7.1 Integration By Parts/7 - Revision history

Kimberlyr70044@students.laalliance.org at 00:19, 30 November 2022

2022-11-30T00:19:54Z

← Older revision		Revision as of 00:19, 30 November 2022
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	\begin{align}		\begin{align}
	&=\int x^2 sin (\pi x)\\[2ex]		&=\int x^2 sin (\pi x)\\[2ex]
	U=x^2 ,\\[x] du= 2xdx ,\\[x] dv= sin(\pi x)dx ,\\[x] v=-\frac{1}{\pi}cos(\pi x) \\[2ex]		U=x^2 ,\\[2ex] du= 2xdx ,\\[2ex] dv= sin(\pi x)dx ,\\[2ex] v=-\frac{1}{\pi}cos(\pi x) \\[2ex]
	&=-\frac{x^2}{\pi}cos(\pi x) +\frac{2}{\pi} \int x cos (\pi x) dx\\[2ex]		&=-\frac{x^2}{\pi}cos(\pi x) +\frac{2}{\pi} \int x cos (\pi x) dx\\[2ex]
	U=x,\\[x] du= dx,\\[x] dv= cos(\pi x) dx,\\[x] v=\frac{1}{\pi}sin (\pi x) \\[2ex]		U=x,\\[2ex] du= dx,\\[2ex] dv= cos(\pi x) dx,\\[2ex] v=\frac{1}{\pi}sin (\pi x) \\[2ex]
	&=-\frac{x^2}{\pi}cos(\pi x)+ \frac{2}{\pi}[\frac{x}{\pi}sin(\pi x) - \frac{1}{\pi}\int sin (\pi x) dx]\\[2ex]		&=-\frac{x^2}{\pi}cos(\pi x)+ \frac{2}{\pi}[\frac{x}{\pi}sin(\pi x) - \frac{1}{\pi}\int sin (\pi x) dx]\\[2ex]
	&= -\frac{x^2}{\pi}cos(\pi x)+ \frac{2}{\pi}[\frac{x}{\pi}sin(\pi x) - \frac{1}{\pi^2}cos(\pi x) ] +c\\[2ex]		&= -\frac{x^2}{\pi}cos(\pi x)+ \frac{2}{\pi}[\frac{x}{\pi}sin(\pi x) - \frac{1}{\pi^2}cos(\pi x) ] +c\\[2ex]
	\end{align}		\end{align}
	</math>		</math>

Kimberlyr70044@students.laalliance.org at 00:18, 30 November 2022

2022-11-30T00:18:42Z

← Older revision		Revision as of 00:18, 30 November 2022
Line 2:		Line 2:
	\begin{align}		\begin{align}
	&=\int x^2 sin (\pi x)\\[2ex]		&=\int x^2 sin (\pi x)\\[2ex]
	U=x^2 ,\\[ex] du= 2xdx ,\\[ex] dv= sin(\pi x)dx ,\\[ex] v=-\frac{1}{\pi}cos(\pi x) \\[2ex]		U=x^2 ,\\[x] du= 2xdx ,\\[x] dv= sin(\pi x)dx ,\\[x] v=-\frac{1}{\pi}cos(\pi x) \\[2ex]
	&=-\frac{x^2}{\pi}cos(\pi x) +\frac{2}{\pi} \int x cos (\pi x) dx\\[2ex]		&=-\frac{x^2}{\pi}cos(\pi x) +\frac{2}{\pi} \int x cos (\pi x) dx\\[2ex]
	U=x,\\[ex] du= dx,\\[ex] dv= cos(\pi x) dx,\\[ex] v=\frac{1}{\pi}sin (\pi x) \\[2ex]		U=x,\\[x] du= dx,\\[x] dv= cos(\pi x) dx,\\[x] v=\frac{1}{\pi}sin (\pi x) \\[2ex]
	&=-\frac{x^2}{\pi}cos(\pi x)+ \frac{2}{\pi}[\frac{x}{\pi}sin(\pi x) - \frac{1}{\pi}\int sin (\pi x) dx]\\[2ex]		&=-\frac{x^2}{\pi}cos(\pi x)+ \frac{2}{\pi}[\frac{x}{\pi}sin(\pi x) - \frac{1}{\pi}\int sin (\pi x) dx]\\[2ex]
	&= -\frac{x^2}{\pi}cos(\pi x)+ \frac{2}{\pi}[\frac{x}{\pi}sin(\pi x) - \frac{1}{\pi^2}cos(\pi x) ] +c\\[2ex]		&= -\frac{x^2}{\pi}cos(\pi x)+ \frac{2}{\pi}[\frac{x}{\pi}sin(\pi x) - \frac{1}{\pi^2}cos(\pi x) ] +c\\[2ex]
	\end{align}		\end{align}
	</math>		</math>

