5.3 The Fundamental Theorem of Calculus/31: Difference between revisions
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Therefore, <math> \int_{0}^\frac{\pi}{4}sec^{2}(t)dt = 1 </math> | |||
Use FTC #2 <math> \int_{a}^{b}f(x)dt = F(b)-F(a) </math> | |||
Revision as of 19:33, 25 August 2022
Evaluate the integral
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 \int_{0}^\frac{\pi}{4}sec^{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 \int_{0}^\frac{\pi}{4}sec^{2}(t)dt= tan(\frac{\pi}{4})-tan(0)=1-0=1 }
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 \int_{0}^\frac{\pi}{4}sec^{2}(t)dt = 1 }
Use FTC #2 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 \int_{a}^{b}f(x)dt = F(b)-F(a) }