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

Latest revision as of 21:25, 6 September 2022

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

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-Evaluate the integral
+<math>
+\int_{0}^{\frac{\pi}{4}}\sec^{2}(t)\,dt = \tan(t)\bigg|_{0}^{\frac{\pi}{4}}=\tan\left(\frac{\pi}{4}\right)-\tan(0)=1-0=1
-<math> \int_{0}^\frac{\pi}{4}sec^{2}(t)dt </math>
-<math>
-\int_{0}^\frac{\pi}{4}sec^{2}(t)dt= tan(\frac{\pi}{4})-tan(0)=1-0=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>