Kimberlyr70044@students.laalliance.org at 00:18, 30 November 2022

2022-11-30T00:18:06Z

← Older revision		Revision as of 00:18, 30 November 2022
Line 2:		Line 2:
	\begin{align}		\begin{align}
	&=\int x^2 sin (\pi x)\\[2ex]		&=\int x^2 sin (\pi x)\\[2ex]
	U=x^2 ,\\[e] du= 2xdx ,\\[e] dv= sin(\pi x)dx ,\\[e] v=-\frac{1}{\pi}cos(\pi x) \\[2ex]		U=x^2 ,\\[ex] du= 2xdx ,\\[ex] dv= sin(\pi x)dx ,\\[ex] v=-\frac{1}{\pi}cos(\pi x) \\[2ex]
	&=-\frac{x^2}{\pi}cos(\pi x) +\frac{2}{\pi} \int x cos (\pi x) dx\\[2ex]		&=-\frac{x^2}{\pi}cos(\pi x) +\frac{2}{\pi} \int x cos (\pi x) dx\\[2ex]
	U=x,\\[e] du= dx,\\[e] dv= cos(\pi x) dx,\\[e] v=\frac{1}{\pi}sin (\pi x) \\[e]		U=x,\\[ex] du= dx,\\[ex] dv= cos(\pi x) dx,\\[ex] v=\frac{1}{\pi}sin (\pi x) \\[2ex]
	&=-\frac{x^2}{\pi}cos(\pi x)+ \frac{2}{\pi}[\frac{x}{\pi}sin(\pi x) - \frac{1}{\pi}\int sin (\pi x) dx]\\[2ex]		&=-\frac{x^2}{\pi}cos(\pi x)+ \frac{2}{\pi}[\frac{x}{\pi}sin(\pi x) - \frac{1}{\pi}\int sin (\pi x) dx]\\[2ex]
	&= -\frac{x^2}{\pi}cos(\pi x)+ \frac{2}{\pi}[\frac{x}{\pi}sin(\pi x) - \frac{1}{\pi^2}cos(\pi x) ] +c\\[2ex]		&= -\frac{x^2}{\pi}cos(\pi x)+ \frac{2}{\pi}[\frac{x}{\pi}sin(\pi x) - \frac{1}{\pi^2}cos(\pi x) ] +c\\[2ex]
	\end{align}		\end{align}
	</math>		</math>

