5.3 The Fundamental Theorem of Calculus/53: Difference between revisions
No edit summary |
No edit summary |
||
| Line 1: | Line 1: | ||
<math>\int_{2x}^{3x}\frac{u^2-1}{u^2+1}du</math> | <math>g(x)=\int_{2x}^{3x}\frac{u^2-1}{u^2+1}du</math> | ||
<math>\frac{d}{dx}\left[\int_{2x}^{3x}\frac{u^2-1}{u^2+1}du\right]=(3*\frac{3x^2-1}{3x^2+1})-(2*\frac{2x^2-1}{2x^2+1})</math> | <math>\frac{d}{dx}g(x) = \frac{d}{dx}\left[\int_{2x}^{3x}\frac{u^2-1}{u^2+1}du\right]=(3*\frac{3x^2-1}{3x^2+1})-(2*\frac{2x^2-1}{2x^2+1})</math> | ||
Revision as of 22:19, 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_{2x}^{3x}\frac{u^2-1}{u^2+1}du}
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_{2x}^{3x}\frac{u^2-1}{u^2+1}du\right]=(3*\frac{3x^2-1}{3x^2+1})-(2*\frac{2x^2-1}{2x^2+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 =(3*\frac{9x^2-1}{9x^2+1})-(2*\frac{4x^2-1}{4x^2+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 =(\frac{3(9x^2-1)}{9x^2+1})-(\frac{2(4x^2-1)}{4x^2+1})}