5.3 The Fundamental Theorem of Calculus/8: Difference between revisions
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\begin{align} | \begin{align} | ||
g(x)=\int_{3}^{x}e^{t^2-t}dt \\ | g(x)=\int_{3}^{x}e^{t^2-t}dt \\ | ||
\frac{d}{dx}\left[g(x)\right] = \frac{d}{dx}\left[\int_{3}^{x}e^{t^2-t}dt\right]=1e^{x^2-x}-0e^{3^2-3}=e^{x^2-x} \\ | |||
\end{align} | \end{align} | ||
Revision as of 20:32, 23 August 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 \begin{align} g(x)=\int_{3}^{x}e^{t^2-t}dt \\ \frac{d}{dx}\left[g(x)\right] = \frac{d}{dx}\left[\int_{3}^{x}e^{t^2-t}dt\right]=1e^{x^2-x}-0e^{3^2-3}=e^{x^2-x} \\ \end{align} }
Failed to parse (Conversion error. Server ("https://en.wikipedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\begin{aligned}u&={\tfrac {1}{\sqrt {2}}}(x+y)\qquad &x&={\tfrac {1}{\sqrt {2}}}(u+v)\\[0.6ex]v&={\tfrac {1}{\sqrt {2}}}(x-y)\qquad &y&={\tfrac {1}{\sqrt {2}}}(u-v)\end{aligned}}}