5.3 The Fundamental Theorem of Calculus/11: Difference between revisions
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<math> g(x)= \int_{x}^{\pi}\sqrt{1+sec(t)}\cdot dt </math> <br><br> | <math> g(x)= \int_{x}^{\pi}\sqrt{1+sec(t)}\cdot dt </math> <br><br> | ||
<math> \frac{d}{dx}\left[g(x)\right]=\frac{d}{dx}\left[\int_\ | <math> \frac{d}{dx}\left[g(x)\right]=\frac{d}{dx}\left[\int_\x^{\pi}\sqrt{1+sec(t)}\cdot dt\right]=0 \cdot \sqrt{1+sec(\pi)} - 1\cdot \sqrt{1+sec(x)} = -\sqrt{1+sec(x)}</math> <br><br> | ||
<math>\text{Therefore, } g'(x)=-\sqrt{1+sec(x)}</math> | <math>\text{Therefore, } g'(x)=-\sqrt{1+sec(x)}</math> | ||
Revision as of 20:06, 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 \frac{d}{dx}\left[g(x)\right]=\frac{d}{dx}\left[\int_\x^{\pi}\sqrt{1+sec(t)}\cdot dt\right]=0 \cdot \sqrt{1+sec(\pi)} - 1\cdot \sqrt{1+sec(x)} = -\sqrt{1+sec(x)}}
