5.5 The Substitution Rule/69: Difference between revisions
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New upper limit: <math> | New upper limit: <math> 1 = e^1 + 1 = e + 1 </math><br> | ||
New lower limit: <math> 0 = e^0 + 0 = 1 </math> | New lower limit: <math> 0 = e^0 + 0 = 1 </math> | ||
Revision as of 15:52, 6 September 2022
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 \int_{0}^{1} \left(\frac{e^z + 1}{e^z + z}\right) }
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} u &= e^z + z \\[2ex] du &= e^z +1 \\[2ex] \end{align} }
New upper limit: Failed to parse (Conversion error. Server ("https://en.wikipedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\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 0 = e^0 + 0 = 1 }
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) }