Kimberlyr70044@students.laalliance.org at 00:17, 30 November 2022

2022-11-30T00:17:19Z

← Older revision		Revision as of 00:17, 30 November 2022
Line 2:		Line 2:
	\begin{align}		\begin{align}
	&=\int x^2 sin (\pi x)\\[2ex]		&=\int x^2 sin (\pi x)\\[2ex]
	& U=x^2 ,\\[~~1ex~~] du= 2xdx ,\\[~~1ex~~] dv= sin(\pi x)dx ,\\[~~1ex~~] v=-\frac{1}{\pi}cos(\pi x) \\[2ex]		U=x^2 ,\\[e] du= 2xdx ,\\[e] dv= sin(\pi x)dx ,\\[e] v=-\frac{1}{\pi}cos(\pi x) \\[2ex]
	&=-\frac{x^2}{\pi}cos(\pi x) +\frac{2}{\pi} \int x cos (\pi x) dx\\[2ex]		&=-\frac{x^2}{\pi}cos(\pi x) +\frac{2}{\pi} \int x cos (\pi x) dx\\[2ex]
	&U=x,\\[~~1ex~~] du= dx,\\[~~1ex~~] dv= cos(\pi x) dx,\\[~~1ex~~] v=\frac{1}{\pi}sin (\pi x) \\[~~2ex~~]		U=x,\\[e] du= dx,\\[e] dv= cos(\pi x) dx,\\[e] v=\frac{1}{\pi}sin (\pi x) \\[e]
	&=-\frac{x^2}{\pi}cos(\pi x)+ \frac{2}{\pi}[\frac{x}{\pi}sin(\pi x) - \frac{1}{\pi}\int sin (\pi x) dx]\\[2ex]		&=-\frac{x^2}{\pi}cos(\pi x)+ \frac{2}{\pi}[\frac{x}{\pi}sin(\pi x) - \frac{1}{\pi}\int sin (\pi x) dx]\\[2ex]
	&= -\frac{x^2}{\pi}cos(\pi x)+ \frac{2}{\pi}[\frac{x}{\pi}sin(\pi x) - \frac{1}{\pi^2}cos(\pi x) ] +c\\[2ex]		&= -\frac{x^2}{\pi}cos(\pi x)+ \frac{2}{\pi}[\frac{x}{\pi}sin(\pi x) - \frac{1}{\pi^2}cos(\pi x) ] +c\\[2ex]
	\end{align}		\end{align}
	</math>		</math>

Kimberlyr70044@students.laalliance.org at 00:15, 30 November 2022

2022-11-30T00:15:21Z

← Older revision		Revision as of 00:15, 30 November 2022
Line 2:		Line 2:
	\begin{align}		\begin{align}
	&=\int x^2 sin (\pi x)\\[2ex]		&=\int x^2 sin (\pi x)\\[2ex]
	& U=x^2 , du= 2xdx , dv= sin(\pi x)dx , v=-\frac{1}{\pi}cos(\pi x) \\[2ex]		& U=x^2 ,\\[1ex] du= 2xdx ,\\[1ex] dv= sin(\pi x)dx ,\\[1ex] v=-\frac{1}{\pi}cos(\pi x) \\[2ex]
	&=-\frac{x^2}{\pi}cos(\pi x) +\frac{2}{\pi} \int x cos (\pi x) dx\\[2ex]		&=-\frac{x^2}{\pi}cos(\pi x) +\frac{2}{\pi} \int x cos (\pi x) dx\\[2ex]
	&U=x, du= dx, dv= cos(\pi x) dx, v=\frac{1}{\pi}sin (\pi x) \\[2ex]		&U=x,\\[1ex] du= dx,\\[1ex] dv= cos(\pi x) dx,\\[1ex] v=\frac{1}{\pi}sin (\pi x) \\[2ex]
	&=-\frac{x^2}{\pi}cos(\pi x)+ \frac{2}{\pi}[\frac{x}{\pi}sin(\pi x) - \frac{1}{\pi}\int sin (\pi x) dx]\\[2ex]		&=-\frac{x^2}{\pi}cos(\pi x)+ \frac{2}{\pi}[\frac{x}{\pi}sin(\pi x) - \frac{1}{\pi}\int sin (\pi x) dx]\\[2ex]
	&= -\frac{x^2}{\pi}cos(\pi x)+ \frac{2}{\pi}[\frac{x}{\pi}sin(\pi x) - \frac{1}{\pi^2}cos(\pi x) ] +c\\[2ex]		&= -\frac{x^2}{\pi}cos(\pi x)+ \frac{2}{\pi}[\frac{x}{\pi}sin(\pi x) - \frac{1}{\pi^2}cos(\pi x) ] +c\\[2ex]
	\end{align}		\end{align}
	</math>		</math>

