5.3 The Fundamental Theorem of Calculus/7: Difference between revisions
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<math>\frac{d}{dx}\left[g(x)\right]=\frac{d}{dx}\left[\int_{1}^{x}\frac{1}{t^3+1}dt\right]= | <math>\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}{(x)^3+1}\right)-(0)\left(\frac{1}{(1)^3+1}\right)=\frac{1}{x^3+1}</math> | (1)\left(\frac{1}{(x)^3+1}\right)-(0)\left(\frac{1}{(1)^3+1}\right)=\frac{1}{x^3+1}</math><Br> | ||
<math>\text{Therefore, } g'(x)=\frac{1}{x^3+1}</math> | <math>\text{Therefore, } g'(x)=\frac{1}{x^3+1}</math> | ||
Revision as of 19:49, 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 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}{(x)^3+1}\right)-(0)\left(\frac{1}{(1)^3+1}\right)=\frac{1}{x^3+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 \text{Therefore, } g'(x)=\frac{1}{x^3+1}}