7.1 Integration By Parts/25 - Revision history

Cynthiac61995@students.laalliance.org at 18:11, 13 December 2022

2022-12-13T18:11:18Z

@@ Line 1: / Line 1: @@
 <math>
-\begin{align}
+\int_{0}^{1}\frac{y}{e^{2y}}dy
 </math>
 <math>

Cynthiac61995@students.laalliance.org at 17:59, 13 December 2022

2022-12-13T17:59:55Z

← Older revision		Revision as of 17:59, 13 December 2022
Line 1:		Line 1:
	<math>		<math>
	\begin{align}		\begin{align}
	\int_{0}^{1}\frac{y}{e^{2y}}dy, \\[2ex]		\int_{0}^{1}\frac{y}{e^{2y}}dy \\[2ex]
	& u=y, -du=dy, \\[2ex]		& u=y, -du=dy, \\[2ex]
	& dv=\frac{dy}{e^{2}y}=e^{-2y}dy, v=\frac{-1}{2}e^{-2y}		& dv=\frac{dy}{e^{2}y}=e^{-2y}dy, v=\frac{-1}{2}e^{-2y}

Cynthiac61995@students.laalliance.org at 17:59, 13 December 2022

2022-12-13T17:59:35Z

← Older revision		Revision as of 17:59, 13 December 2022
Line 1:		Line 1:
	<math>		<math>
	\begin{align}		\begin{align}
	\int_{0}^{1}\frac{y}{e^{2y}}dy,		\int_{0}^{1}\frac{y}{e^{2y}}dy, \\[2ex]
	& u=y, -du=dy,		& u=y, -du=dy, \\[2ex]
	& dv=\frac{dy}{e^{2}y}=e^{-2y}dy, v=\frac{-1}{2}e^{-2y}		& dv=\frac{dy}{e^{2}y}=e^{-2y}dy, v=\frac{-1}{2}e^{-2y}
	\end{align}		\end{align}

Cynthiac61995@students.laalliance.org at 17:58, 13 December 2022

2022-12-13T17:58:55Z

← Older revision		Revision as of 17:58, 13 December 2022
Line 1:		Line 1:
	<math>		<math>
	\int_{0}^{1}\frac{y}{e^{2y}}dy		\begin{align}
	u=y, -du=dy,		\int_{0}^{1}\frac{y}{e^{2y}}dy,
	dv=\frac{dy}{e^{2}y}=e^{-2y}dy, v=\frac{-1}{2}e^{-2y}		& u=y, -du=dy,
			& dv=\frac{dy}{e^{2}y}=e^{-2y}dy, v=\frac{-1}{2}e^{-2y}
			\end{align}
	</math>		</math>

Line 10:		Line 12:
	\int_{0}^{1}\frac{y}{e^{2y}}dy &=\left[\frac{-1}{2}ye^{-2y}\right]\bigg\|_{0}^{1} + \frac{1}{2}\int_{0}{1}e^{-2y}dy \\[2ex]		\int_{0}^{1}\frac{y}{e^{2y}}dy &=\left[\frac{-1}{2}ye^{-2y}\right]\bigg\|_{0}^{1} + \frac{1}{2}\int_{0}{1}e^{-2y}dy \\[2ex]

	&= \left(\frac{-1}{2}e^{-2}+0\right)-\frac{1}{4}\left[e^{-2y}\right]\bigg\|_{0}^(1) \\[2ex]		&= \left(\frac{-1}{2}e^{-2}+0\right)-\frac{1}{4}\left[e^{-2y}\right]\bigg\|_{0}^{1} \\[2ex]
	&= \frac{-1}{2}e^{-2}-\frac{1}{4}e^{-2}+\frac{1}{4} \\[2ex]		&= \frac{-1}{2}e^{-2}-\frac{1}{4}e^{-2}+\frac{1}{4} \\[2ex]
	&= \frac{1}{4}-\frac{3}{4}e^{-2} \\[2ex]		&= \frac{1}{4}-\frac{3}{4}e^{-2} \\[2ex]

Cynthiac61995@students.laalliance.org at 17:55, 13 December 2022

2022-12-13T17:55:06Z

← Older revision		Revision as of 17:55, 13 December 2022
Line 2:		Line 2:
	\int_{0}^{1}\frac{y}{e^{2y}}dy		\int_{0}^{1}\frac{y}{e^{2y}}dy
	u=y, -du=dy,		u=y, -du=dy,
	dv=\frac{dy}{e^{2}y} &=e^{-2y}dy, v=\frac{-1}{2}e^{-2y}		dv=\frac{dy}{e^{2}y}=e^{-2y}dy, v=\frac{-1}{2}e^{-2y}
	</math>		</math>

Cynthiac61995@students.laalliance.org: Created page with " $\int_{0}^{1}\frac{y}{e^{2y}}dy u=y, -du=dy, dv=\frac{dy}{e^{2}y} &=e^{-2y}dy, v=\frac{-1}{2}e^{-2y}$ $\begin{align} \int_{0}^{1}\frac{y}{e^{2y}}dy &=\left[\frac{-1}{2}ye^{-2y}\right]\bigg|_{0}^{1} + \frac{1}{2}\int_{0}{1}e^{-2y}dy \\[2ex] &= \left(\frac{-1}{2}e^{-2}+0\right)-\frac{1}{4}\left[e^{-2y}\right]\bigg|_{0}^(1) \\[2ex] &= \frac{-1}{2}e^{-2}-\frac{1}{4}e^{-2}+\frac{1}{4} \\[2ex] &= \frac{1}{4}-\frac{3}{4}e^{-2} \\[2ex] \end{align}$ "

2022-12-13T17:53:54Z

Created page with "<math> \int_{0}^{1}\frac{y}{e^{2y}}dy u=y, -du=dy, dv=\frac{dy}{e^{2}y} &=e^{-2y}dy, v=\frac{-1}{2}e^{-2y} </math> <math> \begin{align} \int_{0}^{1}\frac{y}{e^{2y}}dy &=\left[\frac{-1}{2}ye^{-2y}\right]\bigg|_{0}^{1} + \frac{1}{2}\int_{0}{1}e^{-2y}dy \\[2ex] &= \left(\frac{-1}{2}e^{-2}+0\right)-\frac{1}{4}\left[e^{-2y}\right]\bigg|_{0}^(1) \\[2ex] &= \frac{-1}{2}e^{-2}-\frac{1}{4}e^{-2}+\frac{1}{4} \\[2ex] &= \frac{1}{4}-\frac{3}{4}e^{-2} \\[2ex] \end{align} </math>"

New page

<math>
\int_{0}^{1}\frac{y}{e^{2y}}dy
u=y, -du=dy,
dv=\frac{dy}{e^{2}y} &=e^{-2y}dy, v=\frac{-1}{2}e^{-2y}
</math>

<math>
\begin{align}

\int_{0}^{1}\frac{y}{e^{2y}}dy &=\left[\frac{-1}{2}ye^{-2y}\right]\bigg|_{0}^{1} + \frac{1}{2}\int_{0}{1}e^{-2y}dy \\[2ex]

&= \left(\frac{-1}{2}e^{-2}+0\right)-\frac{1}{4}\left[e^{-2y}\right]\bigg|_{0}^(1) \\[2ex]
&= \frac{-1}{2}e^{-2}-\frac{1}{4}e^{-2}+\frac{1}{4} \\[2ex]
&= \frac{1}{4}-\frac{3}{4}e^{-2} \\[2ex]

\end{align}
</math>