5.3 The Fundamental Theorem of Calculus/27 - Revision history

Dvaezazizi@laalliance.org: Protected "5.3 The Fundamental Theorem of Calculus/27" ([Edit=Allow only administrators] (indefinite) [Move=Allow only administrators] (indefinite))

2022-09-06T21:17:28Z

Protected "5.3 The Fundamental Theorem of Calculus/27" ([Edit=Allow only administrators] (indefinite) [Move=Allow only administrators] (indefinite))

← Older revision	Revision as of 21:17, 6 September 2022
(No difference)

2022-09-06T21:16:29Z

← Older revision		Revision as of 21:16, 6 September 2022
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	&= \left(\frac{2x^{1+1}}{1+1}+\frac{x^{6+1}}{6+1}\right)\bigg\|_{0}^{2}=\left(x^2+\frac{x^7}{7}\right)\bigg\|_{0}^{2} \\[2ex]		&= \left(\frac{2x^{1+1}}{1+1}+\frac{x^{6+1}}{6+1}\right)\bigg\|_{0}^{2}=\left(x^2+\frac{x^7}{7}\right)\bigg\|_{0}^{2} \\[2ex]

	&= \left((2)^2-\frac{(2)^7}{7}\right)+\left((0)^2+\frac{(0)^7}{7}\right) \\[2ex]		&= \left((2)^2-\frac{(2)^7}{7}\right)-\left((0)^2+\frac{(0)^7}{7}\right) \\[2ex]

	&= \left[4+\frac{2^7}{7}\right]-[0] \\[2ex]		&= \left[4+\frac{2^7}{7}\right]-[0] \\[2ex]

2022-09-06T21:16:15Z

2022-09-06T21:16:00Z

2022-09-06T21:15:48Z

← Older revision		Revision as of 21:15, 6 September 2022
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	\begin{align}		\begin{align}

	\~~int_2~~^0 x(2+x^5)\,dx = \~~int_2~~^0 (2x+x^6)\,dx &= -\int_0^2 (2x+x^6)\,dx \\[2ex]		\int_0^2 x(2+x^5)\,dx = \int_0^2 (2x+x^6)\,dx &= \int_0^2 (2x+x^6)\,dx \\[2ex]


	&= \left(\frac{2x^2}{1+1}+\frac{x^6+1}{6+1}\right)\bigg\|_{0}^{2}=\left(x^2+\frac{x^7}{7}\right)\bigg\|_{0}^{2}		&= \left(\frac{2x^{1+1}}{1+1}+\frac{x^{6+1}}{6+1}\right)\bigg\|_{0}^{2}=\left(x^2+\frac{x^7}{7}\right)\bigg\|_{0}^{2} \\[2ex]
	&= \left((2)^2+\frac{(2)^7}{7}\right)-\left((0)^2+\frac{0^7}{7}\right)
	&= 4+\frac{2^7}{7}		&= \left((2)^2-\frac{(2)^7}{7}\right)+\left((0)^2+\frac{(0)^7}{7}\right) \\[2ex]
	&= \frac{156}{7}
			&= \left[4+\frac{2^7}{7}\right]-[0] \\[2ex]
			&= \frac{156}{7}

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

2022-09-06T21:04:20Z

2022-09-06T21:04:11Z

← Older revision		Revision as of 21:04, 6 September 2022
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	= \left(\frac{2x^2}{1+1}+\frac{x^6+1}{6+1}\right)\bigg\|_{0}^{2}=\left(x^2+\frac{x^7}{7}\right)\bigg\|_{0}^{2}		&= \left(\frac{2x^2}{1+1}+\frac{x^6+1}{6+1}\right)\bigg\|_{0}^{2}=\left(x^2+\frac{x^7}{7}\right)\bigg\|_{0}^{2}
	= \left((2)^2+\frac{(2)^7}{7}\right)-\left((0)^2+\frac{0^7}{7}\right)		&= \left((2)^2+\frac{(2)^7}{7}\right)-\left((0)^2+\frac{0^7}{7}\right)
	= 4+\frac{2^7}{7}		&= 4+\frac{2^7}{7}
	= \frac{156}{7}		&= \frac{156}{7}

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

2022-09-06T21:03:58Z

2022-09-06T21:03:41Z

← Older revision		Revision as of 21:03, 6 September 2022
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	\begin{align}		\begin{align}

	\int_2^0 x(2+x^5)dx = \int_2^0 (2x+x^6)dx		\int_2^0 x(2+x^5)dx = \int_2^0 (2x+x^6)dx &= -\int_0^2 (2x+x^6)dx


	= \left(\frac{2x^2}{1+1}+\frac{x^6+1}{6+1}\right)\bigg\|_{0}^{2}=\left(x^2+\frac{x^7}{7}\right)\bigg\|_{0}^{2}		= \left(\frac{2x^2}{1+1}+\frac{x^6+1}{6+1}\right)\bigg\|_{0}^{2}=\left(x^2+\frac{x^7}{7}\right)\bigg\|_{0}^{2}
	= \left((2)^2+\frac{(2)^7}{7}\right)-\left((0)^2+\frac{0^7}{7}\right)		= \left((2)^2+\frac{(2)^7}{7}\right)-\left((0)^2+\frac{0^7}{7}\right)

2022-09-06T21:03:15Z

← Older revision		Revision as of 21:03, 6 September 2022
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	\int_2^0 x(2+x^5)dx = \int_2^0 (2x+x^6)dx		\int_2^0 x(2+x^5)dx = \int_2^0 (2x+x^6)dx
	= \left(\frac{2x^2}{1+1}+\frac{x^6+1}{6+1}\right)\bigg\|_{0}^{2}=\left(x^2+\frac{x^7}{7}\right)\bigg\|_{0}^{2}~~</math><br>~~		= \left(\frac{2x^2}{1+1}+\frac{x^6+1}{6+1}\right)\bigg\|_{0}^{2}=\left(x^2+\frac{x^7}{7}\right)\bigg\|_{0}^{2}
	= \left((2)^2+\frac{(2)^7}{7}\right)-\left((0)^2+\frac{0^7}{7}\right)~~</math><br>~~		= \left((2)^2+\frac{(2)^7}{7}\right)-\left((0)^2+\frac{0^7}{7}\right)
	= 4+\frac{2^7}{7}~~</math><br>~~		= 4+\frac{2^7}{7}
	= \frac{156}{7}		= \frac{156}{7}

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