5.5 The Substitution Rule/69: Difference between revisions

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<math>  
<math>  
\int_{0}^{1} \left(\frac{e^z + 1}{e^z + z}\right)
</math>
<math>
\begin{align}
\begin{align}
\int_{0}^{1} \left(\frac{e^z + 1}{e^z + z}\right) &= \int_{}^{} \left((e^z +1) (\frac{1}{e^z +z}) \right)


\end{align}
u &= e^z + z \\[2ex]
</math>


<math>
du &= e^z +1 \\[2ex]


u= e^z + z
/end{align}
</math>


du= e^z +1
New upper limit: <math> \1 = e^1 + 1 = e + 1 </math><br>
New lower limit: <math> 0 = e^0 + 0 = 1 </math>


<math>
\begin{align}
\int_{1}^{e+1} \left((e^z +1) (\frac{1}{e^z +z}) \right)
</math>
</math>

Revision as of 15:52, 6 September 2022

Failed to parse (unknown function "\begin{align}"): {\displaystyle \begin{align} u &= e^z + z \\[2ex] du &= e^z +1 \\[2ex] /end{align} }

New upper limit: Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle \1 = e^1 + 1 = e + 1 }
New lower limit:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle \begin{align} \int_{1}^{e+1} \left((e^z +1) (\frac{1}{e^z +z}) \right) }

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