5.3 The Fundamental Theorem of Calculus/31: Difference between revisions

Revision as of 19:16, 26 August 2022

Evaluate the integral

$\int _{0}^{\frac {\pi }{4}}sec^{2}(t)dt$

$\int _{0}^{\frac {\pi }{4}}sec^{2}(t)dt=tan({\frac {\pi }{4}})-tan(0)=1-0=1$

Therefore, $\int _{0}^{\frac {\pi }{4}}sec^{2}(t)dt=1$

(Use FTC #2,) $\int _{a}^{b}f(x)dt=F(b)-F(a)$

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	Therefore, <math> \int_{0}^\frac{\pi}{4}sec^{2}(t)dt = 1 </math>		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>)		(Use FTC #2,) <math> \int_{a}^{b}f(x)dt = F(b)-F(a) </math>