{2} {x^2} + x^2 \right)\bigg\|_{-1}^{0} + \left(\frac{1}{2} {x^2} - x^2 \right)\bigg\|_{0}^{2} \\[2ex]

2022-08-30T19:00:19Z

← Older revision		Revision as of 19:00, 30 August 2022
Line 3:		Line 3:


	\int\limits_{-1}^{2}(x-2\|x\|)dx = \int\limits_{-1}^{0}(x-2(-x))dx + \int\limits_{0}^{2}(x-2(x))dx		\int\limits_{-1}^{2}(x-2\|x\|)dx = \int\limits_{-1}^{0}(x-2(-x))dx + \int\limits_{0}^{2}(x-2(x))dx \\[2ex]

	&= \left(\frac{1}{2} {x^2} + x^2 \right)\bigg\|_{-1}^{0} + \left(\frac{1}{2} {x^2} - x^2 \right)\bigg\|_{0}^{2}		&= \left(\frac{1}{2} {x^2} + x^2 \right)\bigg\|_{-1}^{0} + \left(\frac{1}{2} {x^2} - x^2 \right)\bigg\|_{0}^{2} \\[2ex]

	&= 0- \left(\frac{1}{2} (-1)^2 + (-1)^2 \right) + \left(\frac{1}{2} (2)^2 - (2)^2 \right) - 0		&= 0- \left(\frac{1}{2} (-1)^2 + (-1)^2 \right) + \left(\frac{1}{2} (2)^2 - (2)^2 \right) - 0 \\[2ex]

	&= \left(\frac{1}{2} + 1\right) + \left(\frac{1}{2} 4) - 4\right)		&= \left(\frac{1}{2} + 1\right) + \left(\frac{1}{2} 4) - 4\right) \\[2ex]
	\end{align}		\end{align}
	</math>		</math>