5.3 The Fundamental Theorem of Calculus/7: Difference between revisions
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<math>g(x)=\int_{1}^{x}\frac{1}{t^3+1}dt</math><br> | <math>g(x)=\int_{1}^{x}\frac{1}{t^3+1}dt</math><br> | ||
<math>\frac{d}{dx}\left[g(x)\right]=\frac{d}{dx}\left[\int_{1}^{x}\frac{1}{t^3+1}dt\right]</math>=(1)\left(\frac{1}{1+x^3}\right)</math> | <math>\frac{d}{dx}\left[g(x)\right]=\frac{d}{dx}\left[\int_{1}^{x}\frac{1}{t^3+1}dt\right]</math>=\(1)\left(\frac{1}{1+x^3}\right)</math> | ||
Revision as of 19:38, 25 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_{1}^{x}\frac{1}{t^3+1}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}\left[g(x)\right]=\frac{d}{dx}\left[\int_{1}^{x}\frac{1}{t^3+1}dt\right]}
=\(1)\left(\frac{1}{1+x^3}\right)</math>