5.4 Indefinite Integrals and the Net Change Theorem/23 - Revision history

Dvaezazizi@laalliance.org: Protected "5.4 Indefinite Integrals and the Net Change Theorem/23" ([Edit=Allow only administrators] (indefinite) [Move=Allow only administrators] (indefinite))

2022-09-21T19:40:20Z

Protected "5.4 Indefinite Integrals and the Net Change Theorem/23" ([Edit=Allow only administrators] (indefinite) [Move=Allow only administrators] (indefinite))

← Older revision	Revision as of 19:40, 21 September 2022
(No difference)

Dvaezazizi@laalliance.org at 15:00, 21 September 2022

2022-09-21T15:00:59Z

← Older revision		Revision as of 15:00, 21 September 2022
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	\begin{align}		\begin{align}
	\int_{-1}^{0}(2x-e^x)dx &=\left[\frac{2x^2}{2}-e^x\right]_{-1}^{0} \\[2ex]		\int_{-1}^{0}(2x-e^x)dx &=\left[\frac{2x^2}{2}-e^x\right]_{-1}^{0} \\[2ex]
	&= [0-e^{0}]-[(-1)^2-e^{-1}] \\[2ex]		&= [0^{2}-e^{0}]-[(-1)^2-e^{-1}] \\[2ex]
	&=-1-\left(1-\frac{1}{e}\right) \\[2ex]		&=-1-\left(1-\frac{1}{e}\right) \\[2ex]
	&=\frac{1}{e}-2		&=\frac{1}{e}-2

Dvaezazizi@laalliance.org at 15:00, 21 September 2022

2022-09-21T15:00:44Z

← Older revision		Revision as of 15:00, 21 September 2022
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	<math>		<math>
	\begin{align}		\begin{align}
	\int_{-1}^{0}(2x-e^x)dx &=\left[\frac{2x^2}{2}-e^x\right]~~\bigg\|~~_{-1}^{0} \\[2ex]		\int_{-1}^{0}(2x-e^x)dx &=\left[\frac{2x^2}{2}-e^x\right]_{-1}^{0} \\[2ex]
	&= [0-e^{0}]-[(-1)^2-e^{-1}] \\[2ex]		&= [0-e^{0}]-[(-1)^2-e^{-1}] \\[2ex]
	&=-1-\left(1-\frac{1}{e}\right) \\[2ex]		&=-1-\left(1-\frac{1}{e}\right) \\[2ex]

Dvaezazizi@laalliance.org at 15:00, 21 September 2022

2022-09-21T15:00:26Z

← Older revision		Revision as of 15:00, 21 September 2022
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	<math>		<math>
	\int_{-1}^{0}(2x-e^x)dx =\frac{2x^2}{2}-e^x\bigg\|_{-1}^{0} = -1-(1-\frac{1}{e})=\frac{1}{e}-2		\begin{align}
			\int_{-1}^{0}(2x-e^x)dx &=\left[\frac{2x^2}{2}-e^x\right]\bigg\|_{-1}^{0} \\[2ex]
			&= [0-e^{0}]-[(-1)^2-e^{-1}] \\[2ex]
			&=-1-\left(1-\frac{1}{e}\right) \\[2ex]
			&=\frac{1}{e}-2
			\end{align}

	</math>		</math>

Dvaezazizi@laalliance.org at 14:56, 21 September 2022

2022-09-21T14:56:58Z

← Older revision		Revision as of 14:56, 21 September 2022
Line 1:		Line 1:
	<math>		<math>
	\int_{-1}^{0}(2x-e^x)dx =-1-(1-\frac{1}{e})=\frac{1}{e}-2		\int_{-1}^{0}(2x-e^x)dx =\frac{2x^2}{2}-e^x\bigg\|_{-1}^{0} = -1-(1-\frac{1}{e})=\frac{1}{e}-2
	</math>		</math>

Dvaezazizi@laalliance.org at 14:55, 21 September 2022

2022-09-21T14:55:51Z

@@ Line 1: / Line 1: @@
 <math>
-\int_{-1}^{0}(2x-e^x)dx
+\int_{-1}^{0}(2x-e^x)dx =-1-(1-\frac{1}{e})=\frac{1}{e}-2
 </math>

