5.3 The Fundamental Theorem of Calculus/8: Difference between revisions
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<math>g(x)=\int_{3}^{x}e^{t^2-t}dt</math><br> | <math>g(x)=\int_{3}^{x}e^{t^2-t}dt</math><br> | ||
<math>\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}</math><br> | <math>\frac{d}{dx}g(x) = \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}</math><br> | ||
Therefore, <math>g'(x)=e^{x^2-x}</math> | Therefore, <math>g'(x)=e^{x^2-x}</math> | ||
Revision as of 20:24, 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 g(x)=\int_{3}^{x}e^{t^2-t}dt}
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 \frac{d}{dx}g(x) = \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}}
Therefore, 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 g'(x)=e^{x^2-x}}