5.5 The Substitution Rule/59 - Revision history

Elizabethh58413 at 02:04, 6 September 2022

2022-09-06T02:04:54Z

← Older revision		Revision as of 02:04, 6 September 2022
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	&=-\int_{1}^{\frac{1}{2}}e^u\,du \\[2ex]		&=-\int_{1}^{\frac{1}{2}}e^u\,du \\[2ex]
	&=-e^u\bigg\|_{1}^{\frac{1}{2}} \\[2ex]		&=-e^u\bigg\|_{1}^{\frac{1}{2}} \\[2ex]
			&=-\sqrt{e} - (-e^1) \\[2ex]
	&=e-\sqrt{e}		&=e-\sqrt{e}
	\end {align}		\end {align}
	</math>		</math>

Elizabethh58413 at 02:02, 6 September 2022

2022-09-06T02:02:20Z

← Older revision		Revision as of 02:02, 6 September 2022
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	\begin{align}		\begin{align}
	\int_{1}^{2}\frac{ e^\frac{1}{x}}{x^2}\,dx &=\int_{1}^{2} e^\frac{1}{x}(\frac{1}{x^2}\,dx)		\int_{1}^{2}\frac{ e^\frac{1}{x}}{x^2}\,dx &=\int_{1}^{2} e^\frac{1}{x}(\frac{1}{x^2}\,dx)
	&=\int_{1}^{\frac{1}{2}}e^u\,-du \\[2ex]		&=\int_{1}^{\frac{1}{2}}e^u\,(-du) \\[2ex]
	&=-\int_{1}^{\frac{1}{2}}e^u\,du \\[2ex]		&=-\int_{1}^{\frac{1}{2}}e^u\,du \\[2ex]
	&=-e^u\bigg\|_{1}^{\frac{1}{2}} \\[2ex]		&=-e^u\bigg\|_{1}^{\frac{1}{2}} \\[2ex]

Elizabethh58413 at 02:01, 6 September 2022

2022-09-06T02:01:32Z

← Older revision		Revision as of 02:01, 6 September 2022
Line 10:		Line 10:
	</math>		</math>

	New upper limit:<math> \frac{1}{2} = \frac{1}{(2)} <\math> <br>
	New lower limit: <math> 1 = \frac{1}{(1)} <\math>
			New upper limit:<math>\frac{1}{2} = \frac{1}{(2)} </math><br>
			New lower limit: <math> 1 = \frac{1}{(1)} </math>



	<math>		<math>

Elizabethh58413: Created page with " $\int_{1}^{2}\frac{ e^\frac{1}{x}}{x^2}\,dx$

$\begin{align} u &=\frac{1}{x} \\[2ex] du &=-\frac{1}{x^2}dx \\[2ex] -du &=\frac{1}{x^2}dx \\[2ex] \end{align}$ New upper limit: $\frac{1}{2} = \frac{1}{(2)} <\math>$
New lower limit: $1 = \frac{1}{(1)} <\math> \begin{align} \int_{1}^{2}\frac{ e^\frac{1}{x}}{x^2}\,dx &=\int_{1}^{2} e^\frac{1}{x}(\frac{1}{x^2}\,dx) &=\int_{1}^{\frac{1}{2}}e^u\,-du \\[2ex] &=-\int_{1}..."$

2022-09-06T01:58:01Z

Created page with "<math> \int_{1}^{2}\frac{ e^\frac{1}{x}}{x^2}\,dx </math> <math> \begin{align} u &=\frac{1}{x} \\[2ex] du &=-\frac{1}{x^2}dx \\[2ex] -du &=\frac{1}{x^2}dx \\[2ex] \end{align} </math> New upper limit:<math> \frac{1}{2} = \frac{1}{(2)} <\math> New lower limit: <math> 1 = \frac{1}{(1)} <\math> <math> \begin{align} \int_{1}^{2}\frac{ e^\frac{1}{x}}{x^2}\,dx &=\int_{1}^{2} e^\frac{1}{x}(\frac{1}{x^2}\,dx) &=\int_{1}^{\frac{1}{2}}e^u\,-du \\[2ex] &=-\int_{1}..."

New page

<math> \int_{1}^{2}\frac{ e^\frac{1}{x}}{x^2}\,dx </math> 

<math>
\begin{align}
u &=\frac{1}{x} \\[2ex]
du &=-\frac{1}{x^2}dx \\[2ex]
-du &=\frac{1}{x^2}dx \\[2ex]
\end{align}
</math>

New upper limit:<math> \frac{1}{2} = \frac{1}{(2)} <\math> 
New lower limit: <math> 1 = \frac{1}{(1)} <\math>

<math>
\begin{align}
\int_{1}^{2}\frac{ e^\frac{1}{x}}{x^2}\,dx &=\int_{1}^{2} e^\frac{1}{x}(\frac{1}{x^2}\,dx)
&=\int_{1}^{\frac{1}{2}}e^u\,-du \\[2ex]
&=-\int_{1}^{\frac{1}{2}}e^u\,du \\[2ex]
&=-e^u\bigg|_{1}^{\frac{1}{2}} \\[2ex]
&=e-\sqrt{e}
\end {align}
</math>