Dvaezazizi@laalliance.org at 14:55, 21 September 2022

2022-09-21T14:55:26Z

← Older revision		Revision as of 14:55, 21 September 2022
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	<math>		<math>
	\~~int\limits_~~{-1}^{0}(2x-e^x)dx		\int_{-1}^{0}(2x-e^x)dx
	</math>		</math>

	<math>		<math>
	=\~~int\limits_~~{-1}^{0}2xdx-\int\limits_{-1}^{0}e^xdx		=\int_{-1}^{0}2xdx-\int\limits_{-1}^{0}e^xdx
	</math>		</math>

Jonathanv62005@students.laalliance.org at 19:26, 26 August 2022

2022-08-26T19:26:15Z

← Older revision		Revision as of 19:26, 26 August 2022
Line 8:		Line 8:

	<math>		<math>
	=-1-(-1-\frac{1}{e})=\frac{1}{e}-2		=-1-(1-\frac{1}{e})=\frac{1}{e}-2
	</math>		</math>

Jonathanv62005@students.laalliance.org at 19:22, 26 August 2022

2022-08-26T19:22:44Z

← Older revision		Revision as of 19:22, 26 August 2022
Line 2:		Line 2:
	\int\limits_{-1}^{0}(2x-e^x)dx		\int\limits_{-1}^{0}(2x-e^x)dx
	</math>		</math>

	<math>		<math>
	=\int\limits_{-1}^{0}2xdx-\int\limits_{-1}^{0}e^xdx=-1-(-1-\frac{1}{e})=\frac{1}{e}-2		=\int\limits_{-1}^{0}2xdx-\int\limits_{-1}^{0}e^xdx
			</math>

			<math>
			=-1-(-1-\frac{1}{e})=\frac{1}{e}-2
	</math>		</math>

Jonathanv62005@students.laalliance.org: Created page with " $\int\limits_{-1}^{0}(2x-e^x)dx$ $=\int\limits_{-1}^{0}2xdx-\int\limits_{-1}^{0}e^xdx=-1-(-1-\frac{1}{e})=\frac{1}{e}-2$ "

2022-08-26T19:21:50Z

Created page with "<math> \int\limits_{-1}^{0}(2x-e^x)dx </math> <math> =\int\limits_{-1}^{0}2xdx-\int\limits_{-1}^{0}e^xdx=-1-(-1-\frac{1}{e})=\frac{1}{e}-2 </math>"

New page

<math>
\int\limits_{-1}^{0}(2x-e^x)dx
</math>
<math>
=\int\limits_{-1}^{0}2xdx-\int\limits_{-1}^{0}e^xdx=-1-(-1-\frac{1}{e})=\frac{1}{e}-2
</math>

5.4 Indefinite Integrals and the Net Change Theorem/23 - Revision history

Dvaezazizi@laalliance.org: Protected "5.4 Indefinite Integrals and the Net Change Theorem/23" ([Edit=Allow only administrators] (indefinite) [Move=Allow only administrators] (indefinite))

Dvaezazizi@laalliance.org at 15:00, 21 September 2022

Dvaezazizi@laalliance.org at 15:00, 21 September 2022

Dvaezazizi@laalliance.org at 15:00, 21 September 2022

Dvaezazizi@laalliance.org at 14:56, 21 September 2022

Dvaezazizi@laalliance.org at 14:55, 21 September 2022

Dvaezazizi@laalliance.org at 14:55, 21 September 2022

Jonathanv62005@students.laalliance.org at 19:26, 26 August 2022

Jonathanv62005@students.laalliance.org at 19:22, 26 August 2022

Jonathanv62005@students.laalliance.org: Created page with " \int\limits_{-1}^{0}(2x-e^x)dx =\int\limits_{-1}^{0}2xdx-\int\limits_{-1}^{0}e^xdx=-1-(-1-\frac{1}{e})=\frac{1}{e}-2 "

Jonathanv62005@students.laalliance.org: Created page with " $\int\limits_{-1}^{0}(2x-e^x)dx$ $=\int\limits_{-1}^{0}2xdx-\int\limits_{-1}^{0}e^xdx=-1-(-1-\frac{1}{e})=\frac{1}{e}-2$ "