Kimberlyr70044@students.laalliance.org at 05:57, 29 November 2022

2022-11-29T05:57:12Z

← Older revision		Revision as of 05:57, 29 November 2022
Line 8:		Line 8:
	&= -\frac{x^2}{\pi}cos(\pi x)+ \frac{2}{\pi}[\frac{x}{\pi}sin(\pi x) - \frac{1}{\pi^2}cos(\pi x) ] +c\\[2ex]		&= -\frac{x^2}{\pi}cos(\pi x)+ \frac{2}{\pi}[\frac{x}{\pi}sin(\pi x) - \frac{1}{\pi^2}cos(\pi x) ] +c\\[2ex]
	\end{align}		\end{align}
	<\math>		</math>

Kimberlyr70044@students.laalliance.org: Created page with " $\begin{align} &=\int x^2 sin (\pi x)\\[2ex] & U=x^2 , du= 2xdx , dv= sin(\pi x)dx , v=-\frac{1}{\pi}cos(\pi x) \\[2ex] &=-\frac{x^2}{\pi}cos(\pi x) +\frac{2}{\pi} \int x cos (\pi x) dx\\[2ex] &U=x, du= dx, dv= cos(\pi x) dx, v=\frac{1}{\pi}sin (\pi x) \\[2ex] &=-\frac{x^2}{\pi}cos(\pi x)+ \frac{2}{\pi}[\frac{x}{\pi}sin(\pi x) - \frac{1}{\pi}\int sin (\pi x) dx]\\[2ex] &= -\frac{x^2}{\pi}cos(\pi x)+ \frac{2}{\pi}[\frac{x}{\pi}sin(\pi x) - \frac{1}{\pi^2}cos(\pi x..."$

2022-11-29T05:56:59Z

Created page with "<math> \begin{align} &=\int x^2 sin (\pi x)\\[2ex] & U=x^2 , du= 2xdx , dv= sin(\pi x)dx , v=-\frac{1}{\pi}cos(\pi x) \\[2ex] &=-\frac{x^2}{\pi}cos(\pi x) +\frac{2}{\pi} \int x cos (\pi x) dx\\[2ex] &U=x, du= dx, dv= cos(\pi x) dx, v=\frac{1}{\pi}sin (\pi x) \\[2ex] &=-\frac{x^2}{\pi}cos(\pi x)+ \frac{2}{\pi}[\frac{x}{\pi}sin(\pi x) - \frac{1}{\pi}\int sin (\pi x) dx]\\[2ex] &= -\frac{x^2}{\pi}cos(\pi x)+ \frac{2}{\pi}[\frac{x}{\pi}sin(\pi x) - \frac{1}{\pi^2}cos(\pi x..."

New page

<math>
\begin{align}
&=\int x^2 sin (\pi x)\\[2ex]
& U=x^2 , du= 2xdx , dv= sin(\pi x)dx , v=-\frac{1}{\pi}cos(\pi x) \\[2ex]
&=-\frac{x^2}{\pi}cos(\pi x) +\frac{2}{\pi} \int x cos (\pi x) dx\\[2ex]
&U=x, du= dx, dv= cos(\pi x) dx, v=\frac{1}{\pi}sin (\pi x) \\[2ex]
&=-\frac{x^2}{\pi}cos(\pi x)+ \frac{2}{\pi}[\frac{x}{\pi}sin(\pi x) - \frac{1}{\pi}\int sin (\pi x) dx]\\[2ex]
&= -\frac{x^2}{\pi}cos(\pi x)+ \frac{2}{\pi}[\frac{x}{\pi}sin(\pi x) - \frac{1}{\pi^2}cos(\pi x) ] +c\\[2ex]
\end{align}
<\math>