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

Revision as of 21:23, 6 September 2022

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

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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	<math> \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		<math> \int_{0}^{\frac{\pi}{4}}\sec^{2}(t)\,dt = \int_{0}^\frac{\pi}{4}\sec^{2}(t)dt= tan\left(\frac{\pi}{4}\right)-tan(0)=1-0=1

	</math>